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  4. pH of 0.005 M calcium acetate ( p Ka. of CH 3 COOH =4.74 ) is :

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

AIEEEAIEEE 2002

Solution:

$0.005 \,M$ calcium acetate $(CH_3COO)_2Ca$

$\underset{0.005\,M}{(CH_3COO)_2Ca} \rightarrow Ca^{2+} + \underset{2\times 0.005 = 0.01)}{2CH_3COO^-}$

$\therefore [CH_3COO^-] = 0.01 \,M$

$CH_3COO^- + H_2O \ce{<=>} CH_3COOH + \underset{\text{Alkaline}}{OH^-}$

$pH = 7 + \frac{pK_a}{2} + \frac{log \,C}{2}$

$= 7 + 2.37 + \frac{log\,0.01}{2}$

$= 7 + 2.37 - 1 = 8.37$