Q. $pK_{a}$ of acetic acid is 4.74. The concentration of $CH_{3}COONa\ is\ 0.01\ M$. The pH of $CH_{3}COONa$ is
A) 3.37
B) 4.37
C) 4.74
D) 0.474
Explanation
(a)
$\left\lbrack H^{+} \right\rbrack = C\alpha = \sqrt{K_{a}.C}$
$pH = - \log\left( K_{a}.C \right)^{\frac{1}{2}}$
$= \frac{1}{2}\lbrack - \log{K_{a} - \log{C\rbrack}}$
$= \frac{1}{2}\lbrack 4.74 - \log{10^{- 2}\rbrack}$
$= \frac{1}{2}\lbrack 4.74 + 2\rbrack = 3.37$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.