Answer
pH = 1.7
Work Step by Step
1. Calculate the Ka:
$Ka = 10^{-pKa} = 10^{-2.32} = 4.79 \times 10^{-3}$
2. Drawing the ICE table, we get:
-$[H^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
3. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H^+][A^-]}{ [HA]}$
$Ka = 4.79 \times 10^{- 3}= \frac{x * x}{ 0.1 - x}$
$Ka = 4.79 \times 10^{- 3}= \frac{x^2}{ 0.1 - x}$
$ 4.79 \times 10^{- 4} - 4.79 \times 10^{- 3}x = x^2$
$ 4.79 \times 10^{- 4} - 4.79 \times 10^{- 3}x - x^2 = 0$
Bhaskara:
$\Delta = (- 4.79 \times 10^{- 3})^2 - 4 * (-1) *( 4.79 \times 10^{- 4})$
$\Delta = 2.29 \times 10^{- 5} + 1.91 \times 10^{- 3} = 1.93 \times 10^{- 3}$
$x_1 = \frac{ - (- 4.79 \times 10^{- 3})+ \sqrt { 1.93 \times 10^{- 3}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 4.79 \times 10^{- 3})- \sqrt { 1.93 \times 10^{- 3}}}{2*(-1)}$
$x_1 = - 2.44 \times 10^{- 2} (Negative)$
$x_2 = 1.96 \times 10^{- 2}$
The concentration can't be negative.
$[H^+] = x = 1.96 \times 10^{-2}$
4. Calculate the pH:
$pH = -log[H^+]$
$pH = -log( 1.96 \times 10^{- 2})$
$pH = 1.7$
