A) 11.04
B) 10.24
C) 6.62
D) 5.48
Correct Answer: A
Solution :
From Henderson's equation, \[pOH=p{{K}_{b}}+\log \frac{[salt]}{[base]}\] \[=4.74+\log \frac{0.0025/V}{0.15/V}\] \[=4.74+\log \frac{1}{60}\] \[=4.74-1.78=2.96\] \[pH=14-2.96\] \[=11.04\]You need to login to perform this action.
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