5.1 Classifying Solutions A. The Basics

5.1 Classifying Solutions A. The Basics
Chapter 5: Solutions 5.1 Classifying Solutions A. The Basics classification of matter: All Matter Mixtures Pure Substances Heterogeneous (Mechanical Mixtures and Colloids) Homogeneous (Solutions) Elements Compounds

is any that has matter solid, liquid or gas mass and volume are a type of matter that has pure substances definite fixed composition eg) elements and compounds are combinations of matter that can be mixtures do NOT have definite proportions separated by physical means and

have and the different components are
heterogeneous mixtures (mechanical mixtures) varying composition usually visible are which have solutions homogeneous mixtures and the different components are uniform composition not visible composed of at least one substance dissolved in another

is the and is usually the substance present in the
solvent dissolver, largest quantity (mass, volume, amount) eg) water is what is solute dissolved in the solvent eg) salt

B. Types of Solutions both solvents and solutes can be
solids, liquids or gases you can have various combinations of solute and solvent phases eg) liquid in liquid – ethylene glycol (antifreeze) solid in liquid – Kool Aid gas in liquid – carbonated beverages solid in solid – alloys liquid in solid – mercury amalgam fillings

an solution is any solution in which is the
aqueous water solvent water is called the therefore we will be concerned mainly with “universal solvent” aqueous solutions water dissolves a lot of different solutes because of it’s …this is good because …also bad because dissolve in it easily unique properties 75% of the earth (and our bodies) is water toxins

C. Dissolving and the Forces of Attraction
is a dissolving physical change the of a solid solute are held together by molecules or ions bonds when dissolving occurs, these bonds and the of the solute become break ions or molecules attracted to the solvent particles H2O NaCl

the 3 processes involved in dissolving are:
1. bonds between molecules or ions of broken endothermic (requires energy) solute – always 2. bonds between molecules of broken solvent – always endothermic 3. bonds between molecules or ions of form solute and solvent – always exothermic

overall energy change in dissolving is equal to the of the three steps
sum (law of conservation of energy) if, the overall dissolving process is more energy is released than is required exothermic if, the overall dissolving process is less energy is released than is required endothermic

D. Electrolytes vs Non-Electrolytes
are aqueous solutions that electrolytes (weak and strong) conduct electricity eg) all soluble ionic compounds, very polar molecular compounds (like acids, ammonia) are aqueous solutions that non-electrolytes do not conduct electricity eg) molecular compounds in solution

occurs when when they are dissolved in an
ionic compounds dissociation break apart into their ions aqueous solution are used to show what happens to a substance when it is put into dissociation equations water you can have 4 situations: 1. insoluble ionic or molecular compounds they to any great extent do not dissolve use the for ionic compounds solubility table eg) AgCl(s)  AgCl(s) C25H52(s)  C25H52(s)

2. soluble ionic compounds dissolve to a great extent to form ions in solution are broken ionic bonds use solubility table includes which turn litmus paper bases blue the number of ions balance eg) NaCl(s)  Na+(aq) + Cl(aq) Ca(OH)2(s)  Ca2+(aq) + 2 OH(aq)

3. soluble molecular compounds dissolve to form molecules in solution are broken intermolecular forces (LD, DD, HB)  eg) C12H22O11(s) C12H22O11(aq)

4. acids they are but are molecular compounds very polar dissolve to form in solution ions (ionize) turns litmus paper red the number of ions balance eg) H3PO4(s)  3H+(aq) + PO43(aq)

Examples Write the equations to show what happens to each of the following in water: 1. potassium chloride KCl(s)  K+(aq) + Cl-(aq) 2. carbon dioxide  CO2(g) CO2(aq)

6. gasoline  C8H18(l) C8H18(l) 7. barium sulphate BaSO4(s)  BaSO4(s)

