Buffers/Titration Aqueous Equilibria - I Slide 2 / 113 Review - PDF document

Slide 1 / 113 Buffers/Titration Aqueous Equilibria - I Slide 2 / 113 Review hydrolysis of salts Slide 3 / 113 1 The hydrolysis of a salt of a weak base and a strong acid should give a solution that is __________. weakly basic A neutral B C strongly basic weakly acidic D

Slide 4 / 113 2 A solution of one of the following is acidic. The compound is _________. NaCl A B CH 3 COONa C NH 4 Cl D NH 3 Slide 5 / 113 3 Which salt would undergo hydrolyzes to form an acidic solution? KCl A B NaCl C NH 4 Cl LiCl D Slide 6 / 113 4 Which substance when dissolved in water will produce a solution with a pH greater than 7? A CH 3 COOH NaCl B NaC 2 H 3 O 2 C HCl D

Slide 7 / 113 5 A basic solution would result from the hydrolysis of one of the ions in this compound. The compound is__________. NaNO 3 A B NH 4 Cl C CH 3 COONa D CaCl 2 Slide 8 / 113 6 A water solution of which compound will turn blue litmus red? K 2 CO 3 A B NH 4 Cl NaOH C NaCl D Slide 9 / 113 The Common-Ion Effect · Consider an aqueous solution of acetic acid: CH 3 COOH(aq) + H 2 O(l) ↔ H 3 O + (aq) + CH 3 COO - (aq) · If acetate ion is added to the solution, Le Châtelier 's Principle predicts that the equilibrium will shift to the left.

Slide 10 / 113 The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” In the following Sample Problem, compare the pH 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 The Common-Ion Effect SAMPLE PROBLEM #1 a) Calculate the pH of a 0.20 M HFsolution. The K a for HF is 6.8 ´ 10 −4 . HF(aq) + H 2 O(l) ↔ H 3 O + (aq) + F - (aq) H 3 O + (aq) F − (aq) HF(aq) H 2 O(l) Initial 0.20 M 0 0 Change -x +x +x At Equilibrium 0.20-x x x Slide 12 / 113 The Common-Ion Effect K a = [H 3 O + ][F - ] = 6.8 x 10 -4 [HF] x 2 6.8 x 10 -4 = (0.20) x 2 = (0.20) (6.8 ´ 10 −4 ) x = 0.012 [H + ] = [F - ] = 0.012 So, and pH = 1.93

Slide 13 / 113 The Common-Ion Effect SAMPLE PROBLEM #1 (con't) b) Calculate the pH of a for the same 0.20 M HF solution that also contains 0.10 M NaF. K a for HF is 6.8 ´ 10 −4 . K a = [H 3 O + ][F - ] = 6.8 x 10 -4 [HF] Slide 14 / 113 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. · We use the same equation and the same K a expression, and set up a similar ICE chart. · But because NaF is a soluble salt, it completely dissociates, so F - is not initially zero. HF(aq) + H 2 O(l) ↔ H 3 O + (aq) + F - (aq) H 3 O + (aq) F − (aq) HF(aq) H 2 O(l) 0.20 M 0 0.10 M Initial Change at Equilibrium Slide 15 / 113 The Common-Ion Effect H 3 O + (aq) F − (aq) HF(aq) H 2 O(l) Initial 0.20 M 0 0.10 M Change -x +x +x At Equilibrium 0.20-x x 0.10 + x

Slide 16 / 113 The Common-Ion Effect Therefore, x = [H 3 O + ] = 1.4 x 10 −3 and pH = −log (1.4 x 10 −3 ) pH = 2.87 Slide 17 / 113 The Common-Ion Effect 6.8 x 10 -4 = (0.10)(x) K a = (0.20) (0.20)(6.8 x 10 -4 ) = x (0.10) 1.4 ´ 10 −3 = x Remember what the "x" is that you are solving for! Slide 18 / 113 The Common-Ion Effect HF(aq) + H 2 O(l) ↔ H 3 O + (aq) + F - (aq) · Consider the two solutions we just examined in Sample Problem #1. · Compare the following: Final [H 3 O + ] Solution pH HF 0.12 1.4 x 10 -3 HF and NaF · How do these results support Le Chatelier's Principle?

