Acid-Base Equilibria The reaction of weak acids with water, OR the reaction of weak bases with water, always results in an equilibrium!! The equilibrium constant for the reaction of a weak acid with water is K a 1 Mar. 25

Acid-Base Equilibria eg. HF (aq) + H 2 O (l) º K a = [H 3 O+] [F - ] [HF] H 3 O + (aq) + F - (aq) K eq = ? 2 Mar. 25

Acid-Base Equilibria For any weak acid Why is H 2 O (l) omitted from the K a expression? K a = [H 3 O+] [conjugate base] [weak acid] 3 Mar. 25

Acid-Base Equilibria the equilibrium constant for the reaction of a weak base with water is K b HS - (aq) + H 2 O (l) º K b = H 2 S (aq) + OH - (aq) 4 Mar. 25

Acid-Base Equilibria For any weak base K b = [OH - ] [conjugate acid] [weak base] 5 Mar. 25

eg. Write the expression for K b for S 2- (aq) ANSWER: S 2- (aq) + H 2 O (l) º K b = [OH - ] [HS - ] [S 2- ] HS - (aq) + OH - (aq) 6 Mar. 25

5.a) Use K a to find [H 3 O+] for mol/L HF (aq) HF (aq) + H 2 O (l) º H 3 O + (aq) + F - (aq) K a = 6.6 x Mar. 25

x 2 = (0.100)(6.6 x ) x 2 = 6.6 x x = 8.1 x mol/L 1 st try - Ignore x Mar. 25

2 nd try– Include x x 2 = (0.0919)(6.6 x ) x 2 = x x = 7.8 x mol/L Different than 1 st try: CANNOT IGNORE DISSOCIATION Mar. 28

3 rd try– Include new x x 2 = (0.0922)(6.6 x ) x 2 = x x = 7.8 x mol/L [H 3 O + ] = 7.8 x mol/L Same as 2 nd try: Mar. 28

5.b) find [H 3 O+] for mol/L CH 3 COOH (aq CH 3 COOH (aq) + H 2 O (l) º H 3 O + (aq) + CH 3 COO - (aq) K a = 1.8 x Mar. 28

x 2 = (0.250)(1.8 x ) x 2 = 4.5 x x = 2.1 x mol/L 1 st try - Ignore x Mar. 28

2 nd try– Include x x 2 = (0.2479)(1.8 x ) x 2 = x x = 2.1 x mol/L [H 3 O + ] = 2.1 x mol/L Same as 1 st try: Mar. 28

14 To ignore OR not to ignore: that is the question Mar. 28

pH of a weak acid Step #1: Write a balanced equation Step #2: ICE table OR assign variables Step #3: Write the K a expression Step #4: Check (can we ignore dissociation) Step #5: Substitute into Ka expression 15 Mar. 28

pH of a weak acid eg. Find pH of mol/L HF(aq). Step #1: Write a balanced equation HF (aq) + H 2 O (l) º H 3 O + (aq) + F - (aq) 16 Mar. 30

Step #2: Equilibrium Concentrations Let x = [H 3 O + ] at equilibrium [F - ] = x [HF] = x 17 Mar. 30

Step #3: Write the K a expression K a = [H 3 O+] [F - ] [HF] 18 Mar. 30

Step #4: Check (can we ignore dissociation) dissociation (- x) may be IGNORED = 151 (0.100) 6.6 x Acid dissociation CANNOT be IGNORED in this question. [weak acid] KaKa If > 500 We have to use the – x 19 Mar. 30

Step #5: Substitute into Ka expression x 2 = 6.6 x x x x x x x = 0 a = 1b = 6.6 x c = -6.6 x Quadratic Formula!! 20 Mar. 30

Ignore negative roots 21 Mar. 30

a) Find the [H 3 O + ] in mol/L HCN (aq) Check: 4.0 x 10 8 x = 1.24 x [H 3 O + ] = 1.24 x b) Calculate the pH of mol/L HCOOH (aq) Check: 167 x = 2.24 x pH = Try these: Mar. 31

