Agenda Day 64 – Introduction to Solutions Lesson: PPT

Agenda Day 64 – Introduction to Solutions Lesson: PPT
Handouts: 1. Solution Handout Text: 1. P Solutions HW: 1. Finish all the worksheets, including Textbook questions

MATTER Heterogeneous mixture Homogeneous Pure Substance
No Is it uniform throughout? Yes Heterogeneous mixture Homogeneous No Can it be separated by physical means? yes Pure Substance Homogeneous Mixture (Solution) Can it be decomposed into other substance by a chemical process? No yes Compound Element

States of matter in solution
Example of solutions gas in gas air ( N2, O2 , Ar, CO2 , other gases) gas in liquid soda pop (CO2 in water) liquid in liquid gasoline (a mixture of hydrocarbon compounds) solid in liquid Filtrated sea water ( NaCl and other salts in water) gas in solid H2 in platinum or palladium liquid in solid dental amalgams (mercury in silver) solid in solid alloys ( brass, (Cu/Zn), sol-der (Sn/Pb), Steel (Fe/C ))

Definitions Solutions are made up of at least two components
SOLVENT - The substance that does the dissolving and is usually in greater proportion. ( Often will indicate the phase of the solution). SOLUTE - the substance that is dissolved and is usually in smaller proportion A solution that is composed of a high percentage of solute is said to be concentrated A solution with a low percentage of solute is said to be dilute.

SOLUBILITY The SOLUBILITY of a substance in a solvent is the maximum amount of the solute which will dissolve in a fixed quantity of solvent at a specific temperature The solubility of a substance changes with temperature and is dependent on the nature of the solute and solvent. This is because a high temperature means H2O molecules are moving faster (keeping more solid molecules suspended).

Solutions may be classified on the basis of the amount of solute dissolved in the solvent
UNSATURATED SOLUTION - A solution in which more solute can be dissolved while the temperature remains constant. ( solvent volume is unchanged) SATURATED SOLUTION - A solution in which no more solute can be dissolved while the temperature remains constant. ( in a given volume of solvent) A saturated solution represents an equilibrium: the rate of dissolving is equal to the rate of crystallization. The salt continues to dissolve, but crystallizes at the same rate so that there “appears” to be nothing happening.

Dissolving a salt... A salt is an ionic compound - usually a metal cation bonded to a non-metal anion. The dissolving of a salt is an example of equilibrium. The cations and anions are attracted to each other in the salt. They are also attracted to the water molecules. The water molecules will start to pull out some of the ions from the salt crystal.

At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions. However, soon the ions floating in the water begin to collide with the salt crystal and are “pulled back in” to the salt. (precipitation) Eventually the rate of dissociation is equal to the rate of precipitation. The solution is now “saturated”. It has reached equilibrium.

Solubility Equilibrium: Dissociation = Precipitation
In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker. Concentration of the solution is constant. The rate at which the salt is dissolving into solution equals the rate of precipitation. Na+ and Cl - ions surrounded by water molecules NaCl Crystal Dissolving NaCl in water

SUPERSATURATED SOLUTION
SUPERSATURATED SOLUTION - A solution which contains more dissolved solute than it would normally at a given temperature and specific volume of solvent. Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways: Warm the solvent so that it will dissolve more, then cool the solution. Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

Water Polar substances - are ones with an unequal distribution of charge on the molecule These molecules interact with other polar substances because of dipole interactions “like dissolves like” Water is a polar solvent and is known as the universal solvent since it is able to dissolve a great variety of substances [its solutions are known as aqueous (aq)]

Aqueous Solutions Water acts to dissolve both molecular and ionic substances through intermolecular dipole interactions. In some instances water can also form hydrogen bonding when the solute has potential for hydrogen bonding. Water can cause some molecular substances to ionize by H-bonding to water or by LDF ie HCl (g) H2O (l) > H3O+(aq) Cl – (aq) (hydronium) Water causes ionic substances to dissociate ie NaCl (s) > Na+ (aq) Cl –(aq)

Dissolving process in water
2. Hydration of solute Orientation of water molecules around solute Na+ Cl- 1. Overcome attractive forces in solid

Types of attractive forces
For water: dipole-dipole For hydrated ion: ion-dipole Na+ Cl- For NaCl (s): ion-ion

Aqueous Solutions How do we know ions are present in aqueous solutions? Solutions that form ions are known as electrolytes and will conduct electrical current. HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Solutions that do not form ions are known as non-electrolytes. e.g: sugar, ethanol, ethylene glycol

Types of solutes Strong Electrolyte - 100% dissociation,
high conductivity Strong Electrolyte - 100% dissociation, all ions in solution Na+ Cl-

Types of solutes Weak Electrolyte - partial dissociation,
slight conductivity Weak Electrolyte - partial dissociation, molecules and ions in solution CH3COOH CH3COO- H+

Types of solutes Non-electrolyte - No dissociation,
no conductivity Non-electrolyte - No dissociation, all molecules in solution sugar

Electrolytes in the Body
Carry messages to and from the brain as electrical signals. Maintain cellular function with the correct concentrations electrolytes Make your own 50-70 g sugar One liter of warm water Pinch of salt 200ml of sugar free fruit squash Mix, cool and drink

Solvents Non polar substances have an equal distribution of charge and interact with other nonpolar substances because of London (dispersion) force interactions. Other popular solvents: Alcohol, e.g. I2(al) - antiseptic Acetic Acid, e.g. glues and solvents

SOLUBILITY IN WATER Solids usually have a higher solubility in water at higher temperatures Gases always have a higher solubility in water at lower temperatures. This is because when gas molecules are moving faster they are able to escape from the liquid surface. Think of cold soda vs. warm soda. Halogens and oxygen are only slightly soluble in water but because they are so reactive, even in small concentrations they are often very important in solution reactions

Miscible vs. Immiscible
Nonpolar liquids do not dissolve in water to any large degree but instead form a separate layer. Theses liquids are said to be IMMISCIBLE in water Some liquids made up of small polar molecules with the ability for form hydrogen bonds dissolve completely in water and are said to be MISCIBLE. Polar liquids usually have a higher solubility in water at a higher temperature Elements that do not react with water generally have a low solubility in water. Note: there are always exceptions to these generalized statements

Solubility of Solutes in Water
Most solids (endothermic hydration) All gases Solubility, g/100 mL water Some solids (exothermic hydration) Temperature

Properties of Water 1. Water has a high specific heat. - A large amount of energy is required to change the temperature of water. 2. Water has a high heat of vaporization. - The evaporation of water from a surface causes cooling of that surface. 3. Solid water is less dense than liquid water. - Bodies of water freeze from the top down. 4. Water is a good solvent. - Water dissolves polar molecules and ionic compound.

Properties of Water 5. Water organizes nonpolar molecules. - hydrophilic: “water-loving” - hydrophobic: “water-fearing” Water causes hydrophobic molecules to aggregate or assume specific shapes. 6. Water can form ions. H2O (l)  OH-1 (aq) + H+1(aq) hydroxide ion hydrogen ion

Why oil and water don’t mix
12/10/99 + – + – The non-polar substance is pushed away. If it were moving faster it might break through the attractive forces. Solubility is a balance between speed and attraction. + – Also, the more similar the strength of their dipoles the more likely two compounds are to mix.

Solubility Curves Day 65 – Solubility and Saturation - Solubility Curves Lesson: PPT, Try This Activity page 317 old text demo Handouts: 1. Solution Handout, 2. Solubility Curves Assignment. Text: 1. P solutions/ gases HW: 1. Finish all the worksheets, including Textbook questions

How to determine the solubility of a given substance?
Find out the mass of solute needed to make a saturated solution in 100 cm3 of water for a specific temperature(referred to as the solubility). This is repeated for each of the temperatures from 0ºC to 100ºC. The data is then plotted on a temperature/solubility graph,and the points are connected. These connected points are called a solubility curve.

Each substance has its own unique solubility which can
Solubility Curve Solubility vs. Temperature for Solids 140 KI 130 120 shows the dependence of solubility on temperature gases solids NaNO3 110 100 KNO3 90 Each substance has its own unique solubility which can be displayed on a graph 80 HCl NH4Cl 70 Solubility (grams of solute/100 g H2O) 60 NH3 KCl 50 “Solubility Curves for Selected Solutes” Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC. Basic Concepts The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance. When the temperature of a saturated solution decreases, a precipitate forms. Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases. Teaching Suggestions Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course). Questions Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC. Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature? Why do you think solubilities are only shown between 0 oC and 100 oC? In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder? Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature? According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article? 40 30 Solubility depends on the solute, the solvent, and the temperature. NaCl KClO3 20 10 SO2

Determine if a solution is saturated, unsaturated, or supersaturated.
If the solubility for a given substance places it anywhere on it's solubility curve it is saturated. If it lies above the solubility curve, then it’s supersaturated, If it lies below the solubility curve it's an unsaturated solution. Remember though, if the volume of water isn't 100 cm3 to use a proportion first.

Solubility curve Saturated Supersaturated Unsaturated
“Solubility Curves for Selected Solutes” Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC. Basic Concepts The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance. When the temperature of a saturated solution decreases, a precipitate forms. Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases. Teaching Suggestions Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course). Questions Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC. Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature? Why do you think solubilities are only shown between 0 oC and 100 oC? In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder? Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature? According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article? Unsaturated

Formation of a Saturated Solution
Dynamic equilibrium: rate of crystallization = rate of dissolving

A Supersaturated Solution

Solubility curve Any point on a line represents a saturated solution.
In a saturated solution, the solvent contains the maximum amount of solute. Example At 90oC, 40 g of NaCl(s) in 100g H2O(l) represent a saturated solution.

