Answer
$[H_3O^+] = [F^-] = 3.5 \times 10^{-3}M$
$[HF] = 0.017M$
$[OH^-] = 2.9 \times 10^{-12}M$
$pH = 2.46$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$HF(aq) + H_2O(l) \lt -- \gt F^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $F^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [F^-] = 0 + x = x$
-$[HF] = [HF]_{initial} - x$
For approximation, we are going to consider $[HF]_{initial} = [HF]$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][F^-]}{ [HF]}$
$Ka = 7.2 \times 10^{- 4}= \frac{x * x}{ 0.02}$
$Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 0.02}$
$x^2 = 0.02 \times 7.2 \times 10^{-4} $
$x = \sqrt { 0.02 \times 7.2 \times 10^{-4}} = 3.8 \times 10^{-3} $
Percent dissociation: $\frac{ 3.8 \times 10^{- 3}}{ 0.02} \times 100\% = 19\%$
%dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 0.02- x}$
$ 1.4 \times 10^{- 5} - 7.2 \times 10^{- 4}x = x^2$
$ 1.4 \times 10^{- 5} - 7.2 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 7.2 \times 10^{- 4})^2 - 4 * (-1) *( 1.4 \times 10^{- 5})$
$\Delta = 5.2 \times 10^{- 7} + 5.8 \times 10^{- 5} = 5.8 \times 10^{- 5}$
$x_1 = \frac{ - (- 7.2 \times 10^{- 4})+ \sqrt { 5.8 \times 10^{- 5}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 7.2 \times 10^{- 4})- \sqrt { 5.8 \times 10^{- 5}}}{2*(-1)}$
$x_1 = - 4.2 \times 10^{- 3} (Negative)$
$x_2 = 3.5 \times 10^{- 3}$
- The concentration can't be negative, so $x$ = $x_2$ $= 3.5 \times 10^{-3}$
$[H_3O^+] = [F^-] = x = 3.5 \times 10^{-3}M$
$[HF] = [HF]_{initial} - x = (0.020) - (3.5 \times 10^{-3}) = 0.017M$
3. Calculate the hydroxide ion concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$3.5 \times 10^{-3} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{3.5 \times 10^{-3}}$
$[OH^-] = 2.9 \times 10^{-12}M$
4. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 3.5 \times 10^{- 3})$
$pH = 2.46$
