Given that the mixture contains a weak base ($\text{NH}_4\text{OH}$) and its salt with a strong acid ($\text{NH}_4\text{Cl}$), it acts as a basic buffer. According to Henderson-Hasselbalch equation for a basic buffer: $\text{pOH} = pK_b + \log \frac{[\text{Salt}]}{[\text{Base}]}$ We are given: $\text{pH} = 9.25$ $[\text{Base}] = 0.1 \text{ N} = 0.1 \text{ M}$ (since valency factor is 1) $[\text{Salt}] = 0.1 \text{ N} = 0.1 \text{ M}$

We know that $\text{pH} + \text{pOH} = 14$ at $298 \text{ K}$. $\text{pOH} = 14 - 9.25 = 4.75$

Now, substituting the values into the buffer equation: $4.75 = pK_b + \log \left(\frac{0.1}{0.1}\right)$ $4.75 = pK_b + \log(1)$ Since $\log(1) = 0$, we get: $pK_b = 4.75$

Thus, the $pK_b$ of $\text{NH}_4\text{OH}$ is $4.75$.