JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Equilibrium

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}

H_{2} CO_{3}

are 6.37 and 10.32, respectively ;

lo g 2 = 0.30

]

Answer: 10.02

Solution

First,

H_{2} CO_{3}

reacts with NaOH:

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

Initially: 10 mmol

H_{2} CO_{3}

, 10 mmol NaOH, 10 mmol

NaHCO_{3}

, 10 mmol

Na_{2} CO_{3}

.

After neutralization: 0 mmol

H_{2} CO_{3}

, 0 mmol NaOH, 20 mmol

NaHCO_{3}

, 10 mmol

Na_{2} CO_{3}

.

This forms a buffer of

NaHCO_{3}

(acid) and

Na_{2} CO_{3}

(salt).

Using the Henderson-Hasselbalch equation:

pH = p K_{a 2} + lo g \frac{[ Na _{2} CO _{3} ]}{[ NaHCO _{3} ]} = 10.32 + lo g \frac{10}{20} = 10.32 - 0.30 = 10.02

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer