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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Chemistry
Chapter
Ionic Equilibrium

BITSAT 2022 · Chemistry · Ionic Equilibrium

What is [NH_4^ + ] in a solution that is 0.02 M NH_30.01 M KOH ?...

What is [NH4+]\left[\mathrm{NH}_4^{+}\right] in a solution that is $0.02 \mathrm{M} \mathrm{NH}_30.01 \mathrm{M} \mathrm{KOH} ? \left[K_b \left(\mathrm{NH}_{3}\right)=1.8 \times10^{-5}\right]

Options

  1. A3.6×10−5M3.6 \times 10^{-5} \mathrm{M}
  2. B1.8×10−5M1.8 \times 10^{-5} \mathrm{M}
  3. C0.9×10−5M0.9 \times 10^{-5} \mathrm{M}
  4. D7.2×10−5M7.2 \times 10^{-5} \mathrm{M}

Correct answer

A. 3.6×10−5M3.6 \times 10^{-5} \mathrm{M}

Step-by-step solution

NH4OH⇌NH4++OH−Kb=[NH4+][OH−][NH4OH]1.8×10−5=[NH4+][0.01][0.02]∴[NH4+]=3.6×10−5M\mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-}K_b=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}1.8 \times 10^{-5}=\frac{\left[\mathrm{NH}_4^{+}\right][0.01]}{[0.02]}\therefore \quad\left[\mathrm{NH}_4^{+}\right]=3.6 \times 10^{-5} \mathrm{M}
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