Answer
$pH = 4.27$
Work Step by Step
1. Drawing the ICE table, we get:
-$[H^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
For approximation, we consider: $[HA] = [HA]_{initial}$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H^+][A^-]}{ [HA]}$
$Ka = 3 \times 10^{- 8}= \frac{x * x}{ 9.5 \times 10^{- 2}}$
$Ka = 3 \times 10^{- 8}= \frac{x^2}{ 9.5 \times 10^{- 2}}$
$ 2.85 \times 10^{- 9} = x^2$
$x = 5.33 \times 10^{- 5}$
5% test: $\frac{ 5.33 \times 10^{- 5}}{ 9.5 \times 10^{- 2}} \times 100\% = 0.0561\%$
%ionization < 5% : Right approximation.
- $[H^+] = x = 5.33 \times 10^{-5}M$
3. Calculate the pH:
$pH = -log[H^+]$
$pH = -log( 5.33 \times 10^{- 5})$
$pH = 4.27$
