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 |
\[=7-\frac{1}{2}p{{K}_{b}}-\frac{1}{2}\log C\] |
\[=7-\frac{1}{2}\times 5-\frac{1}{2}\times \log (0.1)\] |
\[=7-2.5+0.5=5\] |
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