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Solution of 0.1 N $NH_4OH$ and 0.1 N $NH_4Cl$ has pH 9.25. Then find out $pK_b$ of $NH_4OH$ : -
A
9.25
B
4.75
C
3.75
D
8.25
Detailed Solution
This is a basic buffer. Henderson equation: $pOH = pK_b + \log\dfrac{[salt]}{[base]}$
Here [salt] = [base] = 0.1, so $\log 1 = 0$ and $pOH = pK_b$.
$pOH = 14 - pH = 14 - 9.25 = 4.75$
So $pK_b = 4.75$.
Here [salt] = [base] = 0.1, so $\log 1 = 0$ and $pOH = pK_b$.
$pOH = 14 - pH = 14 - 9.25 = 4.75$
So $pK_b = 4.75$.