A) 9.43
B) 11.72
C) 8.73
D) 9.08
Correct Answer: A
Solution :
\[pOH=p{{K}_{b}}+\log \frac{[Salt]}{[Base]}\] \[=-\log {{K}_{b}}+\log \frac{[Salt]}{[Base]}\] \[=-\log 1.8\times {{10}^{-5}}+\log \frac{0.20}{0.30}\] \[=5-0.25+(-0.176)\] \[=4.75-0.176=4.57\] \[\therefore \]\[pH=14-4.57=9.43\]You need to login to perform this action.
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