Solubility curve Any point below a line represents an unsaturated solution. In an unsaturated solution, the solvent contains less than the maximum amount of solute. Example At 90oC, 30 g of NaCl(s) in 100g H2O(l) represent an unsaturated solution. 10 g of NaCl(s) have to be added to make the solution saturated.

Solubility curve Any point above a line represents a supersaturated solution. In a supersaturated solution, the solvent contains more than the maximum amount of solute. A supersaturated solution is very unstable and the amount in excess can precipitate or crystallize. Example At 90oC, 50 g of NaCl(s) in 100g H2O(l) represent a supersaturated solution. Eventually, 10 g of NaCl(s) will precipitate.

Solubility curve Any solution can be made saturated, unsaturated, or supersaturated by changing the temperature.

Classify as unsaturated, saturated, or supersaturated.
Solubility vs. Temperature for Solids Solubility (grams of solute/100 g H2O) KI KCl 20 10 30 40 50 60 70 80 90 110 120 130 140 100 NaNO3 KNO3 HCl NH4Cl NH3 NaCl KClO3 SO2 gases solids Classify as unsaturated, saturated, or supersaturated. 80 g 30oC 45 g 60oC 50 g 10oC 70 g 70oC =unsaturated per 100 g H2O =saturated =unsaturated =supersaturated

Solids dissolved in liquids Gases dissolved in liquids
To Sol. To Sol. As To , solubility As To , solubility

How to solve solubility curve problems
Look for the intersection of the solubility and temperature Least soluble = lowest line at temp Most soluble = highest line at temp If given different amount of water Sometimes you'll need to determine how much additional solute needs to be added to an unsaturated solution in order to make it saturated.

For example,30 g of potassium nitrate has been added to 100 cm3 of water at a temperature of 50ºC.
How many additional grams of solute must be added in order to make it saturated? From the graph you can see that the solubility for potassium nitrate at 50ºC is 84 grams If there are already 30 grams of solute in the solution, all you need to get to 84 g is 54 more grams ( 84g-30g )

Solubility vs. Temperature for Solids
Solubility (grams of solute/100 g H2O) KI KCl 20 10 30 40 50 60 70 80 90 110 120 130 140 100 NaNO3 KNO3 HCl NH4Cl NH3 NaCl KClO3 SO2 gases solids So sat. 40oC for 500 g H2O = 5 x 66 g = 330 g 120 g < 330 g unsaturated saturation 40oC for 100 g H2O = 66 g KNO3 Per 500 g H2O, 120 g 40oC

What substance has a solubility of 90 g/100 cm3 of water at a temperature of 25ºC ?

What substance has a solubility of 100 g/50 cm3 of water at a temperature of 90ºC ?

What is the solubility of potassium nitrate at 80ºC ?

At what temperature will sodium nitrate have a solubility of 95 g/100cm3 ?

At what temperature will potassium iodide have a solubility of 115 g/50 cm3 ?

What is the solubility of sodium chloride at 25ºC in 150 cm3 of water ?
From the solubility graph we see that sodium chlorides solubility is 36 g. 36g NaCl 100 cm3 x # g NaCl= 150 cm3 54g NaCl =

Which salt is the least soluble in water at 20C°?
Solubility vs. Temperature for Solids Solubility (grams of solute/100 g H2O) KI KCl 20 10 30 40 50 60 70 80 90 110 120 130 140 100 NaNO3 KNO3 HCl NH4Cl NH3 NaCl KClO3 SO2 gases solids Which salt is the least soluble in water at 20C°?

1. Which salt shows the least change in solubility from 0°C to 100°C? 2. A saturated solution of potassium chlorate is formed from 100 g of water. If the solution is cooled from 80°C to 50°C, how many grams of precipitate are formed? 3. What compound shows a decrease in solubility from 0°C

Describe each situation below.
Unsaturated; all solute dissolves; clear solution. Per 100 g H2O, 100 g 50oC. B. Cool solution (A) very slowly to 10oC. Supersaturated; extra solute remains in solution; still clear. C. Quench solution (A) in an ice bath to 10oC. Saturated; extra solute (20 g) cannot remain in solution, becomes visible

Concentration

Agenda Day 66 – Concentration Lesson: PPT,
Handouts: 1. Concentration& Dilution Handout. 2. Concentration of Solutions Worksheet Text: 1. P Concentration ( %, ppm) HW: 1. Finish all the worksheets, including Textbook questions

Solution Concentration
Concentration = quantity of solute quantity of solution (not solvent) There are 3 basic ways to express concentration: 1) percentages, 2) very low concentrations, and 3) molar concentrations % concentration can be in V/V, W/W, or W/V Like most %s, V/V and W/W need to have the same units on top and bottom. W/V is sort of in the same units; V is mostly water and water’s density is 1 g/mL or 1 kg/L 3 g H2O2/100 mL solution  3 g H2O2/100 g solution

Solution and Concentration
Other ways of expressing concentration Molarity(M): moles solute / Liter solution Mass percent: (mass solute / mass of solution) * 100 Molality* (m) - moles solute / Kg solvent Mole Fraction(A) - moles solute / total moles solution * Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution.

% Concentration % (w/w) = % (w/v) = % (v/v) = [ 3% w/w = 3 g/100 g]
[ 3% w/v = 3 g/100 mL] % (v/v) = [ 3% v/v = 3 mL/100 mL]

Units of Concentrations
amount of solute per amount of solvent or solution g solute g solution x 100 g solute g solute + g solvent x 100 = Percent (by mass) = moles of solute volume in liters of solution Molarity (M) = moles = M x VL

Solution Concentration
Expressing concentrations in parts per million (ppm) requires the unit on top to be 1,000,000 times smaller than the unit on the bottom E.g. 1 mg/kg or g/g Notice that any units expressed as a volume must be referring to a water solution (1L = 1kg)- density of water For parts per billion (ppb), the top unit would have to be 1,000,000,000 times smaller 1 ppm = 1 g/106 mL = 1 g/ 1000 L = 1 mg/L = 1 mg/kg = 1 g/g 1g = 1000mg 1000mg/1000L 1mg = 1000 g 1000g/1000g

Molarity Molar concentration is the most commonly used in chemistry. amount of solute (in moles) Molar concentration = volume of solution (in litres) UNITS: ( mol/L) or (mol .L-1) or M

Concentration: Percentage Examples
10 g / 260 g = 3.8 % W/W 30 mL / 280 mL = 11% V/V (in reality may be off) 8.0 g / 100 g = 8% W/W What is the % W/W of copper in an alloy when 10.0 g of Cu is mixed with 250 g of Zn? What is approximate % V/V if 30.0 mL of pure ethanol is added to 250 mL of water? What is the % W/W if 8.0 g copper is added to enough zinc to produce 100 g of an alloy?

Concentration: Molarity Example
If g of KMnO4 is dissolved in enough water to give 250. mL of solution, what is the molarity of KMnO4? As is almost always the case, the first step is to convert the mass of material to moles. 0.435 g KMnO4 x 1 mol KMnO4 = mol KMnO4 158.0 g KMnO4 Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to L . Molarity KMnO4 = mol KMnO4 = M 0.250 L solution

Step 1: Calculate moles of NiCl2•6H2O
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl2•6H2O Step 2: Calculate Molarity [NiCl2•6 H2O ] = M

Concentration: Mixed Example
A solution of H2O2 is 3% (w/v). Calculate the mass of H2O2 in mL of solution. Calculate the mass of H2O2in 1L of solution. Calculate the number of moles of H2O2 in 1L of solution. State the molar concentration of the solution Calculate the ppm of H2O2 .

Concentration: Mixed Example Answers
a) b) c) d) e)

Making Molar Solutions
From Solids

Preparation of Solutions

1. 0 L of water was used to make 1. 0 L of solution
1.0 L of water was used to make 1.0 L of solution. Notice the water left over.

What are molar solutions?
A molar solution is one that expresses “concentration” in moles per volume. Usually the units are in mol/L mol/L can be abbreviated as M Molar solutions are prepared using: a balance to weigh moles (as grams) a volumetric flask to measure litres L refers to entire volume, not water! Because the units are mol/L, we can use the equation M = n/L Alternatively, we can use the factor label method.

Calculations with molar solutions
Q: How many moles of NaCl are required to make 7.5 L of a 0.10 M solution? M=n/L, n = 0.10 M x 7.5 L = 0.75 mol # mol NaCl = 7.5 L x 0.10 mol NaCl 1 L = 0.75 mol But in the lab we weigh grams not moles, so … Q: How many grams of NaCl are required to make 7.5 L of a 0.10 M solution? # g NaCl = 7.5 L x 0.10 mol NaCl 1 L x g NaCl 1 mol NaCl =43.83 g

More Practice Questions
How many grams of nitric acid are present in 1.0 L of a 1.0 M HNO3 solution? 2. Calculate the number of grams needed to produce 1.00 L of these solutions: a) 1.00 M KNO3 b) 1.85 M H2SO c) 0.67 M KClO3 3. Calculate the # of grams needed to produce each:1 a) 0.20 L of 1.5 M KCl b) L of M HCl c) 0.20 L of 0.09 mol/L AgNO3 d) 250 mL of 3.1 mol/L BaCl2 4. Give the molarity of a solution containing 10 g of each solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)2 63 g 101 g 181 g 82 g a) 22 g b) 1.75 g c) 3 g d) 0.16 kg 5. Describe how 100 mL of a 0.10 mol/L NaOH solution would be made. a) mol/L b) mol/L

Practice making molar solutions
Calculate # of grams required to make 100 mL of a 0.10 M solution of NaOH (see above). Get volumetric flask, plastic bottle, 100 mL beaker, eyedropper. Rinse all with tap water. Fill a beaker with distilled water. Pour mL of H2O from beaker into flask. Weigh NaOH. Add it to flask. Do step 5 quickly. Mix (by swirling) until the NaOH is dissolved. Add distilled H2O to just below the colored line. Add distilled H2O to the line using eyedropper. Place solution in a bottle. Place label (tape) on bottle (name, date, chemical, molarity). Place bottle at front. Rinse & return equipment.

Concentration and Dilution
How can a solution be made less concentrated? More solvent can be added. What is this process called? Dilution This process is used extensively in chemistry... the concentration decreases in dilution, BUT what happens to the moles of the solute? Do they increase? Decrease? Stay the same?

Agenda Day 67 – Dilutions Lesson: PPT,
Handouts: 1. Concentration& DilutionHandout Text: 1. P HW: 1. Finish all the worksheets, including Textbook questions Worksheets,

ACTIVITY PART 1: Preparing Solutions by Dilution
Part 2: Diluting A Standard Solution

Dilution of Solutions Figure 4.19

Dilution of solutions Since moles are constant, the new concentration may be found using the following formula: n 1 = n 2 C1V1 = C2V2 Initial concentration Final volume Initial volume Final concentration

Dilution Example #1 A stock solution of 1.00M of NaCl is available. How many milliliters are needed to make a mL of 0.750M? What we know: the molarity of the stock solution which is 1.00 M, and the two components of the diluted solution which are C2= 0.750M and V2= 100 mL. C1V1 = C2V2

Dilution Example #2 Concentrated HCl is 12M. What volume is needed to
Make 2L of a 1M solution? What we know: the molarity of the stock solution which is 12M, and the two values for the diluted solution which are C2=1M and V2=2L. C1V1 = C2V2

Dilution Example #3 Calculate the final concentration if 2L of 3M of NaCl and 4L of 1.50M of NaCl are mixed. Assume there is no volume contraction upon mixing. For this, you must use the equation: n1 + n2 = nF C1V1 + C2V2= CFVF

Making Molar Solutions
From Liquids (More accurately, from stock solutions)

Making molar solutions from liquids
Not all compounds are in a solid form Acids are purchased as liquids (“stock solutions”). Yet, we still need a way to make molar solutions of these compounds. The Procedure is similar: Use pipette to measure moles (via volume) Use volumetric flask to measure volume Now we use the equation C1V1 = C2V2

Reading a pipette Identify each volume to two decimal places (values tell you how much you have expelled) 5.00

More Practice 1. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution? 2. You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L? mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M? 4. What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl? 5. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl? 6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water? 7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution? 8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out, what is the concentration of the remaining HF?

Dilution problems (1-6) 1. M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M) V1 = L = mL 2. M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L) M2 = 1.2 M 3. M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M) V2 = 0.4 L or 400 mL

Dilution problems (4 - 6) = 0.8 mol + 1.8 mol = 2.6 mol
4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L) = 0.8 mol mol = 2.6 mol # L = 0.4 L L # mol/L = 2.6 mol / 1 L = 2.6 mol/L 5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L) = 1.5 mol mol = 1.9 mol # mol/L = 1.9 mol / 5 L = 0.38 mol/L 6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L) = 1.5 mol + 0 mol = 1.5 mol # mol/L = 1.5 mol / 5 L = 0.3 mol/L Or, using M1V1 = M2V2, M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L

Dilution problems (7, 8) 7. M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M) V2 = 16 L 8. The concentration remains 0.2 M, both volume and moles are removed when the solution is poured out. Remember M is mol/L. Just like the density of a copper penny does not change if it is cut in half, the concentration of a solution does not change if it is cut in half.

Practice making molar solutions
Calculate # of mL of 1 M HCl required to make 100 mL of a 0.1 M solution of HCl Get volumetric flask, pipette, plastic bottle, mL beaker, 50 mL beaker, eyedropper. Rinse all with tap water. Dry 50 mL beaker Place about 20 mL of 1 M HCl in 50 mL beaker Rinse pipette, with small amount of acid Fill flask about 1/4 full with distilled water Add correct amount of acid with pipette. Mix. Add water to line (use eyedropper at the end) Place solution in plastic bottle Label bottle. Place at front of the room. Rinse and return all other equipment.

Solubility Guidelines for Aqueous Solutions

Agenda Day 68 – Solubility Rules & Precipitation Reactions and Net Ionic Equations Lesson: PPT, Handouts: 1. Solubility/ Net Ionic/ Stoichiometry Handout 2. Text: 1. P Memorize Rule 1 and 2 HW: 1. Finish all the worksheets, including Textbook questions

How is a precipitate different from insoluble?
Solute ions that don’t dissociate… ~ insoluble Already dissolved ions get together to form a new compound ~precipitate

What does the formation of a precipitate indicate?
Precipitates indicate a chemical reaction occurred How can we know if an ionic compound will dissolve? Two ways: Mix different solutions together and see what happens, or… 2. Learn from what others have already done.

When two aqueous ionic compounds are mixed together, the compounds will either dissociate , or precipitate If no reaction occurs, write NR. Evidence of a double displacement reaction: Precipitate Gas Water

Precipitate Reactions
Based on the solubility guidelines Example: Na2SO4(aq) + 2AgNO3(aq)  Ag2SO4(S) + 2NaNO3(aq)

Reactions Forming a Gas
Reactions that produce a compound that decomposes into a gas and water Example: K2SO3(aq) + 2 HNO3(aq)  2 KNO3(aq) + SO2(g) + H2O(l) CaCO3(aq) +2 HCl(aq)  CaCl2(aq) CO2 (g) + H2O(l)

Reactions Forming Water
Neutralization reaction between an acid and a base to produce water and a salt Example: H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2 H2O(l)

Solubility Slightly Soluble?
Precipitation refers to the formation of a solid from ions. A precipitate is “insoluble” Soluble and insoluble are general terms to describe how much of a solid dissolves. Solubility can be predicted from rules ( Table 1 on pg.424) You will have to memorize some of these rules, and you will need to know how to use the rules to predict solubility. Slightly Soluble? Consider it insoluble Almost everything dissolves at least a little bit.

General Solubility Rules
Salts are generally more soluble in HOT water (Gases are more soluble in COLD water) Alkali Metal salts are very soluble in water. NaCl, KOH, Li3PO4, Na2SO4 etc... Ammonium salts are very soluble in water. NH4Br, (NH4)2CO3 etc… Salts containing the nitrate ion, NO3-, are very soluble in water. Most salts of Cl-, Br- and I- are very soluble in water - exceptions are salts containing Ag+ and Pb2+. soluble salts: FeCl2, AlBr3, MgI2 etc... “insoluble” salts: AgCl, PbBr2 etc...

Soluble or Insoluble? a) Ca(NO3)2 - Soluble rule (salts containing NO3- are soluble) b) FeCl2 - Soluble rule (all chlorides are soluble) c) Ni(OH)2 - Insoluble rule (all hydroxides are insoluble) d) AgNO3 - Soluble rule (salts containing NO3- are soluble) e) BaSO4 - Insoluble rule (Sulfates are soluble, except … Ba2+) f) CuCO3 - Insoluble rule (containing CO32- are insoluble)

S2– PO43– Na+ 2Ca2+ 3Cl– Al3+ Net ionic equations

Review: forming ions Ionic (i.e. salt) refers to +ve ion plus -ve ion
Usually this is a metal + non-metal or metal + polyatomic ion (e.g. NaCl, NaClO3, Li2CO3) Polyatomic ions are listed in the nomenclature package (aq) means aqueous (dissolved in water) For salts (aq) means the salt exists as ions NaCl(aq) is the same as: Na+(aq) + Cl–(aq) Acids form ions: HCl(aq) is H+(aq) + Cl–(aq), Bases form ions: NaOH(aq) is Na+ + OH– Q - how is charge determined (+1, -1, +2, etc.)? A - via valences (periodic table or see the nomenclature package ) F, Cl gain one electron, thus forming F–, Cl– Ca loses two electrons, thus forming Ca2+

Background: valences and formulas
Charge can also be found via the compound E.g. in NaNO3(aq) if you know Na forms Na+, then NO3 must be NO3– (NaNO3 is neutral) By knowing the valence of one element you can often determine the other valences Q - Write the ions that form from Al2(SO4)3(aq)? Step 1 - look at the formula: Al2(SO4)3(aq) Step 2 - determine valences: Al3 (SO4)2 (Al is 3+ according to the periodic table) Step 3 - write ions: Al3+(aq) + 3SO42–(aq) Note that there are 2 aluminums because Al has a subscript of 2 in the original formula

Practice with writing ions
Q - Write ions for Na2CO3(aq) A - 2Na+(aq) + CO32–(aq) (from the PT Na is 1+. There are 2, thus we have 2Na+. There is only one CO3. It must have a 2- charge) Notice that when ions form from molecules, charge can be separated, but the total charge (and number of each atom) stays constant. Q - Write ions for Ca3(PO4)2(aq) & Cd(NO3)2(aq) A - 3Ca2+(aq) + 2PO43–(aq) A - Cd2+(aq) + 2NO3–(aq) Q - Write ions for Na2S(aq) and Mg3(BO3)2(aq) A - 2Na+(aq) + S2–(aq), 3Mg2+(aq)+ 2BO33–(aq)

Types of chemical equations
Equations can be divided into 3 types 1) Molecular, 2) Ionic, 3) Net ionic Here is a typical molecular equation: Cd(NO3)2(aq) + Na2S(aq)  CdS(s) + 2NaNO3(aq) We can write this as an ionic equation (all compounds that are (aq) are written as ions): Cd2+(aq) + 2NO3–(aq) + 2Na+(aq) + S2–(aq)  CdS(s) + 2Na+(aq) + 2NO3–(aq) To get the NET ionic equation we cancel out all terms that appear on both sides: Net: Cd2+(aq) + S2–(aq)  CdS(s)

Ionic Equations Ionic: Ag+(aq) + NO3-(aq) + NH4+(aq) + Cl-(aq) Note: combine, in your head, the positive and negative ions. If together a pair is insoluble, they will form a precipitate (s). In this case AgCl is insoluble  AgCl(s) + NO3-(aq) + NH4+(aq) Net ionic: Ag+(aq) + Cl-(aq)  AgCl(s) If no solid is formed then write N.R.

