# 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl.  If pKb of ammonia solution is 4.75, the pH of the mixture will be : Option: 1 3.75 Option: 2 4.75 Option: 3 8.25 Option: 4 9.25

Initially, before reaction

mmol of $\mathrm{NH_3}$ = 10

mmol of $\mathrm{HCl}$ = 5

Therefore, after reaction,

mmol of $\mathrm{NH_3}$ = 5

mmol of $\mathrm{NH_4Cl}$ = 5

Now, putting the values of concentrations in the equation

$\mathrm{pOH = pK_b + \log \frac{[Salt]}{[Base]}}$

$\Rightarrow \mathrm{pOH = pK_b \ \ ( \because [Salt]=[Base])}$$\Rightarrow \mathrm{pOH = 4.75}$

Thus, we have

$\Rightarrow \mathrm{pH =14-pOH =14- 4.75=9.25}$

Hence, the correct answer is Option (4)

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