100 mL of 0.2 M HA (K_a=1.0x10^-5) is mixed with 100 mL of 0.2 M HB (K_a=4.0x10^-5). Both are weak monobasic acids. Using [H+]=sqrtK_1c_1+K_2c_2, the pH of the mixture is nearest to (use log5=0.699)

JEE Advanced levelSingle correct optionDifficulty 3 of 3ie092

$100 mL$ of $0.2 M$ $HA$ ( $K_{a} = 1.0 \times 1 0^{- 5}$ ) is mixed with $100 mL$ of $0.2 M$ $HB$ ( $K_{a} = 4.0 \times 1 0^{- 5}$ ). Both are weak monobasic acids. Using $[H X^{+}] = K_{1} c_{1} + K_{2} c_{2}$ , the pH of the mixture is nearest to (use $lo g 5 = 0.699$ )

After mixing each acid is

0.1 M

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

, so

[H X^{+}] = 2.236 e - 03 M

and

pH = 2.65

. The pH of

HA

alone is 3.00, and averaging pH values is invalid.