Answer
pH = 8.08
Work Step by Step
- NaF: Na does not hydrolyze, and $F^-$ is the conjugate base of a weak acid.
$$K_b(F^-) = \frac{1.0 \times 10^{-14}}{6.6 \times 10^{-4}} = 1.5 \times 10^{-11}$$
Now, find the pH of a basic solution of 0.10 M $F^-$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ F^- ]& [ HF ]& [ OH^- ]\\
Initial& 0.10 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.10 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ HF ][ OH^- ]}{[ F^- ]}$$
$$K_b = \frac{(x)(x)}{[ F^- ]_{initial} - x}$$
3. Assuming $ 0.10 \gt\gt x$:
$$K_b = \frac{x^2}{[ F^- ]_{initial}}$$
$$x = \sqrt{K_b \times [ F^- ]_{initial}} = \sqrt{ 1.5 \times 10^{-11} \times 0.10 }$$
$x = 1.2 \times 10^{-6} $
4. Test if the assumption was correct:
$$\frac{ 1.2 \times 10^{-6} }{ 0.10 } \times 100\% = 1.2 \times 10^{-3} \%$$
5. Thus, it is correct to say that $x = 1.2 \times 10^{-6} $
6. $[OH^-] = x = 1.2 \times 10^{-6} $
7. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.2 \times 10^{-6} } = 8.3 \times 10^{-9} \space M$$
$$pH = -log[H_3O^+] = -log( 8.3 \times 10^{-9} ) = 8.08 $$
