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Ionic Equilibrium question

2025 · 7 Apr · Shift 2 · Q24
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Ionic Equilibrium question

2025 · 7 Apr · Shift 2 · Q24

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
One litre buffer solution was prepared by adding 0.10 mol each of NH3\mathrm{NH}_3NH3​ and NH4Cl\mathrm{NH}_4 \mathrm{Cl}NH4​Cl in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ‾\underline{\hspace{2cm}}​×10−2\times 10^{-2}×10−2. (Nearest integer) Given : pKb\mathrm{pK}_{\mathrm{b}}pKb​ of NH3=4.745\mathrm{NH}_3=4.745NH3​=4.745 and log⁡103=0.477\log _{10} 3=0.477log10​3=0.477
Numerical answer
View written solutionFree

Correct answer: 48

  1. Initial buffer composition

In 1 L1\text{ L}1 L solution:

  • NH3\mathrm{NH_3}NH3​ (base) =0.10 mol=0.10\text{ mol}=0.10 mol
  • NH4Cl\mathrm{NH_4Cl}NH4​Cl gives NH4+\mathrm{NH_4^+}NH4+​ (acid) =0.10 mol=0.10\text{ mol}=0.10 mol

This is a basic buffer (NH3/NH4+)\left(\mathrm{NH_3/NH_4^+}\right)(NH3​/NH4+​).

  1. Initial pH of the buffer

For a basic buffer:

pOH=pKb+log⁡[salt][base]\mathrm{pOH}=\mathrm{p}K_b+\log\frac{[\text{salt}]}{[\text{base}]}pOH=pKb​+log[base][salt]​

Since both are equal,

log⁡0.100.10=log⁡1=0\log\frac{0.10}{0.10}=\log 1=0log0.100.10​=log1=0

So,

pOH=4.745\mathrm{pOH}=4.745pOH=4.745

Hence,

pH=14−4.745=9.255\mathrm{pH}=14-4.745=9.255pH=14−4.745=9.255
  1. Effect of adding HCl

HCl\mathrm{HCl}HCl reacts completely with NH3\mathrm{NH_3}NH3​:

NH3+HCl→NH4++Cl−\mathrm{NH_3+HCl\rightarrow NH_4^++Cl^-}NH3​+HCl→NH4+​+Cl−

Added HCl =0.05 mol=0.05\text{ mol}=0.05 mol

So after reaction:

  • NH3\mathrm{NH_3}NH3​ left =0.10−0.05=0.05 mol=0.10-0.05=0.05\text{ mol}=0.10−0.05=0.05 mol
  • NH4+\mathrm{NH_4^+}NH4+​ formed =0.10+0.05=0.15 mol=0.10+0.05=0.15\text{ mol}=0.10+0.05=0.15 mol
  1. New pH after HCl addition

Again using buffer formula:

pOH=pKb+log⁡[salt][base]\mathrm{pOH}=\mathrm{p}K_b+\log\frac{[\text{salt}]}{[\text{base}]}pOH=pKb​+log[base][salt]​ pOH=4.745+log⁡0.150.05\mathrm{pOH}=4.745+\log\frac{0.15}{0.05}pOH=4.745+log0.050.15​ pOH=4.745+log⁡3\mathrm{pOH}=4.745+\log 3pOH=4.745+log3

Given log⁡3=0.477\log 3=0.477log3=0.477, so

pOH=4.745+0.477=5.222\mathrm{pOH}=4.745+0.477=5.222pOH=4.745+0.477=5.222

Thus,

pH=14−5.222=8.778\mathrm{pH}=14-5.222=8.778pH=14−5.222=8.778
  1. Change in pH
ΔpH=9.255−8.778=0.477\Delta \mathrm{pH}=9.255-8.778=0.477ΔpH=9.255−8.778=0.477

Given answer format is ‾×10−2\underline{\hspace{2cm}}\times 10^{-2}​×10−2.

So,

0.477=47.7×10−20.477=47.7\times 10^{-2}0.477=47.7×10−2

Nearest integer =48=48=48.

  1. Final answer
48\boxed{48}48​

The derived answer matches the stored correct answer.

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