Equations must be balanced
There are two conditions for molecular, ionic, and net ionic equations Materials balance Both sides of an equation should have the same number of each type of atom Electrical balance Both sides of a reaction should have the same net charge Q- When NaOH(aq) and MgCl2(aq) are mixed, _______ (s) and NaCl(aq) are produced. Write balanced molecular, ionic & net ionic equations Mg(OH)2

Determining net ionic reactions
30/09/99 Determining net ionic reactions Step 1: Write formula for reactants and products using valences. Products are determined by switching +ve and –ve. Step 2: Determine if any products are insoluble (use solubility rules). Note: all reactants must be soluble (i.e. aq) in order to mix. If all products are aqueous: “no reaction” Step 3: Balance the equation Step 4: Write the ionic equation Step 5: Write the net ionic equation

Step 4: Write the ionic equation
If you look at what we call the IONIC EQUATION, you will note that some of the ions are common to both the reactant and product side of the equation. Such ions are called SPECTATOR IONS, as they play no part in the observed reaction. In fact, these spectator ions may be cancelled from each side of the equation, leaving only the NET IONIC EQUATION. Step 5: Write the net ionic equation

First write the skeleton equation
NaOH(aq) + MgCl2(aq)  Mg(OH)2(s) NaCl(aq) Next, balance the equation 2 2 Ionic equation: 2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq)  Mg(OH)2(s) + 2Na+(aq) + 2Cl-(aq) Net ionic equation: 2OH-(aq) + Mg2+(aq)  Mg(OH)2(s) Write balanced ionic and net ionic equations: CuSO4(aq) + BaCl2(aq)  CuCl2(aq) + BaSO4(s) Fe(NO3)3(aq) + LiOH(aq)  ______(aq) + Fe(OH)3(s) Na3PO4(aq) + CaCl2(aq)  _________(s) + NaCl(aq) Na2S(aq) + AgC2H3O2(aq)  ________(aq) + Ag2S(s) LiNO3 Ca3(PO4)2 NaC2H3O2

Net Ionic Equation for Single Displacement Reaction
Write the net ionic equation for the reaction that occurs when aluminum metal is placed in a solution of copper (II) chloride. Go over Sample Problem 2 on P. 427

Net: SO42–(aq) + Ba2+(aq)  BaSO4(s)
Cu2+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq)  Cu2+(aq) + 2Cl–(aq) + BaSO4(s) Net: SO42–(aq) + Ba2+(aq)  BaSO4(s) Fe3+(aq) + 3NO3–(aq) + 3Li+(aq) + 3OH–(aq)  3Li+(aq) + 3NO3–(aq) + Fe(OH)3(s) Net: Fe3+(aq) + 3OH–(aq)  Fe(OH)3(s) 2Na3PO4(aq) + 3CaCl2(aq) Ca3(PO4)2(s)+ 6NaCl(aq) 6Na+(aq) + 2PO43–(aq) + 3Ca2+(aq) + 6Cl–(aq)  Ca3(PO4)2(s)+ 6Na+(aq) + 6Cl–(aq) Net: 2PO43–(aq) + 3Ca2+(aq)  Ca3(PO4)2(s) 2Na+(aq) + S2–(aq) + 2Ag+(aq) + 2C2H3O2–(aq)  2Na+(aq) + 2C2H3O2–(aq) + Ag2S(s) Net: S2–(aq) + 2Ag+(aq)  Ag2S(s)

Solutions stoichiometry

Day 69 – Solutions Stoichiometry
Lesson: PPT, Handouts: 1. Solubility/ Net Ionic/ StoichiometryHandout Text: 1. P HW: 1. Finish all the worksheets, including Textbook questions

Stoichiometry overview
Recall that in stoichiometry the mole ratio provides a necessary conversion factor: grams (x)  moles (x)  moles (y)  grams (y) molar mass of x molar mass of y mole ratio from balanced equation We can do something similar with solutions: volume (x)  moles(x) moles (y) volume(y) mol/L of x mol/L of y mole ratio from balanced equation

Question 1 # L Ca(OH)2= 0.0250 L Al2(SO4)3 0.125 mol Al2(SO4)3
Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. Calculate the volume of mol/L calcium hydroxide solution that can be completely reacted with 25.0 mL of mol/L aluminum sulfate solution. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) Calculate mol Al2(SO4)3, then mol of Ca(OH)2 then v = n/C # L Ca(OH)2= L Al2(SO4)3 0.125 mol Al2(SO4)3 L Al2(SO4)3 x 3 mol Ca(OH)2 1 mol Al2(SO4)3 x L Ca(OH)2 mol Ca(OH)2 x = L Ca(OH)2

Given: Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
13/12/99 Given: Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) Al2(SO4)3 3Ca(OH)2 2Al(OH)3 3CaSO4(s) Molar Ratio (MR) 1 3 2 Volume (L) 0.025L Concentration(mol/L) 0.125mol/L 0.025mol/L Moles (n) V = n/ C ? = / 0.025M = L n = C XV 1 : 3 = 3/1 x = mol mol

Question 2 H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L ammonia solution if 50.0 mL of sulfuric acid is used? H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) Calculate mol H2SO4, then c= n/V = n/ L # mol H2SO4= 2.20 mol NH3 L NH3 x 1 mol H2SO4 2 mol NH3 x L NH3 mol H2SO4 = C = mol/L = mol H2SO4 / L = mol/L

2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
Question 3 A chemistry teacher wants 75.0 mL of mol/L iron(Ill) chloride solution to react completely with an excess of mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed? 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) # L Na2CO3= L FeCl3 0.200 mol FeCl3 L FeCl3 x 3 mol Na2CO3 2 mol FeCl3 x L Na2CO3 0.250 mol Na2CO3 x = L Na2CO3 = 90.0 mL Na2CO3

Question 4 What mass of precipitate should result when L of mol/L aluminum nitrate solution is mixed with L of 1.50 mol/L sodium hydroxide solution?

5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq)
# g Al(OH)3= L Al(NO3)3 0.500 mol Al(NO3)3 L Al(NO3)3 x 1 mol Al(OH)3 1 mol Al(NO3)3 x 77.98 g Al(OH)3 1 mol Al(OH)3 x 21.4 g Al(OH)3 = # g Al(OH)3= L NaOH 1.50 mol NaOH L NaOH x 1 mol Al(OH)3 3 mol NaOH x 77.98 g Al(OH)3 1 mol Al(OH)3 x 9.36 g Al(OH)3 =

? 0.36mol 1 3 0.550 L 0.240 L 0.500 mol/L 1.50 mol/L m = n x MM
Al(NO3)3 3NaOH Al(OH)3(s) 3NaNO3(aq) Molar Ratio (MR) 1 3 Mass (g) Volume (L) 0.550 L 0.240 L Concentration (mol/L)) 0.500 mol/L 1.50 mol/L Moles (n) m = n x MM ? = 0.12 x 77.96 = 9.36 g n = C XV 0.36mol L. F. 1 : 3 = 1/3 x 0.36 0.275 mol = 0.12mol

Assignment H2SO4 reacts with NaOH, producing water and sodium sulfate. What volume of 2.0 M H2SO4 will be required to react completely with 75 mL of 0.50 mol/L NaOH? How many moles of Fe(OH)3 are produced when 85.0 L of iron(III) sulfate at a concentration of mol/L reacts with excess NaOH? What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chloride solution.

Assignment a) What volume of 0.20 mol/L AgNO3 will be needed to react completely with 25.0 mL of 0.50 mol/L potassium phosphate? b) What mass of precipitate is produced from the above reaction? What mass of precipitate should result when L of mol/L aluminum nitrate solution is mixed with L of 1.50 mol/L sodium hydroxide solution?

Answers # L H2SO4= 0.075 L NaOH 0.50 mol NaOH L NaOH x 1 mol H2SO4
1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq) # L H2SO4= L NaOH 0.50 mol NaOH L NaOH x 1 mol H2SO4 2 mol NaOH x L H2SO4 2.0 mol H2SO4 x = L = 9.4 mL 2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq) # mol Fe(OH)3= 0.600 mol Fe2(SO4)3 L Fe2(SO4)3 x 2 mol Fe(OH)3 1 mol Fe2(SO4)3 x 85 L Fe2(SO4)3 = 102 mol

# g Zn(OH)2= = 6.21 g 0.0500 L NaOH 2.50 mol NaOH L NaOH x
3. 2NaOH(aq) + ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq) # g Zn(OH)2= = 6.21 g L NaOH 2.50 mol NaOH L NaOH x 1 mol Zn(OH)2 2 mol NaOH x 99.40 g Zn(OH)2 1 mol Zn(OH)2 x 4a. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq) # L AgNO3 = = L = 0.19 L L K3PO4 0.50 mol K3PO4 L K3PO4 x 3 mol AgNO3 1 mol K3PO4 x L AgNO3 0.20 mol AgNO3 x 4b. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq) # g Ag3PO4= = 5.2 g L K3PO4 0.50 mol K3PO4 L K3PO4 x 1 mol Ag3PO4 1 mol K3PO4 x g Ag3PO4 1 mol Ag3PO4 x

QUALITATIVE ANALYSIS, ELECTROMAGNETIC SPECTRUM AND FLAME TESTS

Agenda Day 13 - QUALITATIVE ANALYSIS, ELECTROMAGNETIC SPECTRUM AND FLAME TESTS Lesson: PPT, Handouts: 1.PPT Handout; Text: 1. P. 20; Qualitative Analysis Involving Colours HW: 1. Finish all the worksheets, including Textbook questions

Reactions in Aqueous Solutions
Sodium chloride + Silver nitrate  ? Predict what will occur in this reaction. Observe the reaction – record qualitative observations. Explain what occurred – do your observation match the predictions?

