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  4. If the equilibrium constant of the reaction of weak acid HA with strong base is 109, then pH of 0.1 M NaA is

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Q. If the equilibrium constant of the reaction of weak acid HA with strong base is $ {{10}^{9}}, $ then pH of 0.1 M NaA is

MGIMS WardhaMGIMS Wardha 2013

Solution:

$ HA+O{{H}^{-}}{{H}_{2}}O+{{A}^{-}} $ $ K=\frac{[{{A}^{-}}]}{[HA]O{{H}^{-}}]} $ $ HA{{H}^{+}}+{{A}^{-}} $ $ {{K}_{a}}=\frac{[{{H}^{+}}][{{A}^{-}}]}{[HA]} $ $ \frac{{{K}_{a}}}{K}=[{{H}^{+}}][O{{H}^{-}}]={{K}_{w}} $ Or $ {{K}_{a}}={{K}_{2}}K={{10}^{-14}}\times {{10}^{9}}={{10}^{-5}} $ $ p{{K}_{a}}=5 $ $ {{A}^{-}} $ solution is alkaline due to hydrolysis $ \therefore $ $ pH=7+\frac{p{{k}_{a}}}{2}+\frac{\log C}{2}=2+\frac{5}{2}+\frac{\log 0.1}{2}=9 $