JEE Main · 2022 · Shift-IImediumIEQ-075

Class XII students were asked to prepare one litre of buffer solution of pH 8.26 by their chemistry teacher. The amount…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

Class XII students were asked to prepare one litre of buffer solution of pH 8.26 by their chemistry teacher. The amount of ammonium chloride to be dissolved by the student in 0.2 M ammonia solution to make one litre of the buffer is:

[pKb(NH3)=4.74; M(NH3)=17 g mol−1; M(NH4Cl)=53.5 g mol−1][\mathrm{pK_b(NH_3)} = 4.74;\ M(\mathrm{NH_3}) = 17\ \mathrm{g\ mol^{-1}};\ M(\mathrm{NH_4Cl}) = 53.5\ \mathrm{g\ mol^{-1}}]

Options
  1. a

    53.5 g

  2. b

    72.3 g

  3. c

    107 g

    ✓
  4. d

    126 g

Correct Answerc

107 g

Detailed Solution

Step 1 — Find pOH: pOH=14−8.26=5.74\mathrm{pOH} = 14 - 8.26 = 5.74

Step 2 — Henderson-Hasselbalch for base:

pOH=pKb+log⁡[NH4+][NH3]5.74=4.74+log⁡[NH4Cl]0.2log⁡[NH4Cl]0.2=1.00[NH4Cl]0.2=10[NH4Cl]=2.0 M\mathrm{pOH} = \mathrm{pK_b} + \log\frac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} 5.74 = 4.74 + \log\frac{[\mathrm{NH_4Cl}]}{0.2} \log\frac{[\mathrm{NH_4Cl}]}{0.2} = 1.00 \frac{[\mathrm{NH_4Cl}]}{0.2} = 10 [\mathrm{NH_4Cl}] = 2.0\ \mathrm{M}

Step 3 — Mass of NH4Cl\mathrm{NH_4Cl} in 1 L: m=2.0 mol×53.5 g mol−1=107 gm = 2.0\ \mathrm{mol} \times 53.5\ \mathrm{g\ mol^{-1}} = 107\ \mathrm{g}

Answer: Option (3) — 107 g

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