A) 9.25
B) 4.75
C) 3.75
D) 8.25
Correct Answer: B
Solution :
\[pOH=p{{K}_{b}}+\log \frac{[salt]}{[base]}\] \[\because \] \[[N{{H}_{4}}Cl]=[N{{H}_{4}}OH]\] \[\therefore \] \[pOH=p{{K}_{b}}\] \[p{{K}_{b}}=pOH=14-pH\] \[=14-9.25=4.75\]You need to login to perform this action.
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