CHAPTER 13 SOLUTIONS. SOLUTION REVIEW Solutions are homogenous mixtures.  They consist of a larger component called the solvent and one or more smaller.

CHAPTER 13 SOLUTIONS

SOLUTION REVIEW Solutions are homogenous mixtures.  They consist of a larger component called the solvent and one or more smaller components called the solutes.  Can be in the solid, liquid, or gaseous state.

Solution Examples Margarine Tap Water Steel 18 Carat Gold Air Sterling Silver

What is the solvent in air ? Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon,nitrogenoxygenargon 0.038% carbon dioxide, and small amounts of other gases.carbon dioxide Air also contains a variable amount of water vapor,water vapor on average around 1% Solution Examples

What is the solvent in air ? Nitrogen, N 2 Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon,nitrogenoxygenargon 0.038% carbon dioxide, and small amounts of other gases.carbon dioxide Air also contains a variable amount of water vapor,water vapor on average around 1% Solution Examples

What is the solvent in air ? Nitrogen, N 2 What is a solute in air? Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon,nitrogenoxygenargon 0.038% carbon dioxide, and small amounts of other gases.carbon dioxide Air also contains a variable amount of water vapor,water vapor on average around 1% Solution Examples

What is the solvent in air ? Nitrogen, N 2 What is a solute in air? Oxygen, O 2 Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon,nitrogenoxygenargon 0.038% carbon dioxide, and small amounts of other gases.carbon dioxide Air also contains a variable amount of water vapor,water vapor on average around 1% Solution Examples

What is the solvent in 18 ct gold ? Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. Solution Examples

What is the solvent in 18 ct gold ? Gold Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. Solution Examples

What is the solvent in 18 ct gold ? Gold What are the solutes? Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. Solution Examples

What is the solvent in 18 ct gold ? Gold What are the solutes? Silver and Copper Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. Solution Examples

Salt water: what is the solvent in salt water ? Examples of solutions include: Solution Examples

Salt water: what is the solvent in salt water ? Water, H 2 O Examples of solutions include: Solution Examples

Salt water: what is the solvent in salt water ? Water, H 2 O What is a solute in sea water? Examples of solutions include: Solution Examples

Salt water: what is the solvent in salt water ? Water, H 2 O What is a solute in sea water? NaCl (salt) Examples of solutions include: Solution Examples

Some general properties of solutions include:  Solutions may be formed between solids, liquids or gases.  They are homogenous in composition  They do not settle under gravity  They do not scatter light (Called the Tyndall Effect) Solute particles are too small to scatter light and therefore light will go right through a solution like is shown on the next slide. Solution Properties

Tyndall Effect Laser light reflected by a colloid. In a solution you would not see any red light.

Soluble substances are those that can dissolve in a given solvent. Insoluble or immiscible substances are those that cannot dissolve in a given solvent. Which of the following are soluble in water? NaCl, sugar, cooking oil, alcohol, gasoline, motor oil Solubility

Soluble substances are those that can dissolve in a given solvent. Insoluble or immiscible substances are those that cannot dissolve in a given solvent. Which of the following are soluble in water? NaCl, sugar, cooking oil, alcohol, gasoline, motor oil Which of the following are immiscible in cooking oil? NaCl, sugar, alcohol, gasoline, motor oil, water Solubility

The maximum amount of a given solute a solvent can dissolve is called the solubility. The solubility is dependent on the temperature and pressure. Solubility is often expressed in terms of grams of solute per 100 g of solvent but may have other units. When a solvent contains the minimum amount of a solute possible the solutions is said to be unsaturated. Solubility When a solvent contains the maximum amount of a solute possible the solutions is said to be saturated. When a solvent contains more than the maximum amount of a solute possible the solutions is said to be supersaturated.

Solutions form when a soluble solute(s) is dissolved in a solvent. In biological systems aqueous (solutions where water is the solvent) are of particular importance. The solubility of most liquids and solids in water increases with temperature. The effect of pressure on the solubility of liquid or solid solutes in water is negligible. Solubility

Solubility Curves of Various Solutes

By forming a solution at a high temperature then slowly cooling it we can form supersaturated solutions that contain more solute than in a saturated solution. These kinds of solutions are very unstable and tend to separate out the excess solute with the slightest disturbance. http://www.youtube.com/watch?v=uy6eKm8IRdI&NR=1 http://www.youtube.com/watch?v=aC-KOYQsIvU&feature=related Supersaturated Solutions

The solubility of gases in water decreases with temperature.  Are cold carbonated drinks bubblier than warm carbonated drinks? The solubility of many gases in water is directly proportional to the pressure being applied to the solution. i.e. double the pressure, double the solubility  What happens when the cork is removed from a bottle of champagne?  What is the origin of decompression sickness?  Anyone heard of hyperbaric therapy? Solubility

SOLUTION FORMATION When we place an ionic solid in water there will be attractive forces between the ions at the surface of the crystal and the water molecules. These attractive forces are called ion-dipole forces. Water molecules orient such that the positive end of the molecule is oriented towards the negative ions at the surface and vice versa.

