Test Index

Ionic Equilibrium

Section: Chemistry

Question : 2 of 14

Marks: +1, -0

When 30 mL of

0.2\ \mathrm{M}\ \mathrm{NH}_4\mathrm{OH}

2\ \mathrm{M}\ \mathrm{NH}_4\mathrm{Cl}

pK_b

\mathrm{NH}_4\mathrm{OH}

Solution:

pOH of buffer solution

\mathrm{pOH}=14-\mathrm{pH}\Rightarrow \mathrm{pH}=14-8.2=5.8

Moles of

\mathrm{NH}_4\mathrm{OH}=0.2\times 0.030

=0.006\ \mathrm{mol}

Moles of

\mathrm{NH}_4\mathrm{Cl}=2\times 0.030=0.060\ \mathrm{mol}

Total volume

=30\ \mathrm{mL}+30\ \mathrm{mL}=0.060\ \mathrm{L}

Concentration of conjugate base and acid in buffer

[\mathrm{NH}_4\mathrm{OH}]=\frac{0.006\ \mathrm{mol}}{0.060\ \mathrm{L}}=0.1\ \mathrm{M}

[\mathrm{NH}_4\mathrm{Cl}]=\frac{0.060\ \mathrm{mol}}{0.060\ \mathrm{L}}=1\ \mathrm{M}

Using Henderson equation

\mathrm{pOH}=pK_b+\log\frac{[\mathrm{NH}_4\mathrm{Cl}]}{[\mathrm{NH}_4\mathrm{OH}]}

5.8=pK_b+\log\left(\frac{1}{01}\right)

pK_b=4.8

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