Reactions in AqeousSolutions
Sodium chloride + Silver nitrate  ? Recall: Double displacement reactions only occur if one or more of the products form either a solid precipitate, a liquid or a gas. NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s) Net Ionic Equation: Cl-(aq) + Ag+(aq)  AgCl(s)

Identifying Ions in Solution
A variety of qualitative analysis techniques exist to determine the type of ions present in an aqueous solution. 1) Colour of Solution Some ions produce a characteristic colour when dissolved in solution Ex) Copper (II) is blue Ex) Copper (I) is green Ex) Iron (III) is yellow-orange Ex) Permanganate (MnO4-) is purple

A variety of qualitative analysis techniques exist to determine the type of ions present in an aqueous solution. 2) Colour of precipitate Some metal ions produce precipitates with a characteristic colour. Ex) Lead (II) iodide is bright yellow ppt.

3) Sequential Qualitative Chemical Analysis The addition of other specific ions to an unknown solution can help to identify the presence of unknown ions through the observation of a precipitate Mixture of Ions: Ba2+, Ag+, Ca2+ Check the solubility table for an anion that will precipitate with only one of these cations. After precipitation, remove the precipitate and test the remaining solution (the filtrate) for the presence of the other anion.

3) Sequential Qualitative Chemical Analysis The addition of other specific ions to an unknown solution can help to identify the presence of unknown ions through the observation of a precipitate Mixture of Ions: Ba2+, Ag+, Ca2+ Mixture of Ions: Ba2+, Ca2+ Mixture of Ions: Ca2+ Filter, then add Na2SO4(aq) Filter, then add Na3PO4(aq) Add NaCl(aq) AgCl(s) BaSO4(s) Ca3(PO4)2(s)

FLAME TESTS Flame tests rely on the idea that each element has a characteristic set of emissions and will produce a unique and characteristic colour when heated. The sample is identified by comparing the observed flame color against known values from a table or chart. Limitations of the Flame Test The test cannot detect low concentrations of most ions. The brightness of the signal varies from one sample to another. For example, the yellow emission from sodium is much brighter than the red emission from the same amount of lithium. Impurities or contaminants affect the test results. Sodium, in particular, is present in most compounds and will color the flame. Sometimes a blue glass is used to filter out the yellow of sodium. The test cannot differentiate between all elements. Several metals produce the same flame color. Some compounds do not change the color of the flame at all.

4) Flame Test When dissolved metallic ions are placed in a flame, the flame emits a specific colour characteristic of that ion Ion Symbol Colour Lithium Li+ Crimson red Sodium Na+ Yellow-orange Potassium K+ Lavender Cesium Cs+ Blue Calcium Ca2+ Reddish-orange Strontium Sr2+ Bright Red Barium Ba2+ Yellowish-green Copper (II) Cu2+ Bluish-green Lead (II) Pb2+ Bluish-white

Flame Test Colors Symbol Element Color As Arsenic Blue B Boron Bright green Ba Barium Pale/Yellowish Green Ca Calcium Orange to red Cs Cesium Blue Cu(I) Copper(I) Blue Cu(II) Copper(II) non-halide Green Cu(II) Copper(II) halide Blue-green Fe Iron Gold In Indium Blue K Potassium Lilac to red Li Lithium Magenta to carmine Mg Magnesium Bright white Mn(II) Manganese(II) Yellowish green Mo Molybdenum Yellowish green Na Sodium Intense yellow P Phosphorus Pale bluish green Pb Lead Blue Rb Rubidium Red to purple-red Sb Antimony Pale green Se Selenium Azure blue Sr Strontium Crimson Te Tellurium Pale green Tl Thallium Pure green Zn Zinc Bluish green to whitish green

Let’s Review Energy – Radiation of different wavelengths affect matter differently – certain wavelengths (near infrared) may burn your skin with a heat burn, overexposure to X radiation causes tissue damage. These diverse effects are due to differences in the energy of the radiation. Radiation of high frequency and short wavelength are more energetic than radiation of lower frequency and longer wavelength.

While light exhibits many wavelike properties, it can also be thought of as a stream of particles. Each particle of light carries a quantum of energy. Einstein called these particles PHOTONS. A photon is a particle of electromagnetic radiation having zero mass and carrying a quantum of energy. Albert Einstein

The electromagnetic spectrum is continuous showing no breaks between different energy waves.
When looking at visible, ‘white light’ through a spectroscope there is a continuous spectrum of coloured light – R O Y G B V (the colours of the rainbow). Red has the lowest energy with the largest wavelength, smallest frequency and violet has the greatest , with lower wavelength and greater frequency.

The Hydrogen-Atom Line-Emission Spectrum
When investigators passed an electric current through a vacuum tube containing hydrogen gas at low pressure, they observed the emission of a characteristic pinkish glow. When a narrow beam of the emitted light was shined through a prism, it was separated into a series of specific frequencies (and therefore specific wavelengths, c =) of visible light. The bands of light were part of what is known as hydrogen’s LINE-EMISSION SPECTRUM.

When gaseous atoms of elements are heated and the electrons absorb sufficient energy to ‘escape’ their ground state configuration, ‘excited’ electrons jump to higher energy levels ( the greater the distance between the nucleus of the atom and the energy level the greater the energy needed by the electron to travel to that level). ‘Excited electrons’ are unstable and so return to ground state releasing the absorbed energy in discrete energy emissions patterns. In the visible spectrum, these emissions appear as discrete lines of colour in a spectroscope.

Bohr’s Theory of the Atom:

Bohr’s Model of the Atom
The hydrogen atom emits visible light when its electron moves from the third through sixth energy levels to the second energy level. Ultraviolet radiation is emitted when the electron moves from the second through sixth energy levels to the first energy level. Infrared radiation is emitted when the electron moves from the fourth through sixth energy levels to the third energy level. Complete the following diagram so the purple and red arrows were used to represent the transitions that result in ultraviolet and infrared emissions to the diagrams below.

Q: “ Spectra lines are the fingerprints of elements”
Q: “ Spectra lines are the fingerprints of elements”. Explain what is meant by this statement. Q: According to the Bohr theory, what happens to an electron in an atom as it absorbs energy and as it releases energy?

Flame Test Demo- Pg. 439 Compound Cation Colour Lithium chloride Li+
Potassium chloride K+ Copper (II) chloride Cu2+ Sodium chloride Na+ Strontium chloride Sr2+

Acids and Bases

Agenda Day 71 – Strong and Weak Acids and Bases Intro Lesson:
“the acid test” “acid drop” “acid rain” “put the acid on” “do acid” “acid head” Day 71 – Strong and Weak Acids and Bases Intro Lesson: Handouts: 1. Acid/Base Handout Text: 1. P , Dissociation vs Ionization Arrhenius and Bronsted-Lowry Definitions of Acids and Bases HW: 1. Finish all the worksheets, including Textbook questions

Properties of acids and bases
Get 8 test tubes. Rinse all tubes well with water. Add acid to four tubes, base to the other four. Touch a drop of base to your finger. Record the feel in the chart (on the next slide). Wash your hands with water. Repeat for acid. Use a stirring rod, add base to the litmus and pH papers (for pH paper use a colour key to find a number). Record results. Repeat for acid. Into the four base tubes add: a) two drops of phenolphthalein, b) 2 drops of bromothymol, c) a piece of Mg, d) a small scoop of baking soda. Record results. Repeat for acid. Clean up (wash tubes, pH/litmus paper in trash).

Describe the solution in each of the following as:
1) acid 2) base or 3)neutral. A. ___soda B. ___soap C. ___coffee D. ___ wine E. ___ water F. ___ grapefruit

Describe each solution as:
1) acid 2) base or 3) neutral. A. _1_ soda B. _2_ soap C. _1_ coffee D. _1_ wine E. _3_ water F. _1_ grapefruit

Identify each as characteristic of an A) acid or B) base
____ 1. Sour taste ____ 2. Produces OH- in aqueous solutions ____ 3. Chalky taste ____ 4. Is an electrolyte ____ 5. Produces H+ in aqueous solutions

Identify each as a characteristic of an A) acid or B) base
_A_ 1. Sour taste _B_ 2. Produces OH- in aqueous solutions _B_ 3. Chalky taste A, B 4. Is an electrolyte _A_ 5. Produces H+ in aqueous solutions

Properties of Acids They taste sour (don’t try this at home).
They can conduct electricity. Can be strong or weak electrolytes in aqueous solution React with metals to form H2 gas. Change the color of indicators (for example: blue litmus turns to red). React with bases (metallic hydroxides) to form water and a salt.

Properties of Acids They have a pH of less than 7 (more on this concept of pH in a later lesson) They react with carbonates and bicarbonates to produce a salt, water, and carbon dioxide gas How do you know if a chemical is an acid? It usually starts with Hydrogen. HCl, H2SO4, HNO3, etc. (but not water!)

Acids Affect Indicators, by changing their color
Blue litmus paper turns red in contact with an acid (and red paper stays red).

