Question

The $pH$ of $10^{-8}MNaOH$ aqueous solution at $25^{\circ }C$ , is

• A. 7.02
• B. 7.0
• C. 6.89
• D. 6.0

Answer 1

Answer: (A)

$\left[O{H}^{-}\right]$ of $NaOH=10^{-8}M$

Since, the concentration of $O{H}^{-}$ obtained from $NaOH$ is very low, the concentration of $O{H}^{-}$ in water cannot be neglected. Thus,

Total, $\left[O{H}^{-}\right]={\left[O{H}^{-}\right]}_{NaOH}+{\left[O{H}^{-}\right]}_{H_{2}O}$

$=10^{-8}+10^{-7}$

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

$pOH=-\log \left[O{H}^{-}\right]$

$=-\log \left(1.1\times 10^{-7}\right)=6.96$

$\therefore pH=14-6.96=7.04$