A) 8
B) 10
C) 4
D) 5
Correct Answer: C
Solution :
[c] Hydrolysis of a salt is reverse reaction of acid-base neutralization reaction. \[\therefore {{K}_{h}}=\frac{{{K}_{w}}}{{{K}_{a}}}=\frac{{{10}^{-14}}}{{{10}^{-7}}}={{10}^{-7}}\] \[[O{{H}^{-}}]=ch=c\times \sqrt{\frac{{{K}_{h}}}{c}}=\sqrt{c\times {{K}_{h}}}=\sqrt{{{10}^{-8}}}={{10}^{-4}}\] \[\Rightarrow pO{{H}^{-}}=-log\left[ O{{H}^{-}} \right]=-\log [{{10}^{-4}}]=4\]
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