HCN + H 2 O ⇋ H 3 O + + CN - Let x = [H 3 O + ] x = [CN - ] – x = [HCN] Check: 23 K a = [H 3 O + ] [CN - ] [HCN] = 4.0 x x Quadratic NOT needed Mar. 31

24 x = 1.25 x [H 3 O + ] = 1.25 x mol/L pH = x 2 = 1.55 x Mar. 31

HCOOH + H 2 O ⇋ H 3 O + + HCOO x +x +x – x x x Check: 25 K a = [H 3 O + ] [HCOO - ] [HCOOH] = x Quadratic needed Mar. 31

26 A = 1 B = 1.8 x C = -5.4 x x = 2.24 x [H 3 O + ] = 2.24 x mol/L pH = x 2 = 5.4 x x x x x x x = 0 Mar. 31

Practice 1. Formic acid, HCOOH, is present in the sting of certain ants. What is the [H 3 O + ] of a mol/L solution of formic acid? ( mol/L) 2. Calculate the pH of a sample of vinegar that contains 0.83 mol/L acetic acid. ( [H 3 O + ] = 3.87 x pH = ) 3. What is the percent dissociation of the vinegar in 2.? % diss = % Mar. 31

Practice 4. A solution of hydrofluoric acid has a molar concentration of mol/L. What is the pH of this solution? ( [H 3 O + ] = pH = ) 5. The word “butter” comes from the Greek butyros. Butanoic acid, C 3 H 7 COOH, gives rancid butter its distinctive odour. Calculate the [H 3 O + ] of a 1.0 × 10 −2 mol/L solution of butanoic acid. (Ka = 1.51 × 10 −5 ) (Ans: 3.89 x mol/L) Mar. 31

pH of a weak base  same method as acids  ignore dissociation if  to calculate K b (usually given on the exam) 29 [weak base] KbKb > 500 Apr. 4

pH of a weak base Calculate the pH of mol/L Na 2 CO 3(aq) 30 Apr. 4

CO H 2 O ⇋ HCO OH x +x +x – x x x Check: 31 K b = [OH - ] [HCO 3 - ] [CO 3 2- ] = x → Quadratic needed K b = 1.00 x x = 2.13 x Apr. 4

32 x 2 = 2.13 x x x x x x x = 0 A = 1 B = 2.13 x C = x x = 1.36 x [OH - ] = 1.36 x mol/L pOH = ??pH = Apr. 4

pH of a weak base Calculate the pH of mol/L NaNO 2(aq) 33 Apr. 4 Na + NO 2 - H2OH2O

NO H 2 O ⇋ HNO 2 + OH x +x +x – x x x Check: 34 K b = [OH - ] [HCO 3 - ] [CO 3 2- ] = 3.6 x x OK to ignore –x here ie. NO Quadratic K b = 1.00 x x = 1.39 x Apr. 4

35 x 2 = 6.95 x [OH - ] = 2.6 x mol/L pOH = ?? pH = 8.42 x = 2.6 x Apr. 4

Calculating K a from [weak acid] and pH eg.The pH of a mol/L solution of benzoic acid, C 6 H 5 COOH, is Calculate the numerical value of the Ka for this acid. - Equation - Find [H 3 O + ] from pH - Subtract from [weak acid] - Substitute to find K a See p. 591 #6 & 8 36 Apr. 5

C 6 H 5 COOH (aq) + H 2 O (l) º H 3 O + (aq) + C 6 H 5 COO - (aq) [H 3 O + ] = = mol/L [C 6 H 5 COOH] = – = mol/L Find K a K a = ( )( ) ( ) = 6.2 x [C 6 H 5 COO - ] = mol/L 37 Apr. 5

Calculating K a from [weak acid] and pH eg.The pH of a mol/L solution of benzoic acid, C 6 H 5 COOH, is Calculate the % dissociation for this acid. See p. 591 #’s 5 & 6 [H 3 O + ] = = mol/L = 2.9 % 38 Apr. 5

a)0.250 mol/L chlorous acid, HClO 2(aq) ; pH = % b) mol/L cyanic acid, HCNO (aq) ; pH = % c)0.100 mol/L arsenic acid, H 3 AsO 4(aq) ; pH = % d)0.500 mol/L iodic acid, HIO 3(aq) ; pH = % Calculate the acid dissociation constant, K a, and the percent dissociation for each acid: 39 Apr. 5

