A) \[1.8\times {{10}^{-5}}\]
B) \[1.8\times {{10}^{-10}}\]
C) \[5.55\times {{10}^{-5}}\]
D) \[5.55\times {{10}^{-10}}\]
Correct Answer: D
Solution :
Key Idea: \[{{K}_{h}}=\frac{{{K}_{w}}}{{{K}_{b}}}\] where \[{{K}_{w}}=\] ionic product of water \[=1\times {{10}^{-14}}\] \[{{K}_{b}}=\] degree of dissociation of \[N{{H}_{4}}OH\] \[{{K}_{h}}=\frac{1\times {{10}^{-14}}}{1.8\times {{10}^{-5}}}\] \[=0.555\times {{10}^{-9}}\] \[=5.55\times {{10}^{-10}}\]
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