A) \[pH=p{{K}_{b}}\]
B) \[pH=1/2p{{K}_{b}}-1/2\log \] [salt] /[base]
C) \[pH=14-p{{K}_{b}}-\log \] [salt]/[base]
D) \[pH=pOH-p{{K}_{b}}+\log \] [salt]/ [base]
Correct Answer: C
Solution :
For basic buffer, \[pOH=p{{K}_{b}}+\log \,\frac{[salt]}{[base]}\] \[poH=14-pH\] or \[pH=14-pOH\] \[\therefore \] \[pH=14-p{{K}_{b}}-\log \frac{[salt]}{[base]}\]
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