Question

What is the pH of ${10}^7$mol $L^{-1}$ HCl solution at ${25}^{\circ }$C?

• A. 7.05
• B. 6.2
• C. 6.79
• D. 7

Answer 1

The correct option is

C

6.79

For ${10}^{-7}$mol $L^{-1}$HCl, the concentration of $H^{+}$ due to the dissociation of water will not be negligible in comparison to the concentration $H^{+}$ from the acid. Hence, we have

[$H^{+}{]}_{total}$ = [$H^{+}{]}_{acid}$ + [$H^{+}{]}_{water}$

$={10}^{-7}+{10}^{-7}\left[Since\right[H^{+}]inwateris{10}^{-7}mol{L}^{-1}at{25}^{\circ }C$]
$={10}^{-7}(1+1)$
$=2\times {10}^{-7}$
Putting the value of $\left[H^{+}\right]$ in the equation

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

$pH$= - log [2 \times 10^{-7}]\)

$pH=6.79$