Acids React with Active Metals
Acids react with active metals to form salts and hydrogen gas: HCl(aq) + Mg(s) → MgCl2(aq) + H2(g) This is a single-replacement reaction

Acids React with Carbonates and Bicarbonates
HCl + NaHCO3 Hydrochloric acid + sodium bicarbonate NaCl + H2O + CO2 salt + water + carbon dioxide An old-time home remedy for relieving an upset stomach

Effects of Acid Rain on Marble (marble is calcium carbonate)
George Washington: BEFORE acid rain George Washington: AFTER acid rain

Acids Neutralize Bases
HCl + NaOH → NaCl + H2O -Neutralization reactions ALWAYS produce a salt (which is an ionic compound) and water. -Of course, it takes the right proportion of acid and base to produce a neutral salt

Sulfuric Acid = H2SO4 Highest volume production of any chemical in the U.S. (approximately 60 billion pounds/year) Used in the production of paper Used in production of fertilizers Used in petroleum refining; auto batteries

Nitric Acid = HNO3 Used in the production of fertilizers
Used in the production of explosives Nitric acid is a volatile acid – its reactive components evaporate easily Stains proteins yellow (including skin!)

Hydrochloric Acid = HCl
Used in the “pickling” of steel Used to purify magnesium from sea water Part of gastric juice, it aids in the digestion of proteins Sold commercially as Muriatic acid

Phosphoric Acid = H3PO4 A flavoring agent in sodas (adds “tart”)
Used in the manufacture of detergents Used in the manufacture of fertilizers Not a common laboratory reagent

Acetic Acid = HC2H3O2 (also called Ethanoic Acid, CH3COOH)
Used in the manufacture of plastics Used in making pharmaceuticals Acetic acid is the acid that is present in household vinegar

Properties of Bases (metallic hydroxides)
React with acids to form water and a salt. Taste bitter. Feel slippery (don’t try this either). Can be strong or weak electrolytes in aqueous solution Change the color of indicators (red litmus turns blue).

Examples of Bases (metallic hydroxides)
Sodium hydroxide, NaOH (lye for drain cleaner; soap) Potassium hydroxide, KOH (alkaline batteries) Magnesium hydroxide, Mg(OH)2 (Milk of Magnesia) Calcium hydroxide, Ca(OH)2 (lime; masonry)

Bases Affect Indicators
Red litmus paper turns blue in contact with a base (and blue paper stays blue). Phenolphthalein turns purple in a base.

Bases have a pH greater than 7

Bases Neutralize Acids
Milk of Magnesia contains magnesium hydroxide, Mg(OH)2, which neutralizes stomach acid, HCl. 2 HCl + Mg(OH)2 Magnesium salts can cause diarrhea (thus they are used as a laxative) and may also cause kidney stones. MgCl2 + 2 H2O

Acid-Base Theories OBJECTIVES: Compare and contrast acids and bases as defined by the theories of: Arrhenius, Brønsted-Lowry, and Lewis.

Svante Arrhenius He was a Swedish chemist ( ), and a Nobel prize winner in chemistry (1903) One of the first chemists to explain the chemical theory of the behavior of acids and bases Svante Arrhenius ( )

1. Arrhenius Definition - 1887
Acids produce hydrogen ions (H1+) in aqueous solution (HCl → H1+ + Cl1-) Bases produce hydroxide ions (OH1-) when dissolved in water. (NaOH → Na1+ + OH1-) Limited to aqueous solutions. Only one kind of base (hydroxides) NH3 (ammonia) could not be an Arrhenius base: no OH1- produced.

Polyprotic Acids? Some compounds have more than one ionizable hydrogen to release HNO3 nitric acid - monoprotic H2SO4 sulfuric acid - diprotic - 2 H+ H3PO4 phosphoric acid - triprotic - 3 H+ Having more than one ionizable hydrogen does not mean stronger!

Acids Not all compounds that have hydrogen are acids. Water?
Also, not all the hydrogen in an acid may be released as ions only those that have very polar bonds are ionizable - this is when the hydrogen is joined to a very electronegative element

Arrhenius examples... Consider HCl = it is an acid!
What about CH4 (methane)? O (e.g. H2SO4) was originally thought to cause acidic properties. Later, H was implicated, but it was still not clear why CH4 was neutral. CH3COOH (ethanoic acid, also called acetic acid) - it has 4 hydrogens just like methane does…?

Arrhenius’ theory Limitation
Using Arrhenius’ theory the following would be incorrectly classified as neutral 1. Compounds of hydrogen polyatomic ions (NaHCO3(aq)) Oxides of metals and non metals (CaO(aq) and CO2(g)) Bases other than hydroxides (NH3(aq) and Na2CO3(aq)) Acids that do not contain hydrogen (Al(NO3)3(aq))

Revised Arrhenius theory
Arrhenius made the revolutionary suggestion that some solutions contain ions & that acids produce H3O+ (hydronium) ions in solution. The revised Arrhenius theory involves two key ideas not considered by Arrhenius 1. Collisions with water molecules 2. The nature of hydrogen ions Ionization + H O H O Cl Cl H + +

Agenda Day 72 – Conjugate Acids and Bases Lesson: PPT
Handouts: 1. Acid/Base Handout. 2 Conjugate Acid& Base Worksheet Text: 1. [page old text photocopy!] HW: 1. [P.389 # 18, 19 page 392 # 8, 9, 11 old text photocopy!]

Diagnostic Quiz Q1. Identify the acid, base, conjugate acid, conjugate base, and conjugate acid-base pairs. CO32–(aq) + HC2H3O2(aq)  C2H3O2–(aq) + HCO3

2. Brønsted-Lowry - 1923 A broader definition than Arrhenius
Acid is hydrogen-ion donor (H+ or proton); base is hydrogen-ion acceptor. Acids and bases always come in pairs. HCl is an acid. When it dissolves in water, it gives it’s proton to water. HCl(g) + H2O(l) ↔ H3O+(aq) + Cl-(aq) Water is a base; makes hydronium ion.

Johannes Brønsted Thomas Lowry (1879-1947) (1874-1936) Denmark England

Brønsted-Lowry Theory of Acids & Bases Conjugate Acid-Base Pairs
General Equation

Why Ammonia is a Base Ammonia can be explained as a base by using Brønsted-Lowry: NH3(aq) + H2O(l) ↔ NH41+(aq) + OH1-(aq) Ammonia is the hydrogen ion acceptor (base), and water is the hydrogen ion donor (acid). This causes the OH1- concentration to be greater than in pure water, and the ammonia solution is basic

Acids and bases come in pairs
A “conjugate base” is the remainder of the original acid, after it donates it’s hydrogen ion A “conjugate acid” is the particle formed when the original base gains a hydrogen ion. Thus, a conjugate acid-base pair is related by the loss or gain of a single hydrogen ion. Chemical Indicators? They are weak acids or bases that have a different color from their original acid and base

Acids and bases come in pairs
General equation is: HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) Acid + Base ↔ Conjugate acid + Conjugate base NH3 + H2O ↔ NH41+ + OH1- base acid c.a c.b. HCl + H2O ↔ H3O1+ + Cl1- acid base c.a c.b. Amphoteric – a substance that can act as both an acid and base- as water shows

accepts H+ donates H+ When life goes either way
amphoteric (amphiprotic) substances Acting like a base Acting like an acid HCO3- + H+ - H+ H2CO3 CO3-2 accepts H+ donates H+

Brønsted-Lowry Theory of Acids & Bases

Brønsted-Lowry Theory of Acids & Bases
Notice that water is both an acid & a base = amphoteric Reversible reaction

Organic Acids (those with carbon)
Organic acids all contain the carboxyl group, (-COOH), sometimes several of them. CH3COOH – of the 4 hydrogen, only 1 ionizable (due to being bonded to the highly electronegative Oxygen) The carboxyl group is a poor proton donor, so ALL organic acids are weak acids.

Conjugate Acid-Base Pairs

Conjugate Acid- Base Pairs
In other words: When a proton is gained by a Bronsted-Lowry base, the product formed is referred to as the base’s conjugate acid Conjugate Acid Conjugate Base H2O (l) NH4+(aq) HCO3- OH-(aq) H3O+(aq) NH3 (aq) CO3-2 H2CO3 HCO3-

Practice problems Identify the acid, base, conjugate acid, conjugate base, and conjugate acid-base pairs: HC2H3O2(aq) + H2O(l)  C2H3O2–(aq) + H3O+(aq) acid base conjugate base conjugate acid conjugate acid-base pairs OH –(aq) + HCO3–(aq)  CO32–(aq) + H2O(l) base acid conjugate base conjugate acid conjugate acid-base pairs

Answers: question 18 (a) HF(aq) + SO32–(aq)  F–(aq) + HSO3–(aq) acid
base conjugate base conjugate acid conjugate acid-base pairs (b) CO32–(aq) + HC2H3O2(aq)  C2H3O2–(aq) + HCO3–(aq) base acid conjugate base conjugate acid (c) conjugate acid-base pairs H3PO4(aq) + OCl –(aq)  H2PO4–(aq) + HOCl(aq) acid base conjugate base conjugate acid conjugate acid-base pairs

HCO3–(aq) + S2–(aq)  HS–(aq) + CO32–(aq) acid base conjugate acid
conjugate base conjugate acid-base pairs 8b) H2CO3(aq) + OH –(aq)  HCO3–(aq) + H2O(l) acid base conjugate base conjugate acid conjugate acid-base pairs 11a) H3O+(aq) + HSO3–(aq)  H2O(l) + H2SO3(aq) acid base conjugate base conjugate acid conjugate acid-base pairs 11b) OH –(aq) + HSO3–(aq)  H2O(l) + SO32–(aq) base acid conjugate acid conjugate base conjugate acid-base pairs

What is the conjugate base of the following substances?
H2O ________________ NH4+________________ HNO2_______________ HC2H3O2_________________ What is the conjugate acid of the following substances? HCO3-__________________ H2O____________ HPO42-____________ NH3___________

Strengths of Acids and Bases
OBJECTIVES: Define strong acids and weak acids.