3. solid hydrogen nitrate
 HNO3(s) H+(aq) + NO3(aq) 4. aluminum sulphate Al2(SO4)3 (s)  2 Al3+ (aq) + 3 SO42- (aq) 5. sodium phosphate decahydrate Na3PO4  10H2O(s)  3 Na+ (aq) + PO43- (aq)

many chemical reactions (either in our bodies or in the lab) are if the reactants are not dissolved in very slow water dissolving in water allows the solute particles to with other solute particles separate, disperse and collide particle collisions are necessary for to occur reactions

5.2 Solubility A. The Definitions the of a solute is the
amount of a solute that dissolves in a given quantity of solvent solubility at a given temperature eg) has a solubility of NaCl(s) 36 g/100mL at 20C an is a solution that does have the maximum amount of solute dissolved in it unsaturated solution not

a is a solution that contains the
saturated solution maximum amount of a dissolved solute at a given temperature some will be present undissolved solute the solution may still be able to dissolve other solutes

a contains more dissolved solute than its solubility supersaturated solution at a given temperature sodium acetate video

B. Range of Solubility the degree to which a solute is soluble depends on the strength of attraction between: 1. the solute particles 2. the solute particles and the solvent particles solutes can be: 1. less than 0.1 g/100mL, insoluble – although there is still a tiny bit of dissolving 2. slightly soluble – between 0.1 g/100 mL and 1 g/100 mL 3. soluble – greater than 1 g/100 mL

note that these general rules for solubility do not apply to …the numbers for gases are much and still considered gases lower soluble eg) has a solubility of yet this is considered O2(g) 0.009 g/100 mL at 20C soluble

C. A Closer Look once a solute , it appears that the solute-solvent bonds dissolves don’t break when you study a saturated solution, the amount (mass or moles) of undissolved solute at the bottom remains unchanged over time, undissolved particles become dissolved and dissolved particles crystallize a saturated solution is said to be in a state of equilibrium

equilibrium occurs when a process and the reverse process take place at the
(dissolving) (crystallization) same rate crystallization dissolving

⇌ CuSO4(s)  Cu2+(aq) + SO42(aq) is eg)
Cu2+(aq) + SO42(aq)  CuSO4(s) What we do is use a double arrow to show equilibrium: CuSO4(s) Cu2+(aq) + SO42(aq) eg) dissolving crystallization ⇌

D. Temperature and Solubility
the for a substance must be provided with a certain and it depends on the of the solute solubility value temperature state Solids when a dissolves in a , the holding the solid together must be solid liquid bonds broken at , the particles of the solute and solvent have more higher temperatures energy as temperature the solubility of a solid increases, increases

Liquids the particles of a liquid are held together as as the particles in a solid not strongly when a dissolves in a , additional is needed liquid liquid energy not the solubility of most liquids is not affected by temperature

Gases gas particles move and have a great deal of quickly energy when a dissolves in a , the particles must gas liquid lose energy the temperature would the gas particles and make it for them to dissolve increasing speed up harder as temperature the solubility of a gas increases, decreases

E. Pressure and Solubility
a change in has a negligible effect pressure on the solubility of solids or liquids the solubility of are greatly affected by gases pressure as pressure the solubility of a gas increases, increases eg) pop bottles, cracking knuckles, the “bends”

A. Solutions in our Society
5.3 Concentration A. Solutions in our Society solutions are all around us and affect our lives in many ways we have developed many to meet the and needs of humans technologies personal industrial eg) hair products, breathalyzer test, tests to monitor drinking water

in using solutions in our lives, care must be taken to ensure responsible use
eg) – set up in Canada in 1985 to ensure that companies that deal with chemicals are using them in a safe and appropriate manner Responsible Care Program

using solutions can have intended and unintended consequences for
humans and the environment released into the environment are taken in by organisms toxins this can have toxic effects immediate eg) H2S(g)…sour gas will cause a loss of consciousness at 700 ppm

sometimes the of the food chain are
sometimes the of the food chain are by the toxins since the levels are not that high…but the toxins are then passed along to the of the food chain lower levels unaffected upper levels organisms at the of the food chains (like humans!) can end up with of these chemicals top toxic levels eg) mercury, lead, arsenic, PCB’s, DDT etc.

