JEE Advanced 2022 — Paper 1 — CHEM-I solution

JEE Advanced2022Paper 1CHEM-I

A solution is prepared by mixing 0.01 mol each of H $_{2}$ CO $_{3}$ , NaHCO $_{3}$ , Na $_{2}$ CO $_{3}$ , and NaOH in 100 mL of water. pH of the resulting solution is _______.

[Given: $p K_{a_{1}}$ and $p K_{a_{2}}$ of H $_{2}$ CO $_{3}$ are 6.37 and 10.32, respectively; log2 = 0.30]

Your answer

Solution

NaOH (strong base) first neutralizes the strongest acid present, H $_{2}$ CO $_{3}$ :

$H_{2} CO_{3} + NaOH \to NaHCO_{3} + H_{2} O$

Starting moles: 0.01 each. After this step, H $_{2}$ CO $_{3}$ and NaOH are fully consumed and 0.01 mol additional NaHCO $_{3}$ is formed.

Resulting mixture: NaHCO $_{3}$ : 0.01 + 0.01 = 0.02 mol; Na $_{2}$ CO $_{3}$ : 0.01 mol. This is a buffer of the conjugate acid HCO $_{3}^{-}$ and its conjugate base CO $_{3}^{2 -}$ , governed by $p K_{a_{2}}$ .

Henderson–Hasselbalch:

$pH = p K_{a_{2}} + lo g \frac{[ CO _{3}^{2 -} ]}{[ HCO _{3}^{-} ]} = 10.32 + lo g \frac{0.01}{0.02} = 10.32 - lo g 2 = 10.32 - 0.30 = 10.02$ .

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