Question

$pH$ of $0.005M$ calcium acetate $(p{K}_a$ of $C{H}_{3}COOH=4.74$ ) is :

• A. 7.04
• B. 9.37
• C. 9.26
• D. 8.37

Answer 1

Answer: (D)

$0.005M$ calcium acetate $(C{H}_{3}COO{)}_{2}Ca$

$\underset{0.005M}{(C{H}_{3}COO{)}_{2}Ca}\to C{a}^{2+}+\underset{2\times 0.005=0.01)}{2C{H}_{3}CO{O}^{-}}$

$\therefore [C{H}_{3}CO{O}^{-}]=0.01M$

$C{H}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons C{H}_{3}COOH+\underset{\text{Alkaline}}{O{H}^{-}}$

$pH=7+\frac{p{K}_a}{2}+\frac{logC}{2}$

$=7+2.37+\frac{log0.01}{2}$

$=7+2.37-1=8.37$