Answer
$$[H_3O^+] = [F^-] = 0.023 \space M$$
$$[OH^-] = 4.3 \times 10^{-13} \space M$$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HF ]& [ F^- ]& [ H_3O^+ ]\\
Initial& 0.75 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.75 -x& x& x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ F^- ][ H_3O^+ ]}{[ HF ]}$$
$$K_a = \frac{(x)(x)}{[ HF ]_{initial} - x}$$
3. Assuming $ 0.75 \gt\gt x:$
$$K_a = \frac{x^2}{[ HF ]_{initial}}$$
$$x = \sqrt{K_a \times [ HF ]_{initial}} = \sqrt{ 6.8 \times 10^{-4} \times 0.75 }$$
$x = 0.023 $
4. Test if the assumption was correct:
$$\frac{ 0.023 }{ 0.75 } \times 100\% = 3.1 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 0.023 $
6. $$[H_3O^+] = [F^-] = x = 0.023 $$
7. Calculate the hydronium ion concentration:
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.023 } = 4.3 \times 10^{-13} \space M$$
