A buffer solution contains 0.1 M NH4OHKb=1.8×10−5M and 0.2 M NH4Cl.The pH of this buffer solution at 25∘C is

A buffer solution contains $0.1 M {NH}_{4} OH (K_{b} = 1.8 \times 10^{- 5} M)$ and $0.2 M {NH}_{4} Cl$ .The $pH$ of this buffer solution at $25^{\circ} C$ is

For alkaline buffer solution

$pOH = p K_{b}^{\circ} + \log \frac{[salt]}{[base]} = - \log (1.8 \times 10^{- 5}) + \log (\frac{0.2 M}{0.1 M}) = 4.74 + 0.30 = 5.04$

$pH = 14 - 5.04 = 8.96$

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