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JEE Advanced levelSingle correct optionDifficulty 3 of 3ie044

$x mL$ of $HCl$ of pH $1$ is added to $100 mL$ of $NaOH$ of pH $13$ (298 K), and the final pH is $12$ . Volumes are additive. The value of $x$ is

A $50.0 mL$
B $90.0 mL$
C $100.0 mL$
D $81.8 mL$

Answer: D

n (OH X^{-}) = 0.1 \times 100 = 10 mmol

and

n (H X^{+}) = 0.1 x

. Final pOH

= 2

, so

(10 - 0.1 x) / (100 + x) = 0.01

, i.e.

9 = 0.11 x

and

x = 81.8 mL

.

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