More Practice: Weak Acids: pp. 591, 592 #’s 6 -8 Weak Bases: p. 595 #’s (K b ’s on p. 592) 40 Apr. 5

Acid-Base Stoichiometry Solution Stoichiometry (Review) 1. Write a balanced equation 2. Calculate moles given ( OR n = CV) 3. Mole ratios 4. Calculate required quantity OR OR m = nM 41 Apr. 6

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eg.25.0 mL of mol/L H 2 SO 4(aq) was used to neutralize 36.5 mL of NaOH (aq). Calculate the molar concentration of the NaOH solution. H 2 SO 4(aq) + NaOH (aq) → H 2 O (l) + Na 2 SO 4(aq) 22 n H 2 SO 4 = n NaOH = C NaOH = 43 Apr. 6

Acid-Base Stoichiometry pp. 600, 601 – Sample Problems p. 602 #’s Apr. 6

Dilution Given 3 of the four variables Only one solution C i V i = C f V f Stoichiometry Given 3 of the four variables Two different solutions 4 step method 45

Excess Acid or Base To calculate the pH of a solution produced by mixing an acid with a base:  write the B-L equation (NIE)  calculate the moles of H 3 O + and OH -  subtract to determine the moles of excess H 3 O + or OH -  divide by total volume to get concentration  calculate pH 46

eg.20.0 mL of M Ca(OH) 2(aq) is mixed with 10.0 mL of M HCl (aq). Determine the pH of the resulting solution. ANSWER: Species present: Ca 2+ OH - H 3 O + Cl - H 2 O SB SA 47

C = mol/L V = L C = mol/L V = L NIE:OH - + H 3 O + → 2 H 2 O 4.00 x mol OH x mol H 3 O x mol excess OH - 48 n = CV

= mol/L [OH - ] = mol/L pOH = pH =

Indicators An indicator is a weak acid that changes color with changes in pH HIn is the general formula for an indicator To choose an indicator for a titration, the pH of the endpoint must be within the pH range over which the indicator changes color 50 Apr. 13

HIn (aq) + H 2 O (l) º H 3 O + (aq) + In - (aq) Colour #1Colour #2 HIn is the acid form of the indicator. Adding H 3 O + Adding OH - 51 causes colour 1 (LCP) removes the H 3 O + & causes colour #2 Apr. 13

methyl orange HMo (aq) + H 2 O (l) º H 3 O + (aq) + Mo - (aq) red yellow bromothymol blue HBb (aq) + H 2 O (l) º H 3 O + (aq) + Bb - (aq) yellow blue 52 Apr. 13

Indicators: p a) HMv + H 2 O (l) º H 3 O + (aq) + Mv - (aq) b) HBb + H 2 O (l) º H 3 O + (aq) + Bb - (aq) 2. IndicatorpHcolour thymol blue3.0 yellow methyl red7.9 yellow phenolpthalein7.1colourless indigocarmine13.5 yellow 3.a) pH range: 2.8 – 4.5 b) pH range: 8.0 – Apr. 13

Acid-Base Titration (p. 603 → ) A titration is a lab technique used to determine an unknown solution concentration. A standard solution is added to a known volume of solution until the endpoint of the titration is reached. The endpoint occurs when there is a sharp change in colour The equivalence point occurs when the moles of H 3 O + equals the moles of OH - 54 Apr. 13

Acid-Base Titration The colour change is caused by the indicator added to the titration flask. An indicator is a chemical that changes color over a given pH range (See indicator table) A buret is used to add the standard solution standard solution - solution of known concentration 55

Acid-Base Titration primary standard - a standard solution which can be made by direct weighing of a stable chemical. Data from titrations allows us to calculate an unknown solution concentration. 56