How do solutions form? Why do some substances leave one phase and enter the solution and others don’t? How can we use chemistry to predict solubility's? Lets first look at the formation of a solution between an ionic solute and a polar solvent such as H 2 O. Solution Formation

Ionic compounds are composed of oppositely charged ions arranged in a repeating 3-d arrangement. They are held together by attractive forces between oppositely charged ions. Solution Formation

Ionic compounds are composed of oppositely charged ions arranged in a repeating 3-d arrangement. They are held together by attractive forces between oppositely charged ions Why is chloride ion larger than sodium ion? Solution Formation

SOLUTION FORMATION red is the region where electrons are found most often and blue is where electrons are rarely found

SOLUTION FORMATION If the attractive force between the surface ion and the solvent is greater than the forces between the ion and the solid then the ion will enter the solution phase. K+K+ H2OH2O The ion that has left the solid and becomes completed surrounded by water molecules. It has become solvated or hydrated.

SOLUTION FORMATION Note the different orientation of water molecules around the oppositely charged ions. Positive pole of water directed to the negative ions and the negative pole directed to the positive ions

SOLUTION FORMATION In a solution of an ionic compound a solvated ion will occasionally collide with the surface of the solid. Sometimes when this happens the ion will “stick” to the surface and become part of the solid phase again. This will happen more frequently the more concentrated the solution is.

SOLUTION FORMATION When the rate of ions leaving the solid equals the rate of ions going back to the solid the system is at equilibrium and the solution is saturated. When a solution is at equilibrium with its solute macroscopically there will be no change occurring. However, at the molecular level lots is happening, just in equal and opposite directions.

SOLUTION FORMATION Supersaturated solutions can form because there are no sites for solute ions to collide with. When we place a “seed” crystal in a supersaturated solution this provides the needed sites and the excess solute crystallizes very quickly.

SOLUTION FORMATION In the you tube video we watched you can just see the tiny seed crystals on the persons finger.

SOLUTION FORMATION Polar but non-ionic solutes dissolve in water via a similar mechanism as for ionic compounds.

SOLUTION FORMATION A solute will be insoluble in a solvent if: 1.Forces between solute particles are greater than the forces between solute particles and the solvent.

SOLUTION FORMATION A solute will be insoluble in a solvent if: 2.Forces between the solvent particles are stronger than forces between the solvent and the solute. e.g. The only attractive force between oil and water will is dispersion forces. These are weak compared to hydrogen bonds between water molecules.

SOLUTION FORMATION In a polar solvent there will be attraction between the oppositely charged ends of the molecule. Hydrogen bonds are represented by dotted lines between the water molecules. A hydrogen bond is and intermolecular force between hydrogen of one molecule and O, N, or F of another molecule. Hydrogen must be directly attached to O, N, or F in at least one of the two hydrogen bonded molecules.

SOLUTION FORMATION A good “rule of thumb” that works especially well for non- ionic compounds is: “Like dissolves like” i.e. Polar solvents dissolve polar solutes well and non-polar solvents dissolve non-polar solutes well.

SOLUTION RATE The rate of dissolution is dependent upon: 1.The surface area of the solute. i.e. how finely divided it is. Increasing rate

SOLUTION RATE 2.How hot the solution is. i.e. the kinetic energy of solute and solvent. 3.The rate of stirring. Typically when we are preparing a solution in the lab we will both heat and stir.

SOLUTION RATE When a solute dissolves in a solvent heat can be released or absorbed. When heat is absorbed the process is endothermic and the solution becomes cooler. This effect is used in instant cold packs for sporting injuries and first aid.

HEAT OF SOLUTION Solvent temperature 22.2 ° Solvent temperature 11.3 ° Endothermic Solution

HEAT OF SOLUTION More commonly dissolution is an exothermic process and heat is released when a solute is dissolved. Sometimes when we make a solution it will get so hot it boils!! Exothermic Solution

SOLUTION CONCENTRATION The ratio of the amount of solute to amount of solution, or solvent is defined by the concentration. solute = solvent solution Concentration solute = There are various combinations of units that are used in these rations. g solute mL solute g solution mL solution = = = % (w/w) % (w/v) % (v/v) ppt (w/w) ppt (w/v) ppt (v/v) ppm (w/w) ppb (w/w) ppm (w/v)ppb (w/v) ppm (v/v) ppb (v/v) RatioX 10 2 X 10 3 X 10 6 X 10 9

