1. Acid-Base Equilibria and Solubility Equilibria Chapter 16 Dr. Ali Bumajdad

2. Chapter 16 Topics Homogeneous Versus Heterogeneous Solution Equilibria The Common Ion Effect Buffer Solutions Acid-Base Titrations and Indicators Solubility Equilibria Relation between solubility and K sp The Common Ion Effect and Solubility pH and Solubility Acid –Base Equilibria Dr. Ali Bumajdad

3. H 2 O (l) H + (aq) + OH - (aq) AgNO 3 (aq) +NaCl(aq) AgCl(s) + NaNO 3 (aq) Homogeneous Equilibria Heterogeneous Equilibria

4. Common ion effect is the shift in equilibrium to the left (reactant) caused by the addition of a compound (strong electrolyte) having an ion in common with the dissolved substance (weak electrolyte). This is because of the Le Chatelier Principle. The presence of a common ion decreases the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion The Common Ion Effect pH will increase

5. Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] Henderson-Hasselbalch equation pH = pK a + log [conjugate base] [acid] (2) pK a = -log K a where [H + ] = K a [HA] [A - ] (1)

6. Q) What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = 3.77 + log [0.52] [0.30] = 4.01 Mixture of weak acid and conjugate base! (Ka of HCOOH = 1.7  10 -4 ) - Na is spectator ion and do not affect pH

7. Another solution: - Find [H + ] then find pH [H + ] = K a [HCOOH] [HCOO - ] (1) pH=-log [H + ]

8. Buffer solution is a solution that resists change in pH because it contain both acids to neutralize OH - and base to neutralize H + ions Buffer composition: 1.A weak acid + its salt (that is salt of weak conjugate base) 2.A weak base + its salt (that is salt of weak conjugate acid) Add strong acid H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa Buffer solution

9. HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl -

10. Q) Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) HF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is its conjugate acid buffer solution

12. The greater the amount of the conjugate acid-base pair the greater the buffer capacity (that is the grater the resistance of pH change). Some buffer solution have the same pH but their capacity are different Buffer capacity: Q) Which of the following buffer solutions has greater pH? And which has greater buffer capacity? 1)1M HC 2 H 3 O 2 + 1 M NaC 2 H 3 O 2 2) 0.1M HC 2 H 3 O 2 + 0.1M NaC 2 H 3 O 2 pH = pK a + log [conjugate base] [acid] (2)

13. = 9.20 Q) Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? (K a NH 4 + =5.6  10 -10; K b NH 3 = 1.8  10 -5 ) NH 4 + (aq) H + (aq) + NH 3 (aq) pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (moles) end (moles) 0.0290.001 0.024 0.0280.00.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL [NH 3 ] = 0.025 0.10 pH = pK a + log [NH 3 ] [NH 4 + ] (2) (better to use mole then convert to M)

14. Sa. Ex. How many mole of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form Buffer solution with pH =9.00? (assume adding NH 4 Cl do not affect the total volume (K b = 1.8  10 -5 ) pH = pK a + log [conjugate base] [acid] (2) or pOH = 14 - pH [OH - ]= 10 -pOH K b = [NH 4 + ][OH - ] / [NH 3 ] pK a +pK b = pK w =14 For base or acid and its salt

15. pH = pK a + log [conjugate base] [acid] (2) The added strong acid (HCl) will be completely consumed by the reaction with the buffer and that’s way it will not affect the pH much (a) (b) (K a =1.8  10 -5 ) Start molarity: Change : End molarity: CH 3 COO - + H + CH 3 COOH 1.0 0.1 1.0 -0.1 -0.1 +0.1 0.9 0 1.1 Use equation 2

16. Sa. Ex. Buffer made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L solution. The pH of the buffer is 4.74. (a)Calculate the pH of this solution after 0.20 mol of NaOH is added (neglect any volume change) (b) calculate the pH when 0.020 mol of NaOH is added to 1.00L of water (a)OH - will react with the acid and the no. of mole of the acid and the conjugate base will change HC 2 H 3 O 2 + OH - H 2 O + C 2 H 3 O 2 - Before reaction Change After reaction HC 2 H 3 O 2 H + + C 2 H 3 O 2 - pH = pK a + log [conjugate base] [acid] (2)

17. Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point : the point at which the reaction is complete Indicator : substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink)

18. Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l)

19. Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7) because of the hydrolysis:

20. Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7) because of the hydrolysis: H + (aq) + NH 3 (aq) NH 4 Cl (aq)

21. Q) Exactly 100 mL of 0.10 M HNO 2 are titrated with 100 ml of 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 0.05 – x  0.05x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

22. Acid-Base Indicators HIn (aq) H + (aq) + In - (aq)  10 [HIn] [In - ] Color of acid (HIn) predominates  10 [HIn] [In - ] Color of conjugate base (In - ) predominates Indiators are weak organic acid or base End point: the point when the indicator change color

23. pH

24. The titration curve of a strong acid with a strong base. Methyl red and ph.ph are good indicators but Thymol blue is not

25. Q) Which indicator(s) would you use for a titration of HNO 2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein

26. Solubility Equilibria AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 2 - (aq) K sp = [Ag + ] 2 [CO 3 2 - ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 - (aq) K sp = [Ca 2+ ] 3 [PO 4 3 - ] 2 Heterogeneous equilibrium K sp is the product of the concentrations of the ions involve in the equilibrium all rise to a power = the coefficient

28. Molar solubility, s, (mol/L) is the number of moles of solute dissolved in 1 L to form a saturated solution. Solubility (g/L) is the number of grams of solute dissolved in 1 L to form a saturated solution. Relation between Molar solubility and K sp Molar solubility = the highest concentration of solution Molar solubility change upon adding another solutes Molar solubility change upon changing temperature AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag+] [Cl - ] But [Ag+] = solubility of AgCl and [Cl-] = solubility of AgCl Molar solubility  M.m. = solubility (3)  K sp = s s = s 2 (4)  s = (K sp ) 1/2 (5)

30. / M.m.  M.m.

31. Q) What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M [Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = 1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10 -3 g/L K sp = 1.6 x 10 -10

33. K sp of Cu(OH) 2 = 2.2  10 -20

34. Dissolution of an ionic solid in aqueous solution: Q = K sp Saturated solution Q < K sp Unsaturated solution No precipitate Q > K sp Supersaturated solution Precipitate will form Ion concentration not at equilibrium Ion concentration at equilibrium

35. Q) If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? Ksp of Ca(OH) 2 = 8.0 x 10 -6 The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2 ? [Ca 2+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form

36. K sp of BaSO 4 = 1.1  10 -10

37. Q) What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br - and Cl - at a concentration of 0.02 M? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] K sp = 1.6 x 10 -10 AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 K sp = [Ag + ][Br - ] [Ag + ] = K sp [Br - ] 7.7 x 10 -13 0.020 = = 3.9 x 10 -11 M [Ag + ] = K sp [Cl - ] 1.6 x 10 -10 0.020 = = 8.0 x 10 -9 M 3.9 x 10 -11 M < [Ag + ] < 8.0 x 10 -9 M

38. The Common Ion Effect and Solubility The presence of a common ion decreases the solubility of the salt. This due to LeChatelier’s principle Q) What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr? AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 s 2 = K sp s = 8.8 x 10 -7 NaBr (s) Na + (aq) + Br - (aq) [Br - ] = 0.0010 M AgBr (s) Ag + (aq) + Br - (aq) [Ag + ] = s [Br - ] = 0.0010 + s  0.0010 K sp = 0.0010 x s s = 7.7 x 10 -10

39. Solubility not molar solubility

40. pH and Solubility The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 1.2 x 10 -11 K sp = (s)(2s) 2 = 4s 3 4s 3 = 1.2 x 10 -11 s = 1.4 x 10 -4 M [OH - ] = 2s = 2.8 x 10 -4 M pOH = 3.55 pH = 10.45 At pH less than 10.45 Lower [OH - ] OH - (aq) + H + (aq) H 2 O (l) remove Increase solubility of Mg(OH) 2 At pH greater than 10.45 Raise [OH - ] add Decrease solubility of Mg(OH) 2

41. S -2 is a conjugate base of a weak acid Weak base C l- is a conjugate base of strong acid Extremely weak base SO 4 2- is a conjugate base of a weak acid Weak base