Answer:
\[{{N}_{2}}{{H}_{4}}+{{H}_{2}}O\rightleftharpoons
{{N}_{2}}H_{5}^{+}+O{{H}^{-}}\]
\[pOH=14-9.7=4.3\]
\[[O{{H}^{-}}]=Antilog(-4.3)=5.01\times
{{10}^{-5}}M\]
\[[O{{H}^{-}}]=\sqrt{C{{K}_{b}}}\]
\[5.01\times
{{10}^{-5}}=\sqrt{0.004\times {{K}_{b}}}\]
\[{{K}_{b}}=6.27\times
{{10}^{-7}}\]
\[p{{K}_{b}}=-\log (6.27\times {{10}^{-7}})=6.2\]
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