Answer
pH = 2.599
Work Step by Step
1. Draw the ICE table: (Image on the end)
2. Now, use the Ka value and equation to find "x".
$Ka = 6.5 \times 10^{-5} = \frac{(x)*(x)}{0.10-x} $
$6.5 \times 10^{-5} = \frac{(x^2)}{0.10-x} $
$6.5 \times 10^{-6} - 6.5 \times 10^{-5}x = x^2$
$6.5 \times 10^{-6} - 6.5 \times 10^{-5}x - x^2 = 0$
Bhaskara:
$\Delta = (-6.5 \times 10^{-5})^2 - 4 * (-1) * (6.5 \times 10^{-6})$
$\Delta = 4.225 \times 10^{-9} + 2.6 \times 10^{-5}$
$\Delta \approx 2.6 \times 10^{-5}$
$x_1 = \frac{-(-6.5 \times 10^{-5}) + \sqrt {2.6 \times 10^{-5}}}{2*(-1)}$
or
$x_2 = \frac{-(-6.5 \times 10^{-5}) - \sqrt {2.6 \times 10^{-5}}}{2*(-1)}$
$x_1 = -2.582 \times 10^{-3} (Negative)$
$x_2 = 2.517 \times 10^{-3} $
The concentration can't be negative.
$[H^+] = 0 + x = x$
Therefore, the concentration of $[H^+]$ is $2.517 \times 10^{-3}$
3. Calculate the pH:
$pH = -log[H^+] = -log(2.517 \times 10^{-3}) = 2.599$