is the concentration of toxins as you
biomagnification (bioaccumulation) move up the food chain

we must assess the and of using technologies that contain certain substances
risks benefits eg) heavy metals released from mineral processing, power plants – mercury, arsenic

B. Ways of Expressing Concentration
concentration is the amount of solute relative to the amount of solvent dilute = solute, solvent low high concentrated = high solute, solvent low concentration can be expressed in a variety of ways: ppm, ppb, ppt, % mass, % volume, mol/L, mmol/L, mg/dL, %

these expressions can be seen on many products that we use in our daily lives and in industry
eg) vinegar – toothpaste – blood cholesterol levels – breathalyzer - % (g of alcohol per 100 mL of blood) eg. 5% acetic acid by volume 0.243% w/w NaF 250 mg/dL 0.08

1. Concentration as Percent by Mass
% mass (m/m) = mass solute (g) __ × 100 mass solution (solute + solvent) (g) Example An aqueous solution with a mass of 55.5 g contains 2.85 g of solute dissolved in it. What is the percent by mass of the solute in the solution? 2.85 g percent mass (m/m) = × 100 55.5 g = 5.14 %

2. Concentration in Parts per Million
ppm = parts per million ppb = parts per billion mass solute (g) __ ppm = × 106 ppm mass solution (g)

Example Carbon monoxide is deadly at 900 ppm. If there is 8.50 g of carbon monoxide in a room that contains 11.5 kg of air, what is the concentration of CO(g) in ppm? Is the level of carbon monoxide deadly? ppm = mass solute (g) × 106 ppm mass solution (g) = g × 106 ppm g = 739 ppm no, it’s not deadly

3. Molar Concentration molarity =
number of moles of solute per litre of solvent c = n v n = m M where: c = concentration in mol/L n = number of moles in mol v = volume in L m = mass in g M = molar mass in g/mol

Example 1 A sample of mol of NaCl is dissolved to give L of solution. What is the molar concentration of the solution? n = mol v = L c = n v = mol 0.500 L = 1.80 mol/L

Example 2 Calculate the concentration of 100 mL of a solution containing mol of sulphuric acid. n = mol v = L c = n v = mol 0.100 L = 3.00 mol/L

Example 3 Calculate the molar concentration of a 250 mL solution that has 3.2 g of NaCl dissolved in it. m = 3.2 g v = L M = g/mol n = m M = 3.2 g 58.44 g/mol = …mol c = n v = … mol 0.250 L = 0.22 mol/L

Example 4 Calculate the number of moles of Pb(NO3)2 needed to make 500 mL of a 1.25 mol/L solution. c = 1.25 mol/L v= L n = cv = (1.25 mol/L)(0.500 L) = mol

Example 5 How many litres of 4.22 mol/L solution would contain 3.69 mol of BaCl2? c = 4.22 mol/L n = 3.69 mol v = n c = 3.69 mol 4.22 mol/L = L

Example 6 Calculate the mass of the salt required to prepare 1.50 L of a mol/L solution of K3PO4. n = cv = (0.565 mol/L)(1.50 L) = mol c = mol/L v = 1.50 L M = g/mol m = nM = ( mol)( g/mol) = 180 g

D. Molar Concentration of Ions
you may have to calculate the concentration of in solution ions once you have your dissociation equation, you can calculate the concentration of ions in solution using the mole ratio wanted given are anions negative ions are cations positive ions

Calculate the ion concentrations in a 0.050 mol/L solution of KCl.
Example 1 Calculate the ion concentrations in a mol/L solution of KCl. g w w 1 KCl(s)  1 K+(aq) + 1 Cl-(aq) C = mol/L C = mol/L  1 1 C = mol/L  1 1 = mol/L = mol/L

Example 2 Calculate the ion concentrations in a mol/L solution of Al2(SO4)3. g w w 1 Al2(SO4)3(s)  Al3+(aq) 2 + 3 SO42-(aq) C = mol/L C = mol/L  2 1 C = mol/L  3 1 = 0.10 mol/L = 0.15 mol/L