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 1.Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H 2 O + 25.2 g NaCl

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H 2 O + 25.2 g NaCl 100

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H 2 O + 25.2 g NaCl 100 = 43.0 % NaCl

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H 2 O + 25.2 g NaCl 100 = 43.0 % NaCl 44.6 g NaCl 100 g solution

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H 2 O + 25.2 g NaCl 100 = 43.0 % NaCl 44.6 g NaCl 100 g solution 333 g solution

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H 2 O + 25.2 g NaCl 100 = 43.0 % NaCl 44.6 g NaCl 100 g solution 333 g solution = 149 g NaCl

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H 2 O + 25.2 g NaCl 100 = 43.0 % NaCl 44.6 g NaCl 100 g solution 333 g solution = 149 g NaCl Mass of water?

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H 2 O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H 2 O + 25.2 g NaCl 100 = 43.0 % NaCl 44.6 g NaCl 100 g solution 333 g solution = 149 g NaCl Mass of water?333 g solution – 149 g NaCl = 184 g H 2 O

SAMPLE SOLUTION PROBLEMS 3.How many grams of NaCl are required to dissolve in 88.2 g of water to make a 29.2% (w/w) solution. 4.A sugar solution is 35.2%(w/v) find the mass of sugar contained in a 432 mL sample of this sugar solution.

SOLUTION CONCENTRATION The solution concentration can also be defined using moles. The most common example is molarity (M). The molarity of a solution is defined as: “The number of moles of solute in 1 L of solution” and is given the formula: Moles solute Molarity (M) = Liters solution

MOLARITY SAMPLE PROBLEMS 1.A student dissolves 25.8 g of NaCl in a 250 mL volumetric flask. Calculate the molarity of this solution. (picture of volumetric flask is on the next slide) 2.Find the mass of HCl required to form 2.00 L of a 0.500 M solution of HCl. 3.A student evaporates the water form a 333 mL sample of a 0.136 M solution of NaCl. What mass of salt remains? 4.Find the molarity of sodium ions in a 0.841 N solution of Na 2 SO 4.

In the lab we would use a piece of glassware called a volumetric flask to prepare this solution. SOLUTION PREPARATION

VOLUMETRIC FLASK

SOLUTION DILUTION Often we will want to make a dilute solution from a more concentrated one. To determine how to do this we use the formula : C 1 V 1 = C 2 V 2 Where: C 1 = concentration of more concentrated solution V 1 = volume required of more concentrated solution C 2 = concentration of more dilute solution V 2 = volume of more dilute solution We can use any units in this equation but they must be the same on both sides.

DILUTION PROBLEM How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution? ( 7.10 M)V 1 = (3.00 M) (50.0 mL) (7.10 M) C 1 V 1 = C 2 V 2 (7.10 M)V 1 = (3.00 M) (50.0 mL) V 1 = 21.1 mL This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask.

Solution Stoichiometry Consider the following balanced equation: CaCl 2 (aq) + 2 AgNO 3 (aq) → 2 AgCl (s) + Ca(NO 3 ) 2 (aq) 1.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride.

Solution Stoichiometry Consider the following balanced equation: CaCl 2 (aq) + 2 AgNO 3 (aq) → 2 AgCl (s) + Ca(NO 3 ) 2 (aq) 1.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO 3 L solution

Solution Stoichiometry Consider the following balanced equation: CaCl 2 (aq) + 2 AgNO 3 (aq) →2 AgCl (s) + Ca(NO 3 ) 2 (aq) 1.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO 3 L solution 10 3 mL L solution

Solution Stoichiometry Consider the following balanced equation: CaCl 2 (aq) + 2 AgNO 3 (aq) → 2 AgCl (s) + Ca(NO 3 ) 2 (aq) 1.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO 3 L solution 10 3 mL L solution 2moles AgNO 3 2moles AgCl moles AgCl 143.35 g AgCl33.2 mL

Solution Stoichiometry Consider the following balanced equation: CaCl 2 (aq) + 2 AgNO 3 (aq) → 2 AgCl (s) + Ca(NO 3 ) 2 (aq) 1.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO 3 L solution 10 3 mL L solution 2 moles AgNO 3 2 moles AgCl moles AgCl 143.35 g AgCl33.2 mL

Solution Stoichiometry Consider the following balanced equation: CaCl 2 (aq) + 2 AgNO 3 (aq) → 2 AgCl (s) + Ca(NO 3 ) 2 (aq) 1.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO 3 L solution 10 3 mL L solution 2 moles AgNO 3 2 moles AgCl moles AgCl 143.35 g AgCl 33.2 mL

Solution Stoichiometry Consider the following balanced equation: CaCl 2 (aq) + 2 AgNO 3 (aq) → 2 AgCl (s) + Ca(NO 3 ) 2 (aq) 1.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO 3 L solution 10 3 mL L solution 2 moles AgNO 3 2 moles AgCl moles AgCl 143.35 g AgCl 33.2 mL = 0.476 g AgCl

Solution Stoichiometry Consider the following balanced equation: CaCl 2 (aq) + 2 AgNO 3 (aq) →2 AgCl (s) + Ca(NO 3 ) 2 (aq) 1.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 2.Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and 200.0 mL of a 0.200 M solution of calcium chloride solution. 3.Find the volume of the excess reactant.

