A) \[{{K}_{h}}=\frac{{{K}_{\omega }}}{{{K}_{b}}}\]
B) \[[{{H}^{+}}]=\sqrt{\frac{{{K}_{\omega }}\times C}{{{K}_{b}}}}\]
C) \[h=\sqrt{\frac{{{K}_{\omega }}}{{{K}_{b}}\times C}}\]
D) \[pH=\frac{1}{2}{{\,}_{p}}{{K}_{a}}(N{{H}_{4}}OH)\]
Correct Answer: D
Solution :
\[pH=\frac{1}{2}{{\,}_{p}}{{K}_{\omega }}-\frac{1}{2}\,{{\,}_{b}}{{K}_{\omega }}-\frac{1}{2}\log \,C.\]
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