Example 3 Calculate the concentrations of dissolved Na3PO4(s)  10H2O that gives a 0.30 mol/L concentration of Na+(aq) ions. w g 1 Na3PO4(s)  10H2O  3 Na+(aq) + 1 PO43-(aq) C = 0.30 mol/L  1 3 C = 0.30 mol/L = 0.10 mol/L

Example 4 Calculate the concentration of sodium ions in a NaCl(aq) solution made by dissolving 6.33 g of NaCl(s) in 150 mL of water. g w 1 NaCl(s)  1 Cl-(aq) + 1 Na+(aq) m = 6.33 g M = g/mol v = L n = 0.108…mol  1 1 n = m M = 6.33 g 58.44 g/mol = 0.108…mol c = n v = 0.108…mol 0.150 L = mol/L

5.4 Preparing and Diluting Solutions A. Preparing a Standard Solution
a solution of is called a known concentration standard solution there are two ways to make a solution: 1. dissolve a measured amount of in a certain volume of pure solute solvent 2. a standard solution dilute

to prepare a solution of known concentration from a :
solid solute Steps 1. Calculate the of the required to achieve a specific concentration and volume. mass solute n = cv m = nM

2. ______ g (mass) of _______ (solute). Measure

3. the solute in _______ mL of water (half of the volume). Dissolve

4. Transfer solution to a ______ mL volumetric flask.

5. Fill flask to _______ mL (final volume) and mix by inverting.

Example Describe how to prepare 100 mL of a mol/L solution of KMnO4(aq). n = cv = ( mol/L)(0.100L) = mol v = L c = mol/L M = g/mol m = nM = ( mol)( g/mol) = 1.26 g 1. Measure out of 1.26 g KMnO4(s) 2. Dissolve the KMnO4(s) in of distilled H2O(l). 50 mL 3. Transfer the solution to a volumetric flask. 100 mL 4. Fill to and invert to mix. 100 mL

B. Dilution not all solutions are available in the concentrations we need you can make a out of a solution of known concentration by less concentrated solution diluting it = decreasing the of a solution by adding more dilution concentration solvent (water)

the stays constant!!! number of moles ni = nf and n = CV vici = vfcf where: vi = initial volume in L vf = final volume in L ci = initial concentration in mol/L cf = final concentration in mol/L

Note - vi is always smaller than vf - ci is always bigger than cf

Example 1 What volume of 1.0 mol/L NaCl solution do you need to make 250 mL of a 0.20 mol/L NaCl solution? Vf = 250 mL = L Ci = 1.0 mol/L Cf = 0.20 mol/L ViCi = VfCf Vi(1.0 mol/L) = (0.250L)(0.20 mol/L) Vi = L

Example 2 What is the concentration of a 1.50 L solution if it is made by mixing 500 mL of 14.8 mol/L H2SO4(aq) with 1.00 L of water? Vi = L Vf = 1.50 L Ci = 14.8 mol/L ViCi = VfCf (0.500L)(14.8 mol/L) = (1.50L) Cf Cf = 4.93 mol/L

6.1 Theories of Acids and Bases A. Naming Acids and Bases
Chapter 6: Acids & Bases 6.1 Theories of Acids and Bases A. Naming Acids and Bases acids always have as the state and always have (aq) hydrogen Rules 1. hydrogen becomes acid 2. hydrogen becomes acid 3. hydrogen becomes acid ____ide hydr____ic _____ate _____ic ____ite ____ous

Examples: 1. hydrogen iodide = hydroiodic acid phosphoric acid
Change each of the following to the appropriate acid name and give the formula: 1. hydrogen iodide = hydroiodic acid HI(aq) phosphoric acid H3PO4(aq) 2. hydrogen phosphate = 3. hydrogen nitrite = nitrous acid HNO2(aq) 4. hydrogen sulphite = sulphurous acid H2SO3(aq)

sodium hydrogen carbonate
most bases are ionic compounds that are named accordingly Examples: Name each of the following bases: 1. NaOH(aq) = sodium hydroxide 2. NaHCO3(aq) = sodium hydrogen carbonate 3. Mg(OH)2(aq) = magnesium hydroxide 4. NH3(aq) = ammonia

aqueous hydrogen iodide
IUPAC names for acids and bases are simply the word “aqueous” followed by the ionic name Examples: Write the IUPAC name for each of the following acids and bases: 1. hydroiodic acid = aqueous hydrogen iodide 2. magnesium hydroxide = aqueous magnesium hydroxide 3. sulphurous acid = aqueous hydrogen sulphite 4. sodium hydrogen carbonate = aqueous sodium hydrogen carbonate