Strength OBJECTIVES: Define strong acids and weak acids.
Acids and Bases are classified according to the degree to which they ionize in water: Strong are completely ionized in aqueous solution; this means they ionize 100 % Weak ionize only slightly in aqueous solution Strength is very different from Concentration

Strength Strong – means it forms many ions when dissolved (100 % ionization) Mg(OH)2 is a strong base- it falls completely apart (nearly 100% when dissolved). But, not much dissolves- so it is not concentrated

Let’s examine the behavior of an acid, HA, in aqueous solution.
CHM 101 Let’s examine the behavior of an acid, HA, in aqueous solution. Sinex HA What happens to the HA molecules in solution?

100% ionization of HA HA H+ Strong Acid A- Would the solution be conductive?

Strong Acid Ionizes (makes 100 % ions)

Partial ionization of HA
Weak Acid A- Would the solution be conductive?

At any one time, only a fraction of the molecules are ionized.
HA  H A- HA H+ Weak Acid A- At any one time, only a fraction of the molecules are ionized.

Weak Acid Ionzation (only partially ionizes)

Strength of ACIDS 1. Binary or hydrohalic acids – HCl, HBr, and HI “hydro____ic acid” are strong acids. Other binary acids are weak acids (HF and H2S). Although the H-F bond is very polar, the bond is so strong (due to the small F atom) that the acid does not completely ionize.

2. Oxyacids – contain a polyatomic ion a
2. Oxyacids – contain a polyatomic ion a. strong acids (contain 2 or more oxygen per hydrogen) HNO3 – nitric from nitrate H2SO4 - sulfuric from sulfate HClO4 - perchloric from perchlorate

b. weak acids (acids with l less oxygen than the “ic” ending HNO2 – nitrous from nitrite H3PO3 - phosphorous from phosphite H2SO3 - sulfurous from sulfite HClO2 - chlorous from chlorite c. weaker acids (acids with “hypo ous” have less oxygen than the “ous” ending HNO - hyponitrous H3PO2 - hypophosphorus HClO - hypochorous

d. Organic acids – have carboxyl group -COOH - usually weak acids HC2H3O2 - acetic acid C7H5COOH - benzoic acid

Strength of Bases Strong Bases: metal hydroxides of Group I and II metals (except Be) that are soluble in water and dissociate (separates into ions) completely in dilute aqueous solutions Weak Bases: a molecular substance that ionizes only slightly in water to produce an alkaline (basic) solution (ex. NH3)

What is a strong Base? NaOH(s)  Na+ + OH-
A base that is completely dissociated in water (highly soluble). NaOH(s)  Na+ + OH- Strong Bases: Group 1A metal hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) Heavy Group 2A metal hydroxides [Ca(OH)2, Sr(OH)2, and Ba(OH)2]

For the following identify the acid and the base as strong or weak .
a. Al(OH) HCl  b. Ba(OH) HC2H3O2  c. KOH H2SO4  d. NH H2O  Strong acid Weak base Strong base Weak acid Strong acid Strong base Weak base Weak acid

Strength vs. Concentration
The words concentrated and dilute tell how much of an acid or base is dissolved in solution - refers to the number of moles of acid or base in a given volume The words strong and weak refer to the extent of ionization of an acid or base Is a concentrated, weak acid possible?

Summary: Definitions Acids – produce H+ Bases - produce OH-
Acids – donate H+ Bases – accept H+ Acids – accept e- pair Bases – donate e- pair Arrehenius only in water Bronsted-Lowry any solvent Lewis used in organic chemistry, wider range of substances

Acids Bases Arrhenius Acid: donates (or produces) hydronium ions (H3O+) in water or hydrogen ions (H+) in water Bronsted-Lowry Acid: donates a proton (H+) in water, H3O+ has an extra H+, if it donated it to another molecule it would be H2O (page 467) HNO3 + H2O  H+ + NO3- HNO3 + H2O  H3O+ + NO3- HCl + H2O  H+ + Cl- HCl + H2O  H3O+ + Cl- Arrhenius Base: donates (or produces) hydroxide ions (OH-) in water Bronsted – Lowry Base: accepts a proton in water, OH- needs an extra H+ if it accepts one from another molecule it would be H2O (page 468) KOH + H2O  K+ + OH- NH3 + H2O  NH4+ + OH-

Hydrogen Ions and Acidity
OBJECTIVES: Describe how [H1+] and [OH1-] are related in an aqueous solution. Classify a solution as neutral, acidic, or basic given the hydrogen-ion or hydroxide-ion concentration. Convert hydrogen-ion concentrations into pH values and hydroxide-ion concentrations into pOH values. Describe the purpose of an acid-base indicator.

Agenda Day 73 – pH Calculations Lesson: PPT
Handouts: 1. pH Handout, 2. pH Calculations Worksheet Text: 1. HW: 1. Finish all the worksheets, including Textbook questions

Hydrogen Ions from Water
Water ionizes, and forms ions: H2O + H2O↔ H3O1+ + OH1- Called the “self ionization” of water Occurs to a very small extent: [H3O1+ ] = [OH1-] = 1 x 10-7 M Since they are equal, a neutral solution results from water Kw = [H3O1+ ] x [OH1-] = 1 x M2 Kw is called the “ion product constant” for water

Water Equilibrium

How are (H3O+) and (OH-) related?
Does pure water conduct electrical current? Water is a very, very, very weak electrolyte. H2O + H2O  H3O OH- How are (H3O+) and (OH-) related? [H3O+][OH-] = 10-14 For pure water: [H3O+] = [OH-] = 10-7M This is neutrality and at 25oC is a pH = 7. water

Lone Hydrogen ions do not exist by themselves in solution
Lone Hydrogen ions do not exist by themselves in solution. H+ is always bound to a water molecule to form a hydronium ion

Ion Product Constant H2O ↔ H3O1+ + OH1-
Kw is constant in every aqueous solution: [H3O+] x [OH-] = 1 x M2 If [H3O+] > 10-7 then [OH-] < 10-7 If [H3O+] < 10-7 then [OH-] > 10-7 If we know one, other can be determined If [H3O+] > 10-7 , it is acidic and [OH-] < 10-7 If [H3O+] < 10-7 , it is basic and [OH-] > 10-7 Basic solutions also called “alkaline”

The pH concept – from 0 to 14 pH = pouvoir hydrogene (Fr.) “hydrogen power” definition: pH = -log[H3O+] in neutral pH = -log(1 x 10-7) = 7 in acidic solution [H3O+] > 10-7 pH < -log(10-7) pH < 7 (from 0 to 7 is the acid range) in base, pH > 7 (7 to 14 is base range)

pH Scale [ ] brackets mean concentration or Molarity
The pH scale indicates the hydronium ion concentration, [H3O+] or molarity, of a solution. (In other words how many H3O+ ions are in a solution. If there are a lot we assume it is an acid, if there are very few it is a base.)

pH acid rain (NOx, SOx) pH of 4.2 - 4.4 in 0-14 scale for the chemists
3 4 5 6 7 8 9 10 11 12 acidic (H+) > (OH-) 25oC (H+) = (OH-) distilled water basic or alkaline (H+) < (OH-) normal rain (CO2) pH = 5.3 – 5.7 fish populations drop off pH < 6 and to zero pH < 5 natural waters pH =

pH Scale A change of 1 pH unit represents a tenfold change in the acidity of the solution. For example, if one solution has a pH of 1 and a second solution has a pH of 2, the first solution is not twice as acidic as the second—it is ten times more acidic.

Calculating pOH pOH = -log [OH-] [H+] x [OH-] = 1 x 10-14 M2
pH + pOH = 14 Thus, a solution with a pOH less than 7 is basic; with a pOH greater than 7 is an acid Not greatly used like pH is.

pH and Significant Figures
For pH calculations, the hydrogen ion concentration is usually expressed in scientific notation [H1+] = M = 1.0 x 10-3 M, and has 2 significant figures the pH = 3.00, with the two numbers to the right of the decimal corresponding to the two significant figures

Sample Problem: If [H3O+] = 1.4 x 10-6 mol/L, calculate pH.
pH = -log[H3O+] pH = -log[1.4 x 10-6 ] pH = 5.85 Sample Problem: If pH = 6.45 calculate [H3O+]. [H3O+]= 10 –pH [H3O+]= [H3O+]= 3.5 x 10-7 mol/L

1. What is the pH of a 0.001M NaOH solution?
Example Problems: 1. What is the pH of a 0.001M NaOH solution? 1st step: Write a dissociation equation for NaOH NaOH  Na OH- 0.001mol mol Hydroxide will be produced and the [OH-] = 0.001M 2nd step: pOH = -log [0.001] pOH = 3.0 pH = = 11.0

What is acid rain? NO, NO2 + H2O …  HNO3 SO2, SO3 + H2O …  H2SO4
Dissolved carbon dioxide lowers the pH CO2 (g) + H2O  H2CO3  H+ + HCO3- Atmospheric pollutants from combustion NO, NO H2O …  HNO3 both strong acids SO2, SO H2O …  H2SO4 pH < 5.3

Behavior of oxides in water– Group A
105 Db 107 Bh basic amphoteric acidic 8A 1A 3A 4A 5A 6A 7A 2A Group B basic: Na2O + H2O  2NaOH (O H2O  2OH-) acidic: CO H2O  H2CO3

Measuring pH Why measure pH?
Everyday solutions we use - everything from swimming pools, soil conditions for plants, medical diagnosis, soaps and shampoos, etc. Sometimes we can use indicators, other times we might need a pH meter

pH in the Digestive System
Mouth-pH around 7. Saliva contains amylase, an enzyme which begins to break carbohydrates into sugars. Stomach- pH around 2. Proteins are broken down into amino acids by the enzyme pepsin. Small intestine-pH around 8. Most digestion ends. Small molecules move to bloodstream toward cells that use them

pH The biological view in the human body acidic basic/alkaline saliva
1 2 3 4 5 6 7 8 9 10 11 saliva blood urine gastric juice pancreatic juice bile cerebrospinal fluid

How to measure pH with wide-range paper
1. Moisten the pH indicator paper strip with a few drops of solution, by using a stirring rod. 2.Compare the color to the chart on the vial – then read the pH value.

