Question

The pH of ${10}^{-3}MNaOH$ is

• A. $3$
• B. $11$
• C. $4$
• D. $-11$

Answer 1

Answer: (B)

$[NaOH]={10}^{-3}M$ $\therefore$ $[O{H}^{-}]={10}^{-3}M$ or $[H^{+}]=\frac{{10}^{-14}}{{10}^{-3}}={10}^{-11}M$ $pH=-\log [H^{+}]=-\log {10}^{-11}=11$