Answer
$NaF$ to $HF$ ratio = 3.5;
Work Step by Step
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4}$
$[H_3O^+] = 1 \times 10^{- 4}$
2. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][F^-]}{[HF]}$
$3.5 \times 10^{-4} = \frac{1 \times 10^{-4}*[F^-]}{[HF]}$
$\frac{3.5 \times 10^{-4}}{1 \times 10^{-4}} = \frac{[F^-]}{[HF]}$
$3.5 = \frac{[F^-]}{[HF]}$
