A weak acid react with strong base,...
Question

# A weak acid react with strong base, ionisation constant of weak acid is 10-4. Find out equilibrium constant for this reaction :- (2) 1010 (3) 10-9 (4) 109 (1) 10-10

NEET/Medical Exams
Chemistry
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Answer: The equilibrium constant will be ( 1 times 10^{10} ) Explanation: Dissociation of weak acid is given as: ( H A rightleftharpoons H^{+}+A^{-} ) ( K_{a}=frac{left[H^{+}right]left[A^{-}right]}{[H A]}=10^{-4} ) Dissociation of water can be written as: ( H_{2} O rightleftharpoons O H^{-}+H^{+} ) ( K_{w}=left[H^{+}right]left[O H^{-}right] ) ( K_{w}=10^{-14} ) (Constant) Reaction of strong base (BOH) with weak acid can be written a ( H A+O H^{-} rightleftharpoons H_{2} O+A^{-} ) Dissociation constant can be written as: ( left.K=frac{left[A^{-}right]left[H_{2} Oright]}{[H A][O H]}right] ) Combining equations 1,2 and ( 3, ) we get ( K=frac{K_{0}}{K_{5}} ) Putting values, we get ( K=frac{10^{-4}}{10^{-14}} ) ( K=1 times 10^{10} )