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JEE Advanced levelSingle correct optionDifficulty 3 of 3ie047

A solution contains $0.1 M$ HA ( $K_{a} = 1 \times 1 0^{- 5}$ ) and $0.1 M$ HB ( $K_{a} = 4 \times 1 0^{- 5}$ ), both weak monobasic acids. Neglecting water ionisation and the small ionised fractions, the pH is ( $lo g 5 = 0.699$ )

A $3.00$
B $2.30$
C $2.65$
D $3.35$

Answer: C

Each acid contributes

H X^{+}

:

[H X^{+}]^{2} = K_{H A} c_{H A} + K_{H B} c_{H B} = 1 0^{- 6} + 4 \times 1 0^{- 6} = 5 \times 1 0^{- 6}

. So

pH = \frac{1}{2} (6 - 0.699) = 2.65

.

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