Answer
$pH = 3.76$
Work Step by Step
1. Drawing the ICE table, we get:
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
For approximation, we consider: $[HA] = [HA]_{initial}$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 3 \times 10^{- 7}= \frac{x * x}{ 1 \times 10^{- 1}}$
$Ka = 3 \times 10^{- 7}= \frac{x^2}{ 1 \times 10^{- 1}}$
$ 3 \times 10^{- 8} = x^2$
$x = 1.73 \times 10^{- 4}$
5% test: $\frac{ 1.73 \times 10^{- 4}}{ 1 \times 10^{- 1}} \times 100\% = 0.173\%$
%ionization < 5% : Right approximation.
$[H_3O^+] = x = 1.73 \times 10^{-4} $
3. Calculate the pH:
$pH = -log[H^+]$
$pH = -log( 1.73 \times 10^{- 4})$
$pH = 3.76$
