Answer
$[H_3O^+] = x= 7.24 \times 10^{-2}$
$[OH^-] = 1.38 \times 10^{-13}$
$pH = 1.14$
$pOH = 12.86$
Work Step by Step
1. Drawing the ICE table we get:
- $[H_3O^+] = [A^-] = x$
- $[HA] = [HA]_{initial} - x$
approximation: $[HA] \approx [HA]_{initial}$
2. Use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{[HA]}$
$Ka = 1.1 \times 10^{- 2}= \frac{x * x}{ 5.5 \times 10^{- 1}}$
$Ka = 1.1 \times 10^{- 2}= \frac{x^2}{ 5.5 \times 10^{- 1}}$
$ 6.05 \times 10^{- 3} = x^2$
$x = 7.77 \times 10^{- 2}$
5% test: $\frac{ 7.77 \times 10^{- 2}}{ 5.5 \times 10^{- 1}} \times 100\% = 14.1\%$
%ionization > 5% : Not accurate approximation.
- So we need to consider $[HA] = [HA]_{initial} - x$:
$Ka = 1.1 \times 10^{- 2}= \frac{x^2}{ 5.5 \times 10^{- 1}- x}$
$ 6.05 \times 10^{- 3} - 1.1 \times 10^{- 2}x = x^2$
$ 6.05 \times 10^{- 3} - 1.1 \times 10^{- 2}x - x^2 = 0$
$\Delta = (- 1.1 \times 10^{- 2})^2 - 4 * (-1) *( 6.05 \times 10^{- 3})$
$\Delta = 1.21 \times 10^{- 4} + 2.42 \times 10^{- 2} = 2.43 \times 10^{- 2}$
$x_1 = \frac{ - (- 1.1 \times 10^{- 2})+ \sqrt { 2.43 \times 10^{- 2}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.1 \times 10^{- 2})- \sqrt { 2.43 \times 10^{- 2}}}{2*(-1)}$
$x_1 = - 8.34 \times 10^{- 2} (Negative)$
$x_2 = 7.24 \times 10^{- 2}$
- Concentrations can't be negative.
Therefore: $[H_3O^+] = x= 7.24 \times 10^{-2}$
3. Find the pH, pOH and OH.
$pH = -log[H^+]$
$pH = -log( 7.24 \times 10^{- 2})$
$pH = 1.14$
$pOH + pH = 14$
$pOH = 14 - pH = 14 - 1.14$
$pOH = 12.86$
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{-12.86}$
$[OH^-] = 1.38 \times 10^{-13}$