B. Properties of Acids and Bases
are of a substance empirical properties observable properties acids, bases and neutral substances have some properties that distinguish them and some that are the same

Acids Bases Neutral Substances sour taste bitter taste electrolytes
electrolytes, non-electrolytes neutralize bases neutralize acids react with indicators do not react with indicators affect indicators the same way litmus - red litmus - blue bromothymol blue - yellow bromothymol blue - blue phenolphthalein - phenolphthalein - pink colourless react with to produce metals H2(g) pH greater than 7 less than 7 pH pH of 7 eg) HCl(aq), H2SO4(aq) eg) Ba(OH)2(aq) NH3(aq) eg) NaCl(aq), Pb(NO3)2(aq)

C. Arrhenius Definition
first proposed theory on acids and bases Svante Arrhenius his theory was that some compounds form electrically charged particles when in solution his explanation of the properties of acids and bases is called the Arrhenius theory of acids and bases

an Arrhenius is a substance that (because it is molecular) to form
acid ionizes hydrogen ions, H+(aq), in water an will in an aqueous solution acid increase the [H+(aq)] an Arrhenius is a substance that to form in water base dissociates hydroxide ions, OH(aq), a will in an aqueous solution base increase the [OH-(aq)]

D. Modified Arrhenius Definition
the original definition of acids and bases proposed by Arrhenius is good but it has limitations some substances that might be predicted to be are actually neutral basic eg) Na2CO3(aq), NH3(aq) it has been found that not all bases contain the hydroxide ion as part of their chemical formula

an Arrhenius is a substance that in aqueous solution base (modified)
reacts with water to produce OH(aq) ions eg)  NH3(aq) + H2O() NH4+(aq) + OH(aq) (Na2O, MgO etc) form in water metallic oxides bases eg) Step 1: Na2O(s) + H2O()  2 NaOH(aq) (make the ) metal hydroxide Step 2: 2 NaOH(aq)  2 Na+(aq) + 2 OH(aq) ( the metal hydroxide) dissociate

when acids ionize, they produce
H+(aq) eg) HCl(g)  H+(aq) + Cl(aq) it has been found using analytical technology like X-ray crystallography that in an aqueous solution H+(aq) ions do not exist in isolation the hydrogen ion is extremely positive in charge and water molecules themselves are very polar so… it is that would exist in water without being attracted to the of other highly unlikely hydrogen ions negative poles water molecules

this results in the formation of the
hydronium ion + H3O+(aq)

an Arrhenius is a substance that in aqueous solution acid (modified)
reacts with water to produce H3O+(aq) ions eg)  HCl(aq) + H2O() Cl(aq) + H3O+(aq)  H2SO3(aq) + H2O() HSO3(aq) + H3O+(aq) form in water non-metallic oxides (CO2, SO2 etc) acids eg) Step 1: CO2(g) + H2O()  H2CO3(aq) (combine elements to make an ) all acid Step 2: H2CO3(aq) + H2O()  H3O+(aq) + HCO3(aq) ( acid with water) react

6.2 Strong and Weak Acids and Bases
the of a substance depend on two things: acidic and basic properties 1. the of the solution concentration 2. the of the acid or base identity

A. Strong Acids and Weak Acids
an acid that is called a ionizes almost 100% in water strong acid eg) HCl(aq) + H2O()  H3O+(aq) + Cl(aq) 100% of the becomes H3O+(aq) and Cl(aq) HCl(aq) the concentration of the is the as the concentration of the it came from H3O+(aq) same acid strong acids are strong electrolytes and react vigorously with metals

there are 6 strong acids:
perchloric acid HClO4(aq) hydrobromic acid HBr(aq) hydroiodic acid HI(aq) hydrochloric acid HCl(aq) sulfuric acid H2SO4(aq) nitric acid HNO3(aq) ***on your periodic table