Titration Titration is an experimental procedure to determine the concentration of an unknown acid or base. The figure on the left shows the glassware for a titration experiment. A buret clamp holds the buret to a ring stand and below the buret is a flask containing the solution to be titrated, which includes an indicator. The purpose of the indicator is to indicate the point of neutralization by a color change.

The picture on the left shows the tip of a buret, with air bubble, which is not good, and also shows the stop-cock. Note the position of the stop-cock is in the “off” position. This picture shows the color of the phenolphthalein indicator at the end-point. In this experiment a 23.00 mL aliquot of 0.1000 M NaOH titrant is added to 5.00 mL of an unknown HCL solution. The acid solution in the beaker starts out clear and becomes pink when all of the HCL has been consumed. NaOH + HCl  NaCl + HOH

How can we calculate the concentration of acid in the beaker? Titration

How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! Titration

How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH L NaOH solution

How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH L NaOH solution 10 -3 L solution mL solution

How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH L NaOH solution 10 -3 L solution mL solution 23.00 mL soln

How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH L NaOH solution 10 -3 L solution mL solution 23.00 mL soln mole NaOH mole HCl

How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH L NaOH solution 10 -3 L solution mL solution 23.00 mL soln mole NaOH mole HCl 10 -3 L HCl soln. mL HCl soln.

How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH L NaOH solution 10 -3 L solution mL solution 23.00 mL soln mole NaOH mole HCl 10 -3 L HCl soln. mL HCl soln. 5.00 mL

How can we calculate the concentration of acid in the beaker? Normal procedure, yes, a conversion. Steps 1-4, again! 0.100 mole NaOH L NaOH solution 10 -3 L solution mL solution 23.00 mL soln mole NaOH mole HCl 10 -3 L HCl soln. mL HCl soln. 5.00 mL = 0.460 M HCl

Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H + + A - colorless pink 1.Describe the color change when a strong acid is added?

Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H + + A - colorlesspink 1.Describe the color change when a strong acid is added? Less pink

Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H + + A - colorlesspink 1.Describe the color change when a strong acid is added? 2.Describe the color change when a strong base is added? Less pink

Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H + + A - colorless pink 1.Describe the color change when a strong acid is added? 2.Describe the color change when a strong base is added? Less pink Darker pink

Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H + + A - colorlesspink 1.Describe the color change when a strong acid is added? 2.Describe the color change when a strong base is added? 3.Describe the color change when the pH is lowered? Less pink Darker pink

Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H + + A - colorless pink 1.Describe the color change when a strong acid is added? 2.Describe the color change when a strong base is added? 3.Describe the color change when the pH is lowered? Less pink Darker pink Less pink

Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H + + A - colorlesspink 1.Describe the color change when a strong acid is added? 2.Describe the color change when a strong base is added? 3.Describe the color change when the pH is lowered? 4.Describe the color change when the pH is raised? Less pink Darker pink Less pink

Indicators Indicators are weak organic (carbon containing) acids of various colors depending on the formula of the acid. Below is a generic acid. HA  H + + A - colorless pink 1.Describe the color change when a strong acid is added? 2.Describe the color change when a strong base is added? 3.Describe the color change when the pH is lowered? 4.Describe the color change when the pH is raised? Less pink Darker pink Less pink Darker pink

Color versus pH of Many Different indicators

How can we make an indicator?

Step One Red Cabbage Step Two Cook the Cabbage Step Three Filter the Juice

What color is the juice after filtering?

What color is the juice after filtering? The color of pH 6, 7, or 8 Colors of cabbage juice at various pH values

The End Ch#13 Solutions

CHAPTER 13 SOLUTIONS. SOLUTION REVIEW Solutions are homogenous mixtures.  They consist of a larger component called the solvent and one or more smaller.

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CHAPTER 13 SOLUTIONS. SOLUTION REVIEW Solutions are homogenous mixtures.  They consist of a larger component called the solvent and one or more smaller.

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Presentation on theme: "CHAPTER 13 SOLUTIONS. SOLUTION REVIEW Solutions are homogenous mixtures.  They consist of a larger component called the solvent and one or more smaller."— Presentation transcript:

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