The $K_b$ of $NH_4OH$ is $1.8 \times 10^{-5}$. Calculate the $pH$ of a solution containing $0.15 \ M$ $NH_4OH$ and $0.25 \ M$ $NH_4Cl$.
(A) For a basic buffer solution,the $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given: $K_b = 1.8 \times 10^{-5}$,$[Base] = 0.15 \ M$,$[Salt] = 0.25 \ M$.
$pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
$pOH = 4.745 + \log \frac{0.25}{0.15} = 4.745 + \log(1.667) = 4.745 + 0.222 = 4.967$.
Since $pH + pOH = 14$,$pH = 14 - 4.967 = 9.033$.
Given: $K_b = 1.8 \times 10^{-5}$,$[Base] = 0.15 \ M$,$[Salt] = 0.25 \ M$.
$pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
$pOH = 4.745 + \log \frac{0.25}{0.15} = 4.745 + \log(1.667) = 4.745 + 0.222 = 4.967$.
Since $pH + pOH = 14$,$pH = 14 - 4.967 = 9.033$.
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