Answer
$pH = 3.25$
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.
-$[H_3O^+] = [F^-] = x$
-$[HF] = [HF]_{initial} - x = 1 \times 10^{- 3} - x$
For approximation, we consider: $[HF] = 1 \times 10^{- 3}M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][F^-]}{ [HF]}$
$Ka = 7.2 \times 10^{- 4}= \frac{x * x}{ 1\times 10^{- 3}}$
$Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 1\times 10^{- 3}}$
$ 7.2 \times 10^{- 7} = x^2$
$x = 8.485 \times 10^{- 4}$
Percent ionization: $\frac{ 8.485 \times 10^{- 4}}{ 1\times 10^{- 3}} \times 100\% = 84.85\%$
%ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Kb = 7.2 \times 10^{- 4}= \frac{x^2}{ 1 \times 10^{- 3}- x}$
$ 7.2 \times 10^{- 7} - 7.2 \times 10^{- 4}x = x^2$
$ 7.2 \times 10^{- 7} - 7.2 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 7.2 \times 10^{- 4})^2 - 4 * (-1) *( 7.2 \times 10^{- 7})$
$\Delta = 5.184 \times 10^{- 7} + 2.88 \times 10^{- 6} = 3.398 \times 10^{- 6}$
$x_1 = \frac{ - (- 7.2 \times 10^{- 4})+ \sqrt { 3.398 \times 10^{- 6}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 7.2 \times 10^{- 4})- \sqrt { 3.398 \times 10^{- 6}}}{2*(-1)}$
$x_1 = - 1.282 \times 10^{- 3} (Negative)$
$x_2 = 5.617 \times 10^{- 4}$
- The concentration can't be negative.
$[H_3O^+] = x = 5.617 \times 10^{- 4}$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 5.617 \times 10^{- 4})$
$pH = 3.25$
