Question

Calculate the $pH$ of ${10}^{-10}MHCl$ solution.

• A. $4$
• B. $8$
• C. $7$
• D. $1$

Answer 1

The correct option is C $7$

We know that,
$pH=-log\left[H^{+}\right]$
${10}^{-10}MHCl$ gives ${10}^{-10}$ moles of $\left[H^{+}\right]$ ions in $1\text{L}$ of solution
Now,
$H_{2}O\rightleftharpoons \underset{{10}^{-7}}{H^{+}}+\underset{{10}^{-7}}{O{H}^{-}}$
Therefore,
$[H^{+}{]}_{total}=[H^{+}]fromHCl+[H^{+}]from{H}_{2}O$
$pH=-log[{10}^{-10}+{10}^{-7}]$
${10}^{-10}$ is negligible in comparison to ${10}^{-7}$
$\therefore pH=-log\left[{10}^{-7}\right]=7log10=7$