Question

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

Answer 1

$[H+]$ total $=[H+]$ acid $+[H+]$ 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,

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

$=x\left(say\right)$

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

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, this is the answer.