Question bank/JEE Main/Chemistry

The pH of a solution obtained by mixing 100 mL of 0.2 M
CH_3COOH with 100 mL of 0.2 M NaOH would be: (pK_a for CH_3COOH =
4.74)

A4.74B8.87C9.10D8.57

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CH_3COOH + OH^- [IMAGE] CH3COO^- + H_2O
t = 0 20 20
t = eq - - 20
So, [CH_3 COO^-] $= \frac{20}{200}$ = 0.1M
pH = 7 +
$\frac{1}{2}{pK}_{a} + \frac{1}{2}logC = 7 + 2.37 + \frac{1}{2}$
log10^-1 = 7 + 2.37 - 0.5 = 8.87

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx