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
 

Answers (1)

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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