# What is the pH of a solution that is 0.20 M in HF and 0.40 M in NaF? [Ka = 7.2*10^-4] I know the answer is 3.44 but could someone please explain how they got it? Thank you so much!!!! (:

Dec 31, 2017

You get it like this:

#### Explanation:

HF is a weak acid which dissociates:

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

$\textsf{{K}_{a} = \frac{\left[{H}^{+}\right] \left[{F}^{-}\right]}{\left[H F\right]} = 7.2 \times {10}^{- 4}}$

Rearranging:

$\textsf{\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[H F\right]}{\left[{F}^{-}\right]}}$

Putting in the numbers:

$\textsf{\left[{H}^{+}\right] = 7.2 \times {10}^{- 4} \times \frac{0.2}{0.4} = 3.6 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(3.6 \times {10}^{- 4}\right) = 3.44}$

Dec 31, 2017

When I read this, I interpreted it as having initial concentrations of $\text{HF}$ and $\text{NaF}$. From that, we would consider the following dissociation (units of $\text{M}$):

${\text{HF"(aq) rightleftharpoons "H"^(+)(aq) + "F}}^{-} \left(a q\right)$

$\text{I"" "0.20" "" "" "0" "" "" } 0.40$
$\text{C"" "-x" "" "+x" "" "" } + x$
$\text{E"" "0.20-x" "x" "" "" } 0.40 + x$

This would give:

${K}_{a} = 7.2 \times {10}^{- 4} = \left({\left[\text{H"^(+)]_(eq)["F"^(-)]_(eq))/(["HF}\right]}_{e q}\right)$

$= \frac{x \left(0.40 + x\right)}{0.20 - x}$

Taking the $- \log$ of both sides, this can be made to look like the Henderson-Hasselbalch equation for buffers:

$- \log \left({K}_{a}\right) = - \log x - \log \left(\frac{0.40 + x}{0.20 - x}\right)$

Since $x = \left[{\text{H}}^{+}\right]$, this is really:

$\text{pK"_a = "pH} - \log \left(\frac{0.40 + x}{0.20 - x}\right)$

Or, we can then get the Henderson-Hasselbalch equation like so:

${\text{pH" = "pK}}_{a} + \log \left(\frac{0.40 + x}{0.20 - x}\right)$

In this case, we have a ${K}_{a}$ that is fairly small, so we approximate that $0.40 + x \approx 0.40$ and $0.20 - x \approx 0.20$:

$\textcolor{b l u e}{\text{pH}} \approx - \log \left(7.2 \times {10}^{- 4}\right) + \log \left(\frac{0.40}{0.20}\right)$

$= \textcolor{b l u e}{3.44}$

In homework problems and exams, you can just use the Henderson-Hasselbalch equation to begin with, as long as you know that it only works on buffers containing a weak acid and its conjugate weak base.