⇌ a and only a small percentage of the acid forms
weak acid does not ionize 100% ions in solution eg) CH3COOH(aq) + H2O() ⇌ H3O+(aq) + CH3COO(aq) we use the for weak acids equilibrium arrow weak acids are react much less vigorously with metals weak electrolytes and

B. Strong Bases and Weak Bases
a base that dissociates into ions in water is called a 100% strong base are strong bases ionic hydroxides and metallic oxides eg) NaOH(aq)  Na+(aq) + OH(aq) a and only a small percentage of the base forms weak base does not dissociate 100% ions in solution + eg) NH3(aq) + H2O() ⇌ NH4+(aq) OH(aq) we use the for weak bases equilibrium arrow

C. Monoprotic and Polyprotic Acids
acids that have only per molecule that can are called one hydrogen atom ionize monoprotic acids eg) HCl(aq), HF(aq), HNO3(aq), CH3COOH(aq) eg) HNO3(aq) + H2O()  H3O+(aq) + NO3(aq) monoprotic acids can be strong or weak

acids that contain that can are called
two or more hydrogen atoms ionize polyprotic acids eg) H2SO4(aq), H3PO4(aq) acids with are , with are two hydrogens diprotic three hydrogens triprotic

when polyprotic acids ionize, only hydrogen is removed at a time, with each acid becoming
one progressively weaker eg)  H2SO4(aq) + H2O() H3O+(aq) + HSO4(aq) HSO4(aq) + H2O()  H3O+(aq) + SO42(aq)

D. Monoprotic and Polyprotic Bases
bases that are called react with water in only one step to form hydroxide ions monoprotic bases eg) NaOH(s) bases that react with water in are called two or more steps polyprotic bases eg) CO32(aq), PO43(aq)

as with polyprotic acids, only
OH(aq) is formed at a time, and each new base formed is than the last one weaker eg)  CO32(aq) + H2O() OH(aq) + HCO3(aq) HCO3(aq) + H2O()  OH(aq) + H2CO3(aq)

E. Neutralization the reaction between an acid and a base produces an
ionic compound and water acid + base a salt + water → eg) HCl(aq) + KOH(aq) → KCl(aq) + HOH() the products of are both neutralization neutral in a neutralization reaction or between a , the product is always acid-base reaction strong acid and a strong base water H3O+(aq) + OH(aq)  2 H2O()

F. Acid and Base Spills there are many uses for both acids and bases in our households and in industry due to their, special care must be used when they are being reactivity and corrosiveness produced and transported

the two ways to deal with acid or base spills are:
1. dilution: reduce the by adding concentration water 2. neutralization: you always use a for the neutralization so you aren’t left with another hazardous situation weak acid or base

A. Ion Concentration in Water
6.3 Acids, Bases and pH A. Ion Concentration in Water the “self-ionization” of water is very small (only 2 in 1 billion) H2O() + H2O()  H3O+(aq) + OH-(aq) the concentration of and are hydronium ions equal and constant in pure water hydroxide ions [H3O+(aq)]= [OH-(aq)] = 1.0 x 10-7 mol/L 1.0 x 10-7 mol/L

B. The pH Scale in 1909, Soren Sorenson devised the pH scale
it is used because the [H3O+(aq)] is very small at 25C (standard conditions), most solutions have a pH that falls between 0.0 and 14.0 it is possible to have a pH and a pH negative above 14 it is a based on whole numbers that are powers of 10 logarithmic scale

more acidic more basic neutral
there is a for every change in on the pH scale 10-fold change in [H3O+(aq)] 1 eg) a solution with a pH of 11 is times more basic than a solution with a pH of 9 10  10 = 100 pH Scale more acidic more basic 7 14 neutral

