Answer
Percent dissociation (Pure acid): $2.082\%$
Percent dissociation (With the salt):$= 0.065\%$
This difference occurs because, the presence of the conjugate base in the second solution move the equilibrium to the left side, which is based on Le Chatelier's principle.
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer (Top table).
-$[H_3O^+] = [C_7H_5{O_2}^-] = x$
-$[HC_7H_5O_2] = [HC_7H_5O_2]_{initial} - x = 0.15 - x$
For approximation, we consider: $[HC_7H_5O_2] = 0.15M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_7H_5{O_2}^-]}{ [HC_7H_5O_2]}$
$Ka = 6.5 \times 10^{- 5}= \frac{x * x}{ 0.15}$
$Ka = 6.5 \times 10^{- 5}= \frac{x^2}{ 0.15}$
$ 9.75 \times 10^{- 6} = x^2$
$x = 3.122 \times 10^{- 3}$
Percent dissociation: $\frac{ 3.122 \times 10^{- 3}}{ 0.15} \times 100\% = 2.082\%$
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With sodium benzoate:
1. Draw the ICE table (It is in the end of this answer)
2. Calculate 'x' using the $K_a$ expression.
$ 6.5\times 10^{- 5} = \frac{[C_7H_5{O_2}^-][H_3O^+]}{[HC_7H_5O_2]}$
$ 6.5\times 10^{- 5} = \frac{( 0.1 + x )* x}{ 0.15 - x}$
Considering 'x' has a very small value.
$ 6.5\times 10^{- 5} = \frac{ 0.1 * x}{ 0.15}$
$ 6.5\times 10^{- 5} = 0.6667x$
$\frac{ 6.5\times 10^{- 5}}{ 0.6667} = x$
$x = 9.75\times 10^{- 5}$
Percent dissociation: $\frac{ 9.75\times 10^{- 5}}{ 0.15} \times 100\% = 0.065\%$
x = $[H_3O^+]$
