Calculate the pH of a solution of 0.10 M HF that is also 0.15 M HCN if K_a for HF is 1.0xx10^-5 and K_a for HCN is 1.0xx10^-7?

Jun 28, 2018

$\text{pH} \approx 2.09$

In principle, it does not matter much which acid you allow to go first, as long as the smaller ${K}_{a}$ is smaller by at least ${10}^{3}$. You may want to try this in reverse order, doing $\text{HCN}$ first and $\text{HF}$ second, just to see if you get about the same $\text{pH}$ again.

(You should get $x = 9.64 \times {10}^{- 6} \text{M}$ for $\left[{\text{H"_3"O}}^{+}\right]$ from $\text{HCN}$ going first, and $x \approx \text{0.00812 M}$ for $\text{HF}$ going second. However, when $\text{HF}$ goes, the small $x$ approximation can only be done with $\text{0.10 M}$, and you can say that $9.64 \times {10}^{- 6}$ $\text{<<}$ $x$ instead in this case.)

Also, a common pitfall is to forget about the ${\text{H"_3"O}}^{+}$ contributed by the acid that goes first. That should be included in the initial concentration of ${\text{H"_3"O}}^{+}$ for the acid that goes second, because they suppress each other.

Since these are both weak acids that "have" similar ${K}_{a}$ values, the weaker one ($\text{HCN}$) is not supposed to be ignored. But these ${K}_{a}$ values are not correct...

${K}_{a} \left(\text{HF}\right) = 6.6 \times {10}^{- 4}$
${K}_{a} \left(\text{HCN}\right) = 6.2 \times {10}^{- 10}$

So, we'd "have" to do sequential ICE tables, but we should not have to for this problem. We SHOULD be able to just ignore $\text{HCN}$...

For $\text{HF}$:

${\text{HF"(aq) + "H"_2"O"(l) rightleftharpoons "F"^(-)(aq) + "H"_3"O}}^{+} \left(a q\right)$

$\text{I"" "0.10" "" "" "-" "" "0" "" "" "" } 0$
$\text{C"" "-x" "" "" "-" "+x" "" "" } + x$
$\text{E"" "0.10-x" "-" "" "x" "" "" "" } x$

Therefore, using its ${K}_{a}$ (which is actually $6.6 \times {10}^{- 4}$):

$6.6 \times {10}^{- 4} = \left(\left[\text{F"^(-)]["H"_3"O"^(+)])/(["HF}\right]\right)$

$= {x}^{2} / \left(0.10 - x\right)$

We do the small $x$ approximation because that was the intent of the problem, and:

$6.6 \times {10}^{- 4} \approx {x}^{2} / 0.10$

=> x = sqrt(0.10 cdot6.6 xx 10^(-4)) = "0.00812 M" = ["H"_3"O"^(+)]_("HF")

(the true answer was $\text{0.00780 M}$, 4.15% error, which is just within range of good enough.)

That concentration becomes the initial in the next ICE table, which we will find does not change much at all.

For $\text{HCN}$ after $\text{HF}$:

${\text{HCN"(aq) + "H"_2"O"(l) rightleftharpoons "CN"^(-)(aq) + "H"_3"O}}^{+} \left(a q\right)$

$\text{I"" "0.15" "" "" "-" "" "" "0" "" "" "" } 0.00812$
$\text{C"" "-x" "" "" "-" "" "+x" "" "" } + x$
$\text{E"" "0.15-x" "-" "" "" "x" "" "" "" } 0.00812 + x$

For this ${K}_{a}$ (which is actually $6.2 \times {10}^{- 10}$):

$6.2 \times {10}^{- 10} = \left(\left[\text{CN"^(-)]["H"_3"O"^(+)])/(["HCN}\right]\right)$

$= \frac{x \left(0.00812 + x\right)}{0.15 - x}$

Here we see that $6.2 \times {10}^{- 10}$ $\text{<<}$ $0.00812$, or $8.12 \times {10}^{- 3}$, as well as $0.15$.

Therefore, in the presence of $\text{HF}$:

$6.2 \times {10}^{- 10} = \frac{x \left(0.00812\right)}{0.15}$

$= 0.0541 x$

=> x = 1.15 xx 10^(-8) "M" = ["H"_3"O"^(+)]_("HCN")

As a result, the total $\left[{\text{H"_3"O}}^{+}\right]$ would be:

color(blue)(["H"_3"O"^(+)]) = ["H"_3"O"^(+)]_("HF") + ["H"_3"O"^(+)]_("HCN")

= "0.00812 M" + 1.15 xx 10^(-8) "M" ~~ color(blue)("0.00812 M")

(Clearly, nothing much changed, because $\text{HCN}$ is a much weaker acid in reality, by a factor of about $1000000$.)

That makes the $\boldsymbol{\text{pH}}$:

$\textcolor{b l u e}{\text{pH} = - \log \left(0.00812\right) = 2.09}$