Slide 19 / 113 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 . SAMPLE PROBLEM #2 a) Calculate the pH of a 0.30 M acetic acid, CH 3 COOH. b) Calculate the pH of a solution containing 0.30 M acetic acid and 0.10 M sodium acetate. Slide 20 / 113 The Common-Ion Effect The K a for acetic acid, CH 3 COOH, is 1.8 x 10 -5 . SAMPLE PROBLEM #2 - Answers a) Calculate the pH of a 0.30 M acetic acid solution. K a = [H 3 O + ][C 2 H 3 O 2 - ] = 1.8 x 10 -5 [HC 2 H 3 O 2 ] x 2 = 1.8 x 10 -5 (0.30) x = [H 3 O + ] = ______________ and pH = ___________ Slide 21 / 113 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 . SAMPLE PROBLEM #2 - Answers (con't) b) Calculate the pH of a solution containing 0.30 M acetic acid 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 + ]. K a = [H 3 O + ][C 2 H 3 O 2 - ] = 1.8 x 10 -5 [HC 2 H 3 O 2 ] [H 3 O + ] (0.10) = 1.8 x 10 -5 (0.30) [H 3 O + ] = ______________ So now, and pH = ___________

Slide 22 / 113 The Common-Ion Effect - (aq) + H 3 O + (aq) HC 2 H 3 O 2 (aq) + H 2 O (l) ↔ C 2 H 3 O 2 · Consider the two solutions we just examined in Sample Problem #2. · Compare the following: [OH - ] Solution pH HC 2 H 3 O 2 HC 2 H 3 O 2 and NaC 2 H 3 O 2 · How do these results support Le Chatelier's Principle? Slide 23 / 113 The Common-Ion Effect SAMPLE PROBLEM #3 Calculate the pH of the following solutions: a) 0.85 M nitrous acid, HNO 2 b) 0.85 M HNO 2 and 0.10 M potassium nitrite, KNO 2 The K a for nitrous acid is 4.5 x 10 -4 . Answers: a) b) Slide 24 / 113 7 The ionization of HF will be decreased by the addition of: NaCl A NaF B HCl C Both A and B D E Both B and C

Slide 25 / 113 8 The dissociation of Al(OH) 3 will be decreased by the addition of: A KOH AlCl 3 B Mg(OH) 2 C D Both A and B A, B and C E Slide 26 / 113 9 Which of the following substances will not decrease the ionization of H 3 PO 4 ? K 3 PO 4 A B HCl Na 3 PO 4 C D None of them will decrease the ionization All of them will decrease the ionization E Slide 27 / 113 The Common-Ion Effect The Common-ion effect can also be observed with weak bases. Consider the ionization of ammonia, NH 3 , which is a weak base. + H 2 O <---> NH 4 + + OH - NH 3

Slide 28 / 113 The Common-Ion Effect Suppose a salt of the conjugate base is added to a solution of ammonia. What effect would this have on pH? + 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 The Common-Ion Effect SAMPLE PROBLEM #4 Calculate the pH of the following solutions: 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 The K b for NH 3 is 1.8 x 10 -5 . Slide 30 / 113 The Common-Ion Effect SAMPLE PROBLEM #4 - Answers 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 . + H 2 O <---> NH 4 + + OH - NH 3 + ][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 The Common-Ion Effect SAMPLE PROBLEM #4 - Answers (con't) b) Calculate the pH of 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 . + H 2 O <---> NH 4 + + OH - NH 3 + ][OH - ] [NH 4 K b = 1.8 x 10 -5 = [NH 3 ] (0.15) (x) 1.8 x 10 -5 = (0.45) x = [OH - ] = _____________ So, and pH = _________________ Slide 32 / 113 The Common-Ion Effect SAMPLE PROBLEM #5 Calculate the pH of the following solutions: 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 - . The K b for pyridine is 1.7 x 10 -9 . Answers: a) b) Slide 33 / 113 Buffers Buffers, or buffered solutions, are special mixtures that are resistant to large pH changes, even when small amounts of strong acid or strong base are added.

Slide 34 / 113 Buffers · Buffers are able to resist large pH changes because they contain both an acidic component and a basic component. · Buffers are prepared by mixing either: 1) a weak acid and a salt containing its conjugate base OR 2) a weak base and a salt containing its conjugate acid Slide 35 / 113 Buffers Consider a buffer solution composed of HF and NaF. · The acidic component is HF. This component aids in the neutralization of any strong base that is added to the buffer. · The basic component is the fluoride ion, F - . This component aids in the neutralization of any strong acid that is added to the buffer. Slide 36 / 113 Buffers Consider a buffer solution composed of HF and NaF. Buffer with equal Buffer after conc. of weak acid addition of OH - and its conj. base - - - HF F HF F HF F OH - + H H + + F - --> HF F - + H 2 O <-- HF + OH - F - HF If a small amount of KOH is added to this buffer, for example, the HF reacts with the OH − to make __________.

Slide 37 / 113 Buffers Buffer with equal Buffer after conc. of weak acid addition of H + Buffer after addition of OH - and its conj. base HF F - F - HF F - HF - OH H + - + H 2 O <-- HF + OH - + + F - --> HF F H Similarly, if a small amount of strong acid (H + ) is added, the F − reacts with it to form ____________. Slide 38 / 113 10 If a buffer is made of HA (weak acid) and NaA, which species will counteract the addition of a strong base? A HA Na + B A - C Both HA and Na + D Both HA and A - E Slide 39 / 113 11 If a buffer is made of HNO 2 and KNO 2 , which species will counteract the addition of a strong acid? A HNO 2 K + B - NO 2 C Both HNO 2 and K + D Both K + and NO 2 - E