A) \[{{K}_{b}}={{K}_{w}}\]
B) \[{{K}_{b}}\times {{K}_{b}}={{K}_{w}}\]
C) \[\frac{{{K}_{a}}}{{{K}_{b}}}={{K}_{w}}\]
D) \[{{K}_{b}}=\frac{1}{{{K}_{w}}}\]
Correct Answer: B
Solution :
For the reaction \[HF+{{H}_{2}}O\xrightarrow{{{K}_{a}}}\,\,{{H}_{3}}{{O}^{+}}{{F}^{-}}\] The equilibrium constant expression is \[{{K}_{a}}=\frac{[{{H}_{3}}{{O}^{+}}]\,\,[{{F}^{-}}]}{[HF]}\] ?. (i) For the reaction \[{{F}^{-}}+{{H}_{2}}OHF+O{{H}^{-}}\] The equilibrium constant expression is \[{{K}_{b}}=\frac{[HF]\,\,[O{{H}^{-}}]}{[{{F}^{-}}]}\] ?. (ii) Multiply the Eqs. (i) and (ii), we get \[{{K}_{a}}\times {{K}_{b}}=[{{H}_{3}}{{O}^{+}}]\,\,[O{{H}^{-}}]\] \[{{K}_{a}}\times {{K}_{b}}={{K}_{w}}\] \[(\because \,{{K}_{w}}=[{{H}_{3}}{{O}^{+}}]\,\,[O{{H}^{-}}])\]
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