# What is the pH of a 6.2 * 10^-3 M solution of HF?

Jun 5, 2016

$p H = 2.77$

#### Explanation:

The acid dissociation reaction for $H F$ is the following:

$H F \to {H}^{+} + {F}^{-}$

The ${K}_{a}$ expression is: ${K}_{a} = \frac{\left[{H}^{+}\right] \left[{F}^{-}\right]}{\left[H F\right]} = 6.6 \times {10}^{- 4}$

The pH of this solution could be found by: $p H = - \log \left[{H}^{+}\right]$.

We will need to find the $\left[{H}^{+}\right]$ using $I C E$ table:

$\text{ " " " " " " " "HF" " " " " " "->" " " " H^(+) +" " " } {F}^{-}$
$\text{Initial" " "6.2xx10^(-3)M" " " " " " " "0M" " " " " } 0 M$
$\text{Change" " "-xM" " " " " " " " " "+xM" " " " } + x M$
$\text{Equilibrium" ""(6.2xx10^(-3)-x)M" " "xM" " " " " " } x M$

now, we can replace the concentrations by their values in the expression of ${K}_{a}$:

${K}_{a} = \frac{\left[{H}^{+}\right] \left[{F}^{-}\right]}{\left[H F\right]} = 6.6 \times {10}^{- 4} = \frac{x \times x}{6.2 \times {10}^{- 3} - x}$

Solve for $x = 1.7 \times {10}^{- 3} M$

$\implies p H = - \log \left[{H}^{+}\right] = - \log \left(1.7 \times {10}^{- 3}\right) = 2.77$

Acids & Bases | pH of a Weak Acid.