JEE Main 2023 · Chemistry · Ionic Equilibrium · medium

100 mL of 0.10 M NH3 is mixed with 40 mL of 0.10 M HCl. Given Kb(NH3)=1.8×10^-5 (log 1.8=0.255, log 1.5=0.176), the pH of the resulting solution is

Single correct option

100\ \mathrm{mL}

0.10\ \mathrm M

\mathrm{NH_3}

is mixed with

40\ \mathrm{mL}

0.10\ \mathrm M

\mathrm{HCl}

. Given

K_b(\mathrm{NH_3})=1.8\times10^{-5}

(

\log1.8=0.255

\log1.5=0.176

), the pH of the resulting solution is

Show answer and solution

Answer: (A) $9.43$

Moles:

\mathrm{NH_3}=10

mmol,

\mathrm{HCl}=4

mmol. After reaction:

\mathrm{NH_3}=6

mmol,

\mathrm{NH_4^+}=4

mmol (a basic buffer).

\mathrm{pOH}=\mathrm{p}K_b+\log\frac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]}=4.745+\log\frac46=4.745-0.176=4.569

\mathrm{pH}=14-4.569=9.43

. Option (C) is the pOH, not the pH.

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