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  4. In a basic buffer 0.0025 mole of NH 4 Cl and 0.15 mole of NH 4 OH are present. The pH of the solution will be: ( p Kb=4.74)

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Q. In a basic buffer $0.0025$ mole of $NH _{4} Cl$ and $0.15$ mole of $NH _{4} OH$ are present. The $pH$ of the solution will be : $\left( p K_{b}=4.74\right)$

Rajasthan PMTRajasthan PMT 2004

Solution:

From Handerson's equation
$pOH =p K_{b}+\log \frac{[\text{salt} ]}{[\text{base}]}$
$=4.74+\log \frac{0.0025 / V}{0.15 / V}$
$=4.74+\log \frac{1}{60}$
$=4.74-1.78=2.96$
$\therefore pH =14-2.96=11.04$