Question

pH of a $1.0\times {10}^8$ M solution of HCl is:

• A. $7.02$
• B. $6.958$
• C. $7.4$
• D. $6.8$

Answer 1

Answer: (B)

$6.958$

$HCl$ being strong acid dissolves completely.

$1\times {10}^{-8}M$ is very dilute solution of $HCl$ thereby producing less $H^{+}$ ion concentration than water due to which dissociation of water is to be considered. i.e. $\left[H^{+}\right]$ from water cannot be neglected.

$HCl⟶H^{+}+C{l}^{-}$

${10}^{-8}M$

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

initial step ${10}^{-8}$ $0$

$-x$ $x$ $x$

at $e{q}^n$ ${10}^{-8}+x$ $x$

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

${10}^{-14}=[{10}^{-8}+x]\left[x\right]$

${10}^{-14}=x^2+{10}^{-8}x$

$x\simeq {10}^{-7}$

Total $\left[H^{+}\right]={10}^{-7}+{10}^{-8}$

$=1.1\times {10}^{-7}$

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

$=-\log (1.1\times {10}^{-7})$

$=-\log \left(1.1\right)-\log {10}^{-1}$

$=-0.04139+7$

$=6.95861$