# Given 100.0 mL of a buffered solutions which is 0.30 M HF and 0.30 M NaF and the Ka of HF to be 7.2 x 10^4, how do you find the pH of this buffer?

Oct 20, 2016

I assume that the value of $\textsf{{K}_{a}}$ is $\textsf{7.2 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$.

$\textsf{p H = 3.14}$

#### Explanation:

$\textsf{H F}$ dissociates:

$\textsf{H F r i g h t \le f t h a r p \infty n s {H}^{+} + {F}^{-}}$

For which:

sf(K_a=([H^+][F^-])/([HF])

These are equilibrium concentrations.

Rearranging gives:

sf([H^+]=K_axx([HF])/([F^(-)])

Assuming we have been given equilibrium concentrations these will be the same so:

sf([H^+]=K_(a)xx(cancel([HF]))/(cancel([F^(-)]))

$\therefore$$\textsf{\left[{H}^{+}\right] = 7.2 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(7.2 \times {10}^{- 4}\right) = 3.14}$