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 \rightarrow \, Na^+ +OH^-$
$\therefore \, \, \, \, [OH^-]=[NaOH]=\frac{M}{100}=10^{-2}M$
Now,as $ \, \, \, \, \, \, [H_3O^+][OH^-]=10^{-14}$
$or \, \, \, \, \, \, \, \, \, \, \, \, \, \, [H_3O^+] =\frac{10^{-14}}{[OH^-]}=\frac{10^{-14}}{10^{-2}}=10^{-12}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, pH=-log[H_3O^+]$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =-log \, 10^{-12}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, pH=12$