Slide 1 / 113 Slide 2 / 113 Buffers/Titration Review hydrolysis of salts Aqueous Equilibria - I Slide 3 / 113 Slide 4 / 113 1 The hydrolysis of a salt of a weak base and a strong acid 2 A solution of one of the following is acidic. The should give a solution that is __________. compound is _________. weakly basic NaCl A A B neutral B CH 3 COONa strongly basic C NH 4 Cl C weakly acidic D NH 3 D Slide 5 / 113 Slide 6 / 113 3 Which salt would undergo hydrolyzes to form an acidic 4 Which substance when dissolved in water will produce solution? a solution with a pH greater than 7? A KCl A CH 3 COOH NaCl NaCl B B C NH 4 Cl C NaC 2 H 3 O 2 LiCl HCl D D
Slide 7 / 113 Slide 8 / 113 5 A basic solution would result from the hydrolysis of one 6 A water solution of which compound will turn blue of the ions in this compound. The compound litmus red? is__________. K 2 CO 3 A A NaNO 3 B NH 4 Cl NH 4 Cl B NaOH C CH 3 COONa C NaCl D CaCl 2 D Slide 9 / 113 Slide 10 / 113 The Common-Ion Effect The Common-Ion Effect · Consider an aqueous solution of acetic acid: “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong CH 3 COOH(aq) + H 2 O(l) ↔ H 3 O + (aq) + CH 3 COO - (aq) electrolyte that has an ion in common with the weak electrolyte.” · If acetate ion is added to the solution, Le Châtelier 's Principle predicts that the equilibrium In the following Sample Problem, compare the pH will shift to the left. of a 0.20 M solution of HF with the pH of the same solution after some NaF is added to it. Slide 11 / 113 Slide 12 / 113 The Common-Ion Effect The Common-Ion Effect K a = [H 3 O + ][F - ] SAMPLE PROBLEM #1 = 6.8 x 10 -4 [HF] a) Calculate the pH of a 0.20 M HFsolution. The K a for HF is 6.8 ´ 10 −4 . x 2 6.8 x 10 -4 = HF(aq) + H 2 O(l) ↔ H 3 O + (aq) + F - (aq) (0.20) x 2 = (0.20) (6.8 ´ 10 −4 ) HF(aq) H 2 O(l) H 3 O + (aq) F − (aq) x = 0.012 Initial 0.20 M 0 0 Change -x +x +x [H + ] = [F - ] = 0.012 So, At Equilibrium 0.20-x x x and pH = 1.93
Slide 13 / 113 Slide 14 / 113 The Common-Ion Effect The Common-Ion Effect b) Calculate the pH of a for the same 0.20 M HF solution that also contains 0.10 M NaF. SAMPLE PROBLEM #1 (con't) · We use the same equation and the same K a expression, b) Calculate the pH of a for the same 0.20 M HF and set up a similar ICE chart. solution that also contains 0.10 M NaF. · But because NaF is a soluble salt, it completely dissociates, so F - is not initially zero. K a for HF is 6.8 ´ 10 −4 . HF(aq) + H 2 O(l) ↔ H 3 O + (aq) + F - (aq) HF(aq) H 2 O(l) H 3 O + (aq) F − (aq) 0 0.10 M 0.20 M Initial K a = [H 3 O + ][F - ] = 6.8 x 10 -4 Change [HF] at Equilibrium Slide 15 / 113 Slide 16 / 113 The Common-Ion Effect The Common-Ion Effect Therefore, x = [H 3 O + ] = 1.4 x 10 −3 HF(aq) H 2 O(l) H 3 O + (aq) F − (aq) and Initial 0.20 M 0 0.10 M Change -x +x +x pH = −log (1.4 x 10 −3 ) At Equilibrium 0.20-x x 0.10 + x pH = 2.87 Slide 17 / 113 Slide 18 / 113 The Common-Ion Effect The Common-Ion Effect HF(aq) + H 2 O(l) ↔ H 3 O + (aq) + F - (aq) 6.8 x 10 -4 = (0.10)(x) · Consider the two solutions we just examined in K a = (0.20) Sample Problem #1. (0.20)(6.8 x 10 -4 ) · Compare the following: = x (0.10) Final [H 3 O + ] Solution pH HF 0.12 1.4 ´ 10 −3 = x 1.4 x 10 -3 HF and NaF · How do these results support Le Chatelier's Principle? Remember what the "x" is that you are solving for!