C. Calculating pH and pOH
pH =  log [H3O+(aq)] ***New sig dig rule: when reporting pH or pOH values, only the numbers to the count as significant right of the decimal place Try These: 1. [H3O+(aq)] = 1 x mol/L pH = 2. [H3O+(aq)] = 1.0 x 10-2 mol/L pH = 3. [H3O+(aq)] = 6.88 x 10-3 mol/L pH = 4. [H3O+(aq)] = 9.6 x 10-6 mol/L pH = 10.0 2.00 2.162 5.02

 H3O+(aq) + NO3-(aq) HNO3(aq) + pH = -log[H3O+(aq)]
Example 6.30 g of HNO3 is dissolved in 750 mL of water. What is the pH ?  H3O+(aq) + NO3-(aq) HNO3(aq) + H2O() m = 6.30 g M = g/mol V = L c = 0.133…mol/L x 1/1 = 0.133…mol/L n = m M = 6.30 g 63.02 g/mol = …mol pH = -log[H3O+(aq)] = -log[0.133… mol/L] = 0.875 c = n V = …mol 0.750 L = 0.133…mol/L

just as deals with deals with
pH [H3O+(aq)], pOH [OH(aq)] ***p just means log at SATP… pH + pOH = 14 pH 1 3 5 7 9 11 13 14 14 13 11 9 7 5 3 1 pOH

to calculate the use the same formulas as pH but substitute the
pOH, [OH(aq)] pOH =  log[OH(aq)] Try These: 1. [OH(aq)] = 1.0  mol/L pOH = 2. [OH(aq)] =  10-2 mol/L pOH = 3. [OH(aq)] =  10-6 mol/L pOH = 4. [OH(aq)] = 2  10-6 mol/L pOH = 11.00 1.206 5.0264 5.7

you could also be given the pH or pOH and asked to calculate the
[H3O+(aq)] or [OH-(aq)] [H3O+(aq)] = 10-pH [OH(aq)]= 10-pOH

Try These: 1. pH [H3O+(aq)] = 2. pH [H3O+(aq)] = 3. pH [H3O+(aq)] = 4. pH 7 [H3O+(aq)] = 5. pOH 1.0 [OH(aq)] = 6. pOH 13.2 [OH(aq)] = 7. pOH 6.9 [OH(aq)] = 8. pOH [OH(aq)] = 1 x 10-4 mol/L 6.2 x mol/L 3.98 x mol/L 10-7 mol/L 0.1 mol/L 6  mol/L 1  mol/L mol/L

9. Complete the following table:
[H3O+(aq)] [OH(aq)] pH pOH Acid/Base/ Neutral 4.0 x 10-6 mol/L 2.5 x 10-9 mol/L 5.40 8.60 acid 9.500 4.500 base 3.16 x mol/L 3.16 x 10-5 mol/L 3.30 10.70 5.0 x 10-4 mol/L 2.0  1011 mol/L acid 10 mol/L 1.0 x mol/L -1.00 15.00 acid base 1.0 x mol/L 10 mol/L 15.00 -1.00 1.36 base 2.3 x mol/L 0.044 mol/L 12.64

D. Measuring pH Indicators pH can be measured using :
1. acid-base indicators 2. pH meter Indicators an is any chemical that in an acidic or basic solution acid-base indicator changes colour they can be dried onto strips of paper eg) litmus paper, pH paper

they can be solutions eg) bromothymol blue, universal indicator, indigo carmine etc they can be made from natural substances eg) tea, red cabbage juice, grape juice

each indicator has a where it will
specific pH range change colour you can use to approximate the two or more indicators pH of a solution

pH Meters using a pH meter is the most way of measuring precise pH
it has an that compares the [H3O+(aq)] in the solution to a and it will give a of the pH electrode standard digital readout

E. Diluting an Acid or Base
when you to an , you change the add water acid or base [H3O+(aq)] or the [OH(aq)] diluting an acid will the until a pH of is reached decrease [H3O+(aq)] 7.0 diluting a base will the until a pH of is reached decrease [OH-(aq)] 7.0

5.1 Classifying Solutions A. The Basics

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