# What is the pH of 3.1 times 10^-3 M HCl?

Apr 3, 2017

There are two ways you can do this. The easy way is to realize that $\text{HCl}$ is a strong acid, so its dissociation is considered complete, and $\left[{\text{HCl"] = ["H}}^{+}\right]$.

EASY WAY

Recall:

"pH" = -log["H"^(+)]

From the knowledge that "pH" = -log["H"^(+)] = -log["HCl"], we can say:

$\textcolor{b l u e}{\text{pH") = -log(3.1 xx 10^(-3) "M}} = {2.50}_{8638306}$

$= \textcolor{b l u e}{2.51}$

HARD WAY

The hard way is to look up the ${K}_{a}$ of $\text{HCl}$ and find that it is around $1.3 \times {10}^{6}$ and do an ICE Table as if it were an actually not-heavily-skewed equilibrium. I would never advise any chemistry student to do this...

${\text{HCl"(aq) " "" " -> " "" ""H"^(+)(aq) + "Cl}}^{-} \left(a q\right)$

$\text{I"" "3.1 xx 10^(-3)"M"" "" "" "" ""0 M"" "" ""0 M}$
$\text{C"" "-x" "" "" "" "" "" "" "+x" "" } + x$
$\text{E"" "(3.1 xx 10^(-3) -x)"M"" "" "x" "" "" } x$

${K}_{a} = 1.3 \times {10}^{6} = {x}^{2} / \left(3.1 \times {10}^{- 3} - x\right)$

We could simply make the approximation that $3.1 \times {10}^{- 3} \approx x$, but we are supposing we don't know that yet. So, let's just solve for the quadratic equation.

${x}^{2} + {K}_{a} x - 3.1 \times {10}^{- 3} {K}_{a} = 0$

The quadratic formula gives, with $a = 1$, $b = {K}_{a}$, $c = - 3.1 \times {10}^{- 3} {K}_{a}$:

$x = 0.003099999993$,

and we reject the root $x \approx - 1.3 \times {10}^{6}$ since ${\left[\text{HCl}\right]}_{e q}$ would have become $3.1 \times {10}^{- 3} - \left(- 1.3 \times {10}^{6}\right) \approx 1.3 \times {10}^{6}$, which would falsely imply that $\text{HCl}$ doesn't dissociate at all.

Thus, we get that

${K}_{a} = {\left(3.099999993 \times {10}^{- 3}\right)}^{2} / \left(3.1 \times {10}^{- 3} - 3.099999993 \times {10}^{- 3}\right)$

$= 1.37 \times {10}^{6} \approx 1.3 \times {10}^{6}$,

as it was when we started, and

${\left[\text{HCl}\right]}_{e q} = 3.1 \times {10}^{- 3} - 3.099999993 \times {10}^{- 3} \approx 0$,

and

${\left[{\text{H}}^{+}\right]}_{e q} = 3.099999993 \times {10}^{- 3}$ $\text{M}$ $\approx$ $3.1 \times {10}^{- 3}$ $\text{M}$,

as expected. Finding ${K}_{a}$ is still valid, however, since the denominator isn't exactly zero... So, if we really wanted the "exact" answer:

"pH" = -log["H"^(+)] = 2.50_(8638307) ~~ 2.51

as before, only differing past the 8th decimal place (so who cares?).