Q. pH of 0.005 M calcium acetate is ($pK_{a}of\ CH_{3}COOH = 4.74$)
A) 7.04
B) 9.37
C) 9.26
D) 8.2195
Explanation
(d)
$pH = \frac{1}{2}pK_{w} + \frac{1}{2}\log{C + \frac{1}{2}pK_{a}}$
$= \frac{14}{2} + \frac{1}{2}\log{\left( 5 \times 10^{- 3} \right) + \frac{1}{2} \times 4.74}$
$\ = 7 + \frac{1}{2}\lbrack 0.6990 - 3\rbrack + 2.3$
$pH = 8.2195$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.