Question

A solution of $HCl$ has a $pH=5$. If $1mL$ of it is diluted to $1$ litre, $pH$ of resulting solution will be:

Answer 1

When an acid solution is diluted, the hydronium ion concentration for the diluted solution is obtained by using the formula $M_1\times V_1=M_2\times V_2$

pH = -log[$H^{+}$]

5 = -log[$H^{+}$]
$\left[H^{+}\right]={10}^{-5}$
$M_1={10}^{-5}$ as pH = 5 for HCl, $V_1=1ml,V_2=1l=1000ml$
$M_1{V}_1=M_2{V}_2$
${10}^{-5}\times 1=M_2\times 1000$
$M_2={10}^{-8}$
$[H{]}^{+}={10}^{-8}$

Since the value of the hydronium ion concentration obtained from hydrochloric acid is very small, you should also consider the hydronium ion concentration obtained from the autoionization of water.

Ionisation constant for water is, $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]=\left[{10}^{-7}\right]\left[{10}^{-7}\right]$

To the above value, add hydronium ion concentration obtained from the autoionization of water.
$\left[H^{+}\right]={10}^{-7}M+{10}^{-8}M$

$pH=-log\left[H^{+}\right]$=−log$({10}^{-7}M+{10}^{-8}M)$

$pH=-log\left[{10}^{-8}\right(1+10\left)\right]=log[11\times {10}^{-8}]$
$pH=-[log11+(-8\left)log10\right]=6.9$