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Problem Set 4 Solutions

# Problem Set 4 Solutions - C hemistr y 2101 I ntroductor y...

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Chemistry 2101 Introductory Analytical Chemistry Fall 2009 Solutions Problem Set 4 Chapter 8 Problem 8-6 K sp = [La 3+ ] [OH - ] 3 = 2 10 -21 with [La 3+ ] = 0.010 => [OH - ] = 5.8 10 -7 M => pH = 7.8 Problem 8-7 Strong acids: HCl, HBr, HI, H 2 SO 4 (but not HSO 4 - ), HNO 3 , HClO 4 Strong bases: LiOH, NaOH, KOH, CsOH, R 4 NOH Problem 8-8 Weak acids: RCOOH, R 3 NHX; weak bases: R 3 N, RCOOM (where M is a metal ion) Problem 8-9 (a) [H + ] = 0.01 M => pH = - log [H + ] = 2.00 (b) [OH - ] = 0.035 M => [H + ] = K w / [OH - ] = 2.8 6 10 -13 M => pH = 12.54 (c) [H + ] = 0.030 M => pH = - log [H + ] = 1.52 (d) [H + ] = 3.0 M => pH = - log [H + ] = -0.48 (e) [OH - ] = 0.010 M => [H + ] = K w / [OH - ] = 1.0 10 -12 M => pH = 12.00 Problem 8-10 (a) Cl 3 COOH  Cl 3 COO - + H + Cu 2+ H 2 O  Cu(OH) + + H + C 6 H 5 NH 3 +  C 6 H 5 NH 2 + H + (b) K a for trichloroacetic acid (0.3) is greater than K a for Cu 2+ (3 x 10E-8) and for anilinium ion (2.51 x 10E-5), so trichloroacetic acid is the strongest acid. Problem 8-12 HOSO 2 OH + HOSO 2 OH  HOSO(OH) 2 + HOSO 2 O - or 2 H 2 SO 4  H 3 SO 4 + + HSO 4 - Problem 8-17 HA  A - + H + HA tot = [HA] + [A - ] with [H + ] = [A - ] gives HA tot = [HA] + [H + ] K a = [A - ] [H + ] / [HA] => K a = [H + ] [H + ] / (HA tot - [H + ]) where K a = 1.00 10 -4

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Solving the quadratic equation gives pH = - log [H + ] = 3.02 Fraction of dissociation = (10E3.02)/(0.0100) = 9.51 x 10E-2
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Problem Set 4 Solutions - C hemistr y 2101 I ntroductor y...

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