question_answer

500 ml of 0.2 M BOH (weak base) is mixed with 500 ml of 0.1 M HCl. Resulting buffer has pH = 9. What will be the pH of 0.1 M BCl solution -

A) 1

B) 3    

C) 5

D) 9

Correct Answer: C

Solution :

\[pOH=p{{K}_{b}}+\log \frac{[salt]}{[base]}\]

\[14-9=p{{K}_{b}}+\log \frac{(50/1000ml)}{[50/1000ml]}\]

\[p{{K}_{b}}=5\]

Now pH of BCl solution

\[=7-\frac{1}{2}p{{K}_{b}}-\frac{1}{2}\log C\]

\[=7-\frac{1}{2}\times 5-\frac{1}{2}\times \log (0.1)\]

\[=7-2.5+0.5=5\]