Question

The $pH$ of a $0.02M$ ammonia solution which is $5\%$ ionized will be:

• A. $2$
• B. $5$
• C. $7$
• D. $11$

Answer 1

Answer: (D)

$11$

$C=0.02MN{H}_3=Base$

$\alpha =5$%$=0.05$

$\left({OH}^{-}\right)=C\times \alpha$

$\left({OH}^{-}\right)=0.02\times 0.05$

$={10}^{-3}$

$\therefore pOH=-log\left(OH\right)=-log\left({10}^{-3}\right)$

$=-(-3)$

$=3$

$weknow,pH+pOH=14$

$pH=14-pOH$

$=14-3=11$

Here answer is option D.