Slide 19 / 113 Slide 20 / 113 The Common-Ion Effect The Common-Ion Effect The K a for acetic acid, CH 3 COOH, is 1.8 x 10 -5 . The K a for acetic acid, CH 3 COOH or HC 2 H 3 O 2 , is 1.8 x 10 -5 . SAMPLE PROBLEM #2 - Answers SAMPLE PROBLEM #2 a) Calculate the pH of a 0.30 M acetic acid solution. a) Calculate the pH of a 0.30 M acetic acid, CH 3 COOH. K a = [H 3 O + ][C 2 H 3 O 2 - ] = 1.8 x 10 -5 [HC 2 H 3 O 2 ] b) Calculate the pH of a solution containing 0.30 M acetic acid and 0.10 M sodium acetate. x 2 = 1.8 x 10 -5 (0.30) x = [H 3 O + ] = ______________ and pH = ___________ Slide 21 / 113 Slide 22 / 113 The Common-Ion Effect The Common-Ion Effect The K a for acetic acid, CH 3 COOH or HC 2 H 3 O 2 , is 1.8 x 10 -5 . - (aq) + H 3 O + (aq) HC 2 H 3 O 2 (aq) + H 2 O (l) ↔ C 2 H 3 O 2 SAMPLE PROBLEM #2 - Answers (con't) · Consider the two solutions we just examined in b) Calculate the pH of a solution containing 0.30 M acetic acid Sample Problem #2. and 0.10 M sodium acetate. We use the same K a expression, except now the acetate ion concentration is not the same as [H 3 O + ]. · Compare the following: [OH - ] K a = [H 3 O + ][C 2 H 3 O 2 - ] Solution pH = 1.8 x 10 -5 [HC 2 H 3 O 2 ] HC 2 H 3 O 2 [H 3 O + ] (0.10) HC 2 H 3 O 2 and = 1.8 x 10 -5 NaC 2 H 3 O 2 (0.30) · How do these results support Le Chatelier's Principle? [H 3 O + ] = ______________ So now, and pH = ___________ Slide 23 / 113 Slide 24 / 113 The Common-Ion Effect 7 The ionization of HF will be decreased by the addition of: SAMPLE PROBLEM #3 A NaCl Calculate the pH of the following solutions: NaF B a) 0.85 M nitrous acid, HNO 2 HCl C b) 0.85 M HNO 2 and 0.10 M potassium nitrite, KNO 2 Both A and B D The K a for nitrous acid is 4.5 x 10 -4 . Both B and C E Answers: a) b)
Slide 25 / 113 Slide 26 / 113 8 The dissociation of Al(OH) 3 will be decreased by 9 Which of the following substances will not the addition of: decrease the ionization of H 3 PO 4 ? KOH K 3 PO 4 A A B AlCl 3 B HCl C Mg(OH) 2 Na 3 PO 4 C Both A and B D None of them will decrease the ionization D A, B and C All of them will decrease the ionization E E Slide 27 / 113 Slide 28 / 113 The Common-Ion Effect The Common-Ion Effect The Common-ion effect can also be observed with weak Suppose a salt of the conjugate base is added to a bases. solution of ammonia. What effect would this have on pH? Consider the ionization of ammonia, NH 3 , which is a + H 2 O <---> NH 4 + + OH - weak base. NH 3 + H 2 O <---> NH 4 + + OH - NH 3 As with the previous Sample Problems, let us calculate the pH of the following: a) a 0.45 M solution of NH 3 b) a 0.45 M solution of NH 3 that also contains 0.15 M NH 4 Cl, ammonium chloride -- a soluble salt that readily + ions. yields NH 4 Slide 29 / 113 Slide 30 / 113 The Common-Ion Effect The Common-Ion Effect SAMPLE PROBLEM #4 - Answers SAMPLE PROBLEM #4 Calculate the pH of the following solutions: a) Calculate the pH of a 0.45 M solution of NH 3. The K b for NH 3 is 1.8 x 10 -5 . a) a 0.45 M solution of NH 3 + H 2 O <---> NH 4 + + OH - NH 3 b) a 0.45 M solution of NH 3 that also contains 0.15 M NH 4 Cl The K b for NH 3 is 1.8 x 10 -5 . + ][OH - ] [NH 4 K b = 1.8 x 10 -5 = [NH 3 ] x 2 1.8 x 10 -5 = (0.45) x = [OH - ] = _____________ pH = _________________
Slide 31 / 113 Slide 32 / 113 The Common-Ion Effect The Common-Ion Effect SAMPLE PROBLEM #4 - Answers (con't) SAMPLE PROBLEM #5 b) Calculate the pH of a 0.45 M solution of NH 3 that also contains Calculate the pH of the following solutions: 0.15 M NH 4 Cl. The K b for NH 3 is 1.8 x 10 -5 . + H 2 O <---> NH 4 + + OH - NH 3 a) 0.0750 M pyridine, C 5 H 5 N b) 0.0750 M C 5 H 5 N and 0.0850 M pyridinium chloride, C 5 H 5 NHCl Note that C 5 H 5 NHCl,( salt) dissociates into C 5 H 5 NH + and Cl - . + ][OH - ] [NH 4 K b = 1.8 x 10 -5 = [NH 3 ] The K b for pyridine is 1.7 x 10 -9 . (0.15) (x) 1.8 x 10 -5 = (0.45) Answers: a) b) x = [OH - ] = _____________ So, and pH = _________________ Slide 33 / 113 Slide 34 / 113 Buffers Buffers · Buffers are able to resist large pH changes because they contain both an acidic component and a basic component. Buffers, or buffered solutions, are special · Buffers are prepared by mixing either: mixtures that are 1) a weak acid and a salt containing its conjugate base resistant to large pH OR changes, even when 2) a weak base and a salt containing its conjugate acid small amounts of strong acid or strong base are added. Slide 35 / 113 Slide 36 / 113 Buffers Buffers Consider a buffer solution composed of HF and NaF. Buffer with equal Buffer after Consider a buffer solution composed of HF and NaF. conc. of weak acid addition of OH - and its conj. base · The acidic component is HF. This component aids in the - - - HF F neutralization of any strong base that is added to the buffer. HF F HF F OH - H + · The basic component is the fluoride ion, F - . This F - + H 2 O <-- HF + OH - H + + F - --> HF component aids in the neutralization of any strong acid that is - F added to the buffer. HF If a small amount of KOH is added to this buffer, for example, the HF reacts with the OH − to make __________.
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