Question

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} )

# 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

Solution