Question

Concentration of a weak acid is $0.1$N and $K_b={10}^{-5}$ then $pH$ will be:

• A. $4$
• B. $3$
• C. $2$
• D. $5$

Answer 1

Answer: (D)

$5$

Given $K_b={10}^{-5}$

$p^{K_b}=-\log {10}^{-5}=5$

$\Rightarrow p^{K_a}+p^{K_b}=14$

$\Rightarrow p^{K_a}+5=14$

$\Rightarrow p^{K_a}=14-5=9$

$\Rightarrow K_a={10}^{-9}$

$\Rightarrow \left[H^{+}\right]=\sqrt{K_a.C_{\text{weak acid}}}\sqrt{{10}^{-9}0.1}=\sqrt{{10}^{-10}}={10}^{-5}$

$\therefore pH=-\log \left({10}^{-5}\right)=-(-5\log 10)=5$