Answer
(a) $[C_6H_5CH_2{CO_2}^-] = 3.01 \times 10^{-3}M$
(b) pH = 2.614
Work Step by Step
1. Drawing the ICE table we get:
$[H_3O^+] = [C_6H_5CH_2{CO_2}^-] = x$
$[C_6H_5CH_2CO_2H] = [C_6H_5CH_2CO_2H]_{initial} - x$
(a)
2. Now, use the Ka value and equation to find the 'x' value.
- Approximating $0.186M - x \approx 0.186M$ we get:
$Ka = 4.9 \times 10^{- 5}= \frac{x * x}{ 1.86 \times 10^{- 1}}$
$Ka = 4.9 \times 10^{- 5}= \frac{x^2}{ 1.86 \times 10^{- 1}}$
$ 9.11 \times 10^{- 6} = x^2$
$x = 3.01 \times 10^{- 3}$
5% test: $\frac{ 3.01 \times 10^{- 3}}{ 1.86 \times 10^{- 1}} \times 100\% = 1.623\%$
%ionization < 5% : Right approximation.
$[C_6H_5CH_2{CO_2}^-] = x = 3.01 \times 10^{-3}M$
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(b)
2. Now, use the Ka value and equation to find the 'x' value.
- Approximating $0.121M - x \approx 0.121M$
$Ka = 4.9 \times 10^{- 5}= \frac{x * x}{ 1.21 \times 10^{- 1}}$
$Ka = 4.9 \times 10^{- 5}= \frac{x^2}{ 1.21 \times 10^{- 1}}$
$ 5.92 \times 10^{- 6} = x^2$
$x = 2.43 \times 10^{- 3}$
5% test: $\frac{ 2.43 \times 10^{- 3}}{ 1.21 \times 10^{- 1}} \times 100\% = 2.012\%$
%ionization < 5% : Right approximation.
Therefore: $[H^+] = x = 2.43 \times 10^{-3} $
3. Find the pH value.
$pH = -log[H^+]$
$pH = -log( 2.43 \times 10^{- 3})$
$pH = 2.614$
