Ionic EquilibriumMedium
Question
The pH at the equivalence point when a solution of 0.01 M-CH3COOH is titrated with a solution of 0.01 M-NaOH, is (pKa of CH3COOH = 4.7, log 5 = 0.7)
Options
A.8.2
B.9.4
C.8.35
D.10.5
Solution
Equal volumes of both will consume and hence, $\left\lbrack CH_{3}COONa \right\rbrack = \frac{0.01}{2} = 0.005$
Now, $P^{H} = 7 + \frac{1}{2}\left( P^{K_{a}} + \log C \right) = 7 + \frac{1}{2}\left( 4.7 + \log 0.005 \right) = 8.2$
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