Answer
$[H_3O^+] = 2.55 \times 10^{- 3}M $
pH = 2.594
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.
-$[H_3O^+] = [Conj. Base] = x$
-$[Benzoic Acid] = [Benzoic Acid]_{initial} - x = 0.1 - x$
For approximation, we consider: $[Benzoic Acid] = 0.1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Benzoic Acid]}$
$Ka = 6.5 \times 10^{- 5}= \frac{x * x}{ 0.1}$
$Ka = 6.5 \times 10^{- 5}= \frac{x^2}{ 0.1}$
$ 6.5 \times 10^{- 6} = x^2$
$x = 2.55 \times 10^{- 3}$
Percent ionization: $\frac{ 2.55 \times 10^{- 3}}{ 0.1} \times 100\% = 2.55\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = [Conj. Base] = x = 2.55 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[Benzoic Acid] \approx 0.1M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.55 \times 10^{- 3})$
$pH = 2.594$
