Question

Calculate the pH of ${10}^{-8}\text{M}$ $HCl$.

• A. 8
• B. 6
• C. 7
• D. 6.98

Answer 1

The correct option is D 6.98

On using this relation, $pH=–log\left[H_3{O}^{+}\right]$, we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7.

It may be noted that in very dilute acidic solution when $H^{+}$ concentrations from acid and water are comparable, the concentration of $H^{+}$ from water cannot be neglected.

Therefore,

$\left[H^{+}\right]$ total = $\left[H^{+}\right]$ acid + $\left[H^{+}\right]$ water

Since HCl is a strong acid and is completely ionized.

$\left[H^{+}\right]HCl=1.0\times {10}^{-8}$

The concentration of $H^{+}$ from ionization is equal to the $\left[O{H}^{–}\right]$ from water,

$\left[H^{+}\right]H_{2}O=\left[O{H}^{–}\right]H_{2}O$ = x (say)

$\left[H^{+}\right]$ total $=1.0\times {10}^{-8}+x$

But,

$\left[H^{+}\right]\left[O{H}^{–}\right]=1.0\times {10}^{-14}$

$(1.0\times {10}^{-8}+x)\left(x\right)=1.0\times {10}^{-14}$

$x^2+{10}^{-8}x–{10}^{-14}=0$

Solving for x, we get $x=9.5\times {10}^{-8}$

Therefore,

$\left[H^{+}\right]=1.0\times {10}^{-8}+9.5\times {10}^{-8}$

$=10.5\times {10}^{-8}$

$=1.05\times {10}^{-7}$

$pH=–log\left[H^{+}\right]$

$=–log(1.05\times {10}^{-7})=6.98$

Hence, option $D$ is correct.