Question

The pH of M/100 NaOH solution is :

• A. 2
• B. 10
• C. 6
• D. 12

Answer 1

Answer: (D)

NaOH completely ionises as
$NaOH\to N{a}^{+}+O{H}^{-}$
$\therefore [O{H}^{-}]=[NaOH]=\frac{M}{100}=10^{-2}M$
Now,as $[H_3{O}^{+}][O{H}^{-}]=10^{-14}$
$or[H_3{O}^{+}]=\frac{10^{-14}}{[O{H}^{-}]}=\frac{10^{-14}}{10^{-2}}=10^{-12}$
$\therefore pH=-log[H_3{O}^{+}]$
$=-log10^{-12}$
$pH=12$