Answer
pH = 2.59
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.
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x = 0.1M - x$
For approximation, we consider: $[HA] = [HA]_{initial} = 0.1M$
** HA = Acid, A- = Conjugate base.
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 6.5 \times 10^{- 5}= \frac{x * x}{0.1}$
$Ka = 6.5 \times 10^{- 5}= \frac{x^2}{ 0.1}$
$ 6.5 \times 10^{- 6} = x^2$
$x = 2.54 \times 10^{- 3}$
5% test: $\frac{ 2.54 \times 10^{- 3}}{0.1} \times 100\% = 2.54\%$
%ionization < 5% : Right approximation.
$[H_3O^+] = x = 2.54 \times 10^{-3}$
3. Find the pH:
$pH = -log[H_3O^+]$
$pH = -log( 2.54 \times 10^{- 3})$
$pH = 2.59$
