JEE MAIN - Chemistry (2022 - 26th July Evening Shift - No. 3)
Class XII students were asked to prepare one litre of buffer solution of $$\mathrm{pH} \,8.26$$ by their Chemistry teacher: The amount of ammonium chloride to be dissolved by the student in $$0.2\, \mathrm{M}$$ ammonia solution to make one litre of the buffer is :
(Given: $$\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=4.74$$
Molar mass of $$\mathrm{NH}_{3}=17 \mathrm{~g} \mathrm{~mol}^{-1}$$
Molar mass of $$\mathrm{NH}_{4} \mathrm{Cl}=53.5 \mathrm{~g} \mathrm{~mol}^{-1}$$ )
53.5 g
72.3 g
107.0 g
126.0 g
Explanation
For basic Buffer, $$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { salt }]}{[\text { Base }]}$$
$$\mathrm{pOH}=14-8.26=5.74$$
$$5.74=4.74+\log \frac{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}{0.2}$$
$$\left[\mathrm{NH}_{4} \mathrm{Cl}\right]=2 \mathrm{M}$$
Moles of $$\mathrm{NH}_{4} \mathrm{Cl}=2 \times 1=2$$ moles
Weight of $$\mathrm{NH}_{4} \mathrm{Cl}=2 \times 53.5=107 \mathrm{~g}$$
$$\mathrm{pOH}=14-8.26=5.74$$
$$5.74=4.74+\log \frac{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}{0.2}$$
$$\left[\mathrm{NH}_{4} \mathrm{Cl}\right]=2 \mathrm{M}$$
Moles of $$\mathrm{NH}_{4} \mathrm{Cl}=2 \times 1=2$$ moles
Weight of $$\mathrm{NH}_{4} \mathrm{Cl}=2 \times 53.5=107 \mathrm{~g}$$
Comments (0)
