Question

The pH of $10^{-8}MNaOH$ is:

• A. 8.0
• B. 6.0
• C. 6.96
• D. 7.04

Answer 1

Answer: (D)

For $10^{-8}MNaOH$,
$\left[O{H}^{-}\right]=10^{-8}+10^{-7}$ (from water)
$=11\times 10^{-8}$
$\therefore pOH=-\log \left[O{H}^{-}\right]=-\log 11\times 10^{-8}$
$=8-1.04=6.96$
$pH=14-6.96=7.04$