Answer
(a) pH = 11.128
(b) pH = 8.965
Work Step by Step
(a)
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [N{H_4}^+] = x$
-$[NH_3] = [NH_3]_{initial} - x = 0.1 - x$
For approximation, we consider: $[NH_3] = 0.1M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.1}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{0.1}$
$ 1.8 \times 10^{- 6} = x^2$
$x = 1.342 \times 10^{- 3}$
Percent ionization: $\frac{ 1.342 \times 10^{- 3}}{ 1 \times 10^{- 1}} \times 100\% = 1.342\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = x = 1.342 \times 10^{- 3} $
3. Convert $[OH^-]$ into pH.
$pOH = -log[OH^-]$
$pOH = -log( 1.342 \times 10^{- 3})$
$pOH = 2.872$
$pH + pOH = 14$
$pH + 2.872 = 14$
$pH = 11.128$
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(b)
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [C_5H_5NH^+] = x$
-$[C_5H_5N] = [C_5H_5N]_{initial} - x = 0.05 - x$
For approximation, we consider: $[C_5H_5N] = 0.05M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][C_5H_5NH^+]}{ [C_5H_5N]}$
$Ka = 1.7 \times 10^{- 9}= \frac{x * x}{ 5 \times 10^{- 2}}$
$Ka = 1.7 \times 10^{- 9}= \frac{x^2}{ 5 \times 10^{- 2}}$
$ 8.5 \times 10^{- 11} = x^2$
$x = 9.22 \times 10^{- 6}$
Percent ionization: $\frac{ 9.22 \times 10^{- 6}}{ 5 \times 10^{- 2}} \times 100\% = 0.01844\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = x = 9.22 \times 10^{- 6} $
3. Convert $[OH^-]$ into pH.
$pOH = -log[OH^-]$
$pOH = -log( 9.22 \times 10^{- 6})$
$pOH = 5.035$
$pH + pOH = 14$
$pH + 5.035 = 14$
$pH = 8.965$
