Answer
The pH value is approximately 3.07.
Work Step by Step
1. Calculate the Ka:
$Ka = 10^{-pKa}$
$Ka = 10^{-4.84}$
$Ka = 1.45 \times 10^{-5}$
2. Now, find $[H^+]$ using the Ka formula:
$Ka = \frac{[H^+][A^-]}{[HA]}$
$[H^+] = [A^-] = x$
$1.45 \times 10^{-5} = \frac{x^2}{0.05-x}$
*Since $(0.05-x) \approx (0.05)$
$1.45 \times 10^{-5} \approx \frac{x^2}{0.05}$
$7.25 \times 10^{-7} \approx x^2$
$x \approx 8.51 \times 10^{-4}$
3. Calculate the pH:
$pH = -log(8.51 \times 10^{-4})$
$pH \approx 3.07$
