20 mL of $0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ is added to 30 mL of $0.2 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}$ solution. The pH of the resultant mixture is [$\mathrm{pk}_{\mathrm{b}}$ of $\mathrm{NH}_{4} \mathrm{OH}=4.7$]

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Equilibrium

20 mL of

0.1 M H_{2} SO_{4}

is added to 30 mL of

0.2 M NH_{4} OH

solution. The pH of the resultant mixture is [

pk_{b}

NH_{4} OH = 4.7

]

Answer: (B)

Count the millimoles carefully.

20

mL of

0.1

H X_{2} SO X_{4}

supplies

0.1 \times 20 = 2

mmol of acid, and being diprotic that is

4

mmol of

H X^{+}

;

30

mL of

0.2

NH X_{4} OH

gives

6

mmol of base. Neutralisation consumes

4

mmol of the base to form

4

mmol of

NH X_{4} X^{+}

, leaving

6 - 4 = 2

mmol of

NH X_{4} OH

. That leftover weak base with its conjugate acid is a buffer, so

pO H = p K_{b} + lo g \frac{[ NH X _{4} X ^{+} ]}{[ NH X _{4} OH ]} = 4.7 + lo g \frac{4}{2} = 4.7 + 0.3 = 5.0

, and

p H = 14 - 5.0 = 9.0

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option