Titration Calculation eg. Clem Student performed a titration by adding mol/L HCl (aq) to 10.0 mL samples of Ca(OH) 2(aq). Use the data below to determine the molar concentration of Ca(OH) 2(aq). Trial Final volume 8.48 mL mL mL mL Initial volume 1.05 mL 8.48 mL mL mL Volume HCl (aq) used mL7.23 mL7.21 mL7.22 mL Omit first trial OVERSHOT the endpoint

Equation: 2 HCl (aq) + Ca(OH) 2(aq) → CaCl 2(aq) + 2 H 2 O (l) n HCl = n Ca(OH) 2 = C = 58 C = mol/L V ave = L C = ? mol/L V ave = L

Acid-Base Titration Titration Lab – pp. 606,

Multi-Step Titrations (p ) Polyprotic acids donate their protons one at a time when reacted with a base. eg. Write the equations for the steps that occur when H 3 PO 4(aq) is titrated with NaOH (aq) H 3 PO 4(aq) + OH - (aq) H 2 PO 4 - (aq) + OH - (aq) HPO 4 2- (aq) + OH - (aq) 60

Multi-Step Titrations H 3 PO 4(aq) + OH - (aq) → H 2 PO 4 - (aq) + H 2 O (l) H 2 PO 4 - (aq) + OH - (aq) → HPO 4 2- (aq) + H 2 O (l) HPO 4 2- (aq) + OH - (aq) º PO 4 3- (aq) + H 2 O (l) H 3 PO 4(aq) + 3 OH - (aq) º PO 4 3- (aq) + 3 H 2 O (l) 61

Multi-Step Titrations Write the balanced net ionic equations, and the overall equation, for the titration of Na 2 S (aq) with HCl (aq). p. 611 #’s 21.b), 22, & 23 LAST TOPIC!! Titration CurvesTitration Curves 62

Properties / Operational Definitions Acid-Base Theories and Limitations  Arrhenius – H-X and X-OH  Modified – react with water → hydronium  BLT – proton donor/acceptor (CA and CB) Writing Net Ionic Equations (BLT) Strong vs. Weak pH & pOH calculations Equilibria (K w, K a, K b ) Titrations/Indicators/Titration Curves Dilutions and Excess Reagent questions Acids and Bases 63

Step #2: ICE table [ HF] [H 3 O + ] [F - ] I C E mol/L0 0 - x + x + x x x x 64

CO 3 2- (aq) + H 2 O (l) º HCO 3 - (aq) + OH - (aq) mol/L CO 2 3- (aq) [PO 4 3- ] [HPO 4 2- ] [OH - ] I C E mol/L0 0 - x + x + x x x x 65

Acid-Base Stoichiometry Solution stoichiometry (4 question sheet) Excess reagent problems (use NIE) Titrations Titration curves Indicators STSE: Acids Around Us 66

A primary standard is a pure substance that is stable enough to be stored indefinitely without decomposition, can be weighed accurately without special precautions when exposed to air, and will undergo an accurate stoichiometric reaction in a titration. 67

15.pH of mol/L HOCl HOCl (aq) + H 2 O (l) º H 3 O + (aq) + OCl - (aq) Let x = [H 3 O + ] at equilibrium [OCl - ] = x [HOCl] = x K a = [H 3 O+] [OCl - ] [HOCl] 68

Check: dissociation (- x) may be IGNORED = 1.02 x 10 7 (0.297) 2.9 x X = 9.28 x pH =

0.484 mol/L L mol/L L 16. NIE: OH - + H 3 O + → 2 H 2 O mol OH mol H 3 O mol excess OH - [OH - ] = mol/LpOH = pH =

17. Ignore dissociation [OH - ] = mol/L % diss = 2.92 % 18. V ave = mL n NaOH = mol n H 2 SO 4 = mol C = mol/L 19. K b = 3.93 x % diss =

c) 2.50 mol/L NaCN (aq) K b = 1.61 x Check: 1.5 x 10 5 x = 6.34 x pOH = 2.20pH = d) mol/L K 2 S (aq) K b = Check: 1.3 x = pOH = 1.24pH =