Question

$pH$ of $0.005 M$ calcium acetate $\left( p K_{a}\right.$ of $CH _{3} COOH =4.74$ ) is :

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

Answer 1

The Correct Option is D

$0.005 \,M$ calcium acetate $(CH_3COO)_2Ca$ $\underset{0.005\,M}{(CH_3COO)_2Ca} \rightarrow Ca^{2+} + \underset{2\times 0.005 = 0.01)}{2CH_3COO^-}$ $\therefore [CH_3COO^-] = 0.01 \,M$ $CH_3COO^- + H_2O {<=>} CH_3COOH + \underset{\text{Alkaline}}{OH^-}$ $pH = 7 + \frac{pK_a}{2} + \frac{log \,C}{2}$ $= 7 + 2.37 + \frac{log\,0.01}{2}$ $= 7 + 2.37 - 1 = 8.37$