A) \[{{K}_{b}}=K\omega \]
B) \[{{K}_{b}}=1/K\omega \]
C) \[{{K}_{a}}\times {{K}_{b}}={{K}_{\omega }}\]
D) \[Ka/{{K}_{b}}=K\omega \]
Correct Answer: C
Solution :
\[HF+{{H}_{2}}O\xrightarrow{{{K}_{a}}}{{H}_{3}}{{O}^{+}}+{{F}^{-}}\] \[{{K}_{a}}=\frac{[{{H}_{3}}{{O}^{+}}]\times [{{F}^{-}}]}{[HF]\times [{{H}_{2}}O]}\] ??(1) \[{{F}^{-}}+{{H}_{2}}O\xrightarrow{kb}HF+O{{H}^{-}}\] \[{{K}_{b}}=\frac{[HF]\times [O{{H}^{-}}]}{[{{F}^{-}}][{{H}_{2}}O]}\] Multiply of equations (i) and (ii). \[{{K}_{a}}\times {{K}_{b}}=\frac{[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]}{{{[{{H}_{2}}O]}^{2}}}={{K}_{\omega }}\] \[{{K}_{a}}\times {{K}_{b}}={{K}_{\omega }}\]
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