Answer
Percent ionization = $9.512 \%$
pH = 2.022
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x = 0.1 - x$
For approximation, we consider: $[Acid] = 0.1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = 1 \times 10^{- 3}= \frac{x * x}{ 0.1}$
$Ka = 1 \times 10^{- 3}= \frac{x^2}{ 0.1}$
$ 1 \times 10^{- 4} = x^2$
$x = 1 \times 10^{- 2}$
Percent ionization: $\frac{ 1 \times 10^{- 2}}{ 0.1} \times 100\% = 10\%$
%ionization > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 1 \times 10^{- 3}= \frac{x^2}{ 0.1- x}$
$ 1 \times 10^{- 4} - 1 \times 10^{- 3}x = x^2$
$ 1 \times 10^{- 4} - 1 \times 10^{- 3}x - x^2 = 0$
Bhaskara:
$\Delta = (- 1 \times 10^{- 3})^2 - 4 * (-1) *( 1 \times 10^{- 4})$
$\Delta = 1 \times 10^{- 6} + 4 \times 10^{- 4} = 4.01 \times 10^{- 4}$
$x_1 = \frac{ - (- 1 \times 10^{- 3})+ \sqrt { 4.01 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1 \times 10^{- 3})- \sqrt { 4.01 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 1.051 \times 10^{- 2} (Negative)$
$x_2 = 9.512 \times 10^{- 3}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
- Percent ionization = $\frac{9.512 \times 10^{-3}}{0.1} \times 100\% = 9.512\%$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 9.512 \times 10^{- 3})$
$pH = 2.022$
