Question

Calculate pH of a $1.0\times {10}^{-8}M$ solution of $HCl$.

Answer 1

Step 1. Reactions involved

The total $\left[H_3{O}^{+}\right]$ is from ionization of $HCl$ and $H_{2}O$ both according to the following reactions-

$2H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$

$HCl\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

Step 2. Total $\left[H_3{O}^{+}\right]$

Assume $x=\left[O{H}^{-}\right]=\left[H_3{O}^{+}\right]$ from $H_{2}O$ and $y=\left[H^{+}\right]={10}^{-8}$ from $HCl$ as $HCl$ being strong acid, it completely ionizes.

So, total $\left[H_3{O}^{+}\right]={10}^{-8}+x$ and total $\left[O{H}^{-}\right]=x$ .

We know, $K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right]={10}^{–14}$

$\Rightarrow K_w=({10}^{-8}+x)\left(x\right)={10}^{-14}$

or $x^2+{10}^{-8}x–{10}^{-14}=0$

$\Rightarrow x=9.5\times {10}^{-8}$

So, total $H^{+}=9.5\times {10}^{-8}+{10}^{-8}=10.5\times {10}^{-8}$

Step 3. pH calculation

We know, $pH=-\log \left[H^{+}\right]$

$pH=-\log (10.5\times {10}^{-8})=6.98$


Final answer: $6.98$