JEE Main 2018 · Chemistry · Ionic Equilibrium · medium

100 mL of 0.2 M NH3 (pKb=4.75) is mixed with 50 mL of 0.3 M HCl, and then 10 mL of 0.1 M NaOH is added. The pH of the final solution is approximately

Single correct option

100\ \mathrm{mL}

0.2\ M

\mathrm{NH_3}

(

pK_b=4.75

) is mixed with

50\ \mathrm{mL}

0.3\ M

\mathrm{HCl}

, and then

10\ \mathrm{mL}

0.1\ M

\mathrm{NaOH}

is added. The pH of the final solution is approximately

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Answer: (B) $8.88$

Moles:

\mathrm{NH_3}=20

mmol,

\mathrm{HCl}=15

mmol, giving

\mathrm{NH_3}=5

mmol and

\mathrm{NH_4^+}=15

mmol (a basic buffer). Adding

1

mmol

\mathrm{OH^-}

converts

1

mmol

\mathrm{NH_4^+}

\mathrm{NH_3}

\mathrm{NH_3}=6

\mathrm{NH_4^+}=14

mmol.

pOH=pK_b+\log\dfrac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]}=4.75+\log\dfrac{14}{6}=5.118

, so pH

=14-5.118=8.88

. Volume cancels in the ratio. The value

8.77

is the buffer before NaOH is added.

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake