A) 4
B) 5
C) 9
D) 11
Correct Answer: B
Solution :
\[pOH=-\log {{K}_{b}}+{{\log }_{10}}\frac{N{{H}_{4}}Cl}{N{{H}_{4}}OH}\] \[=-\log {{10}^{-10}}+{{\log }_{10}}\left[ \frac{1}{10} \right]=10-1=9\] \[pH=14-pOH=5\]You need to login to perform this action.
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