Some of the many pH Indicators and their pH range

Acid-Base Indicators Although useful, there are limitations to indicators: usually given for a certain temperature (25 oC), thus may change at different temperatures what if the solution already has a color, like paint? the ability of the human eye to distinguish colors is limited

Acid-Base Indicators A pH meter may give more definitive results
some are large, others portable works by measuring the voltage between two electrodes; typically accurate to within 0.01 pH unit of the true pH Instruments need to be calibrated

Mixing Acids and Bases

Neutralization Reactions
OBJECTIVES: Define the products of an acid-base reaction. Explain how acid-base titration is used to calculate the concentration of an acid or a base. Explain the concept of equivalence in neutralization reactions.

Demo- Discrepant Events
500 mL of 2M HCl mL of 2M NaOH Calculate the amount of water formed. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 1000ml Volumetric flask, Retort stand and ring Funnel Ions are tiny – once put together they occupy more volume ( water is bigger, intermolecular forces) tiek lim

Agenda Day 74 – Acid & Base Titration - Stoichiometry/pH Calculations
Lesson: PPT Handouts: 1. Titration Handout 2. Titration Problems Worksheet Text: 1. P Titration HW: 1. Finish all the worksheets, including Textbook questions

Activity Reviewing Acids and Bases

Acid-Base Reactions Acid + Base → Water + Salt
Properties related to every day: antacids depend on neutralization farmers adjust the soil pH formation of cave stalactites human body kidney stones from insoluble salts

Acid – Base reactions Each salt listed in this table can be formed by the reaction between an acid and a base.

Acid-Base Reactions Neutralization Reaction - a reaction in which an acid and a base react in an aqueous solution to produce a salt and water: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2 H2O(l) According to the Bronsted-Lowry theory, in a neutralization reaction a proton is transferred from the strongest acid to the strongest base

Acid – Base Reactions A reaction between an acid and a base is called neutralization. An acid-base mixture is not as acidic or basic as the individual starting solutions.

Titration- Stoichiometry
Titration is the process of adding a known amount of solution of known concentration to determine the concentration of another solution Remember? - a balanced equation is a mole ratio The equivalence point is when the moles of hydronium ions is equal to the moles of hydroxide ions (= neutralized!)

Titration The concentration of acid (or base) in solution can be determined by performing a neutralization reaction An indicator is used to show when neutralization has occurred Often we use phenolphthalein- because it is colorless in neutral and acid; turns pink in base

Steps - Neutralization reaction
#1. A measured volume of acid of unknown concentration is added to a flask #2. Several drops of indicator added #3. A base of known concentration is slowly added, until the indicator changes color; measure the volume.

Neutralization The solution of known concentration is called the standard solution added by using a burette Figure 1, page 476 Continue adding until the indicator changes color called the “end point” of the titration Go over Sample Problem1 and 2 , page 482

Writing neutralization equations
When acids and bases are mixed, a salt forms NaOH + HCl  H2O + NaCl base + acid  water + salt Ca(OH) H2SO4  2H2O + CaSO4 Question: Write the chemical reaction when lithium hydroxide is mixed with carbonic acid. Step 1: write out the reactants LiOH(aq) + H2CO3(aq)  Step 2: determine products … H2O and Li1(CO3)2 LiOH(aq) + H2CO3(aq)  Li2CO3(aq) + H2O(l) Step 3: balance the equation 2LiOH(aq) + H2CO3(aq)  Li2CO3(aq) + 2H2O(l) lithium hydroxide + carbonic acid  lithium carbonate + water

Assignment Write balanced chemical equations for these neutralization reactions. Under each compound give the correct IUPAC name. iron(II) hydroxide + phosphoric acid Ba(OH)2(aq) + HCl(aq) calcium hydroxide + nitric acid Al(OH)3(aq) + H2SO4(aq) ammonium hydroxide + hydrosulfuric acid KOH(aq) + HClO2(aq)

a) 3Fe(OH)2(aq) + 2H3PO4(aq)  Fe3(PO4)2(aq) + 6H2O(l) iron(II) hydroxide + phosphoric acid  iron (II) phosphate b) Ba(OH)2(aq) + 2HCl(aq)  BaCl2 (aq) + 2H2O(l) barium hydroxide + hydrochloric acid  barium chloride c) Ca(OH)2(aq) + 2HNO3(aq)  Ca(NO3)2(aq) + 2H2O(l) calcium hydroxide + nitric acid  calcium nitrate d) 2Al(OH)3(aq) + 3H2SO4(aq)  Al2(SO4)3(aq) + 6H2O(l) aluminum hydroxide + sulfuric acid  aluminum sulfate e) 2NH4OH(aq) + H2S(aq)  (NH4)2S(aq) + 2H2O(l) ammonium hydroxide+ hydrosulfuric acid ammonium sulfide f) KOH(aq) + HClO2(aq)  KClO2(aq) + H2O(l) potassium hydroxide + chlorous acid  potassium chlorite

1 x Ma Va = 1 x Mb Vb rearranges to Ma = Mb Vb / Va
Sample Problem: Suppose mL of hydrochloric acid was required to neutralize mL of 0.52 M NaOH. What is the molarity ( concentration) of the acid? HCl + NaOH  H2O + NaCl 1 x Ma Va = 1 x Mb Vb rearranges to Ma = Mb Vb / Va so Ma = (0.52 M) (22.50 mL) / (75.00 mL) = 0.16 M Now you try: If mL of M HNO3 neutralized mL of KOH, what is the molarity of the base? Mb = mol/L

TITRATION- Sample Problem: If mL of M LiOH neutralized 40.50 mL of H2SO4, what is the molarity of the acid? 2 LiOH + H2SO4  Li2SO4 + 2 H2O 1. First calculate the moles of base: L LiOH (0.543 mol/1 L) = mol LiOH 2. Next calculate the moles of acid: mol LiOH (1 mol H2SO4 / 2 mol LiOH)= mol H2SO4 3. Last calculate the Molarity: Ma = n/V = mol H2SO4 / L = M

Demo- Precipitation & Conductivity
About 75mL of saturated Ba(OH)2 ( 39 g/L of ocahydrate) and 9M H2SO4 Add acid drop by drop Use magnetic stirrer Ba(OH)2 (aq) + H2SO4 (aq)  BaSO4 (s) + H2O (l) Add extra drop ( conductivity back) H2SO4 (aq) + H2O (l)  H3O +(aq) + SO4 (aq)

Now you try it: If 20. 42 mL of Ba(OH)2 solution was used to titrate29
Now you try it: If mL of Ba(OH)2 solution was used to titrate29.26 mL of M HCl, what is the molarity of the barium hydroxide solution? Mb = mol/L

Titration problems What volume of 0.10 mol/L NaOH is needed to neutralize 25.0 mL of 0.15 mol/L H3PO4? 25.0 mL of HCl(aq) was neutralized by 40.0 mL of 0.10 mol/L Ca(OH)2 solution. What was the concentration of HCl? A truck carrying sulfuric acid is in an accident. A laboratory analyzes a sample of the spilled acid and finds that 20 mL of acid is neutral-ized by 60 mL of 4.0 mol/L NaOH solution. What is the concentration of the acid? What volume of 1.50 mol/L H2S will neutral-ize a solution containing 32.0 g NaOH?

Titration problems 1. (3)(0.15 M)(0.0250 L) = (1)(0.10 M)(VB)
VB= (3)(0.15 M)( L) / (1)(0.10 M) = 0.11 L 2. (1)(MA)( L) = (2)(0.10 M)(0.040 L) MA= (2)(0.10 M)(0.040 L) / (1)( L) = 0.32 M 3. Sulfuric acid = H2SO4 (2)(MA)(0.020 L) = (1)(4.0 mol/L)(0.060 L) MA = (1)(4.0 M)(0.060 L) / (2)(0.020 L) = 6.0 M 4. mol NaOH = 32.0 g x 1 mol/40.00 g = (2)(1.50 mol/L)(VA) = (1)(0.800 mol) VA= (1)(0.800 mol) / (2)(1.50 mol/L) = L

Molarity and Titration
A student finds that mL of a M NaOH solution is required to titrate a mL sample of hydr acid solution. What is the molarity of the acid? A student finds that mL of a M NaHCO3 solution is required to titrate a mL sample of sulfuric acid solution. What is the molarity of the acid? The reaction equation is: H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2

1. How many milliliters of 1. 25 M LiOH must be added to neutralize 34
1. How many milliliters of 1.25 M LiOH must be added to neutralize 34.7 mL of M HNO3? 2. What mass of Sr(OH)2 will be required to neutralize mL of M HBr solution? How many mL of M H2SO4 must be added to neutralize 47.9 mL of M KOH? How many milliliters of 1.25 M LiOH must be added to neutralize 34.7 mL of M HNO3? What mass of Sr(OH)2 will be required to neutralize mL of M HBr solution? How many milliliters of 0.75 M KOH must be added to neutralize 50.0 mL of 2.50 M HCl 10.8 mL g 29.6 mL 10.8 mL g 29.6 mL

Agenda Day 64 – Introduction to Solutions Lesson: PPT

Similar presentations

Presentation on theme: "Agenda Day 64 – Introduction to Solutions Lesson: PPT"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Agenda Day 64 – Introduction to Solutions Lesson: PPT

Similar presentations

Presentation on theme: "Agenda Day 64 – Introduction to Solutions Lesson: PPT"— Presentation transcript:

Similar presentations

About project

Feedback