Chapter 19 ionic equilibria in aqueous system

Chapter 19 ionic equilibria in aqueous system - Chapter...

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Chapter 19 Ionic Equilibria in Aqueous Systems 19.1 To maintain a relatively constant pH in a solution. 19.2 The weak acid in the system reacts with added base and the weak base in the system reacts with added acid. The acid and base need to be of comparable acid/base strength so that they don't neutralize one another. 19.3 The presence of an ion in common between two solutes will cause any equilibrium involving either of them to shift in accordance with LeChatelier's principle. For example, addition of NaF to a solution of HF will cause the equilibrium HF( aq ) + H 2 O( l ) h H 3 O + ( aq ) + F - ( aq ) to shift to the left, away from the excess of F - . 19.4 When H 3 O + is added to any system, the pH will drop. However, the change in pH will be minimized (and perhaps not measurable) with a buffer solution. 19.5 A high buffer capacity results when the weak acid and weak base are both present at high concentration. Addition of 0.01 mole of HCl to a high-capacity buffer will cause a smaller change in pH than with a low-capacity buffer, since the ratio [HA]/[A - ] will change less. 19.6 Only (c) and (e) affect the buffer capacity. In theory, any conjugate pair (of any p K a ) can be used to make a high capacity buffer. With proper choice of components, it can be at any pH. The buffer range changes along with the buffer capacity, but does not determine it. A high-capacity buffer will result when comparable quantities (i.e., buffer-component ratio < 10:1) of weak acid and weak base are dissolved so that their concentrations are relatively high. 19.7 When the buffer-component ratio is 1:1, the capacity of the buffer for "absorbing" added base and acid is equal, and the buffer range will be symmetrical around the p K a . As the ratio deviates more from this, the buffer range will be unsymmetrical (e.g., from p K a - 0.5 to p K a + 1). 19.8 p K a (formic) = 3.74; p K a (acetic) = 4.74. Formic acid would be the buffer choice, since its p K a is closer to the desired pH. If acetic acid were used, the buffer component ratio would be far from 1:1 and the buffer's effectiveness would be lower (see 18.7). The NaOH serves to partially neutralize the acid and produce its conjugate base. 19.9 a) pH would increase; ratio would increase. b) pH would decrease; ratio would decrease. c) pH would increase; ratio would increase. d) pH would decrease; ratio would decrease. 19.10 a) pH would increase by a small amount. b) pH would decrease by a small amount. c) pH would increase by a very small amount. d) pH would increase by a large amount. 19.11 pH = p K a + log [A - ]/[HA] = 4.89 + log (0.25)/(0.15) = 5.11 and [H 3 O + ] = 7.7 x 10 - 6 M 19.12 pH = p K a + log [A - ]/[HA] 103
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= 4.20 + log (0.28)/(0.33) = 4.13 and [H 3 O + ] = 7.4 x 10 - 5 M 104
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19.13 pH = p K a + log[A - ]/[HA] = 3.15 + log (0.65)/(0.50) = 3.26 and [H 3 O + ] = 5.5 x 10 - 4 M 19.14 pH = p K a + log[A - ]/[HA] = 3.17 + log(0.25)/(0.20) = 3.27 and [H 3 O + ] = 5.4 x 10 - 4 M 19.15 pH = p K a + log [A - ]/[HA] = 3.74
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This note was uploaded on 01/21/2010 for the course CHEM 1A taught by Professor Kobiashi during the Spring '07 term at Ventura College.

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Chapter 19 ionic equilibria in aqueous system - Chapter...

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