Question

Weak acid react with strong base ionisation constant of weak acid is 10 power minus 4 find out equilibrium constant for this reaction

Answer 1

Answer: The equilibrium constant will be $1\times 10^{10}$.

Explanation: Dissociation of weak acid is given as:

$HA\rightleftharpoons H^++A^-$

$K_a=\frac{[H^+][A^-]}{[HA]}=10^{-4}$          .....(1)

Dissociation of water can be written as:

$H_2O\rightleftharpoons OH^-+H^+$

$K_w=[H^+][OH^-]$            .....(2)

$K_w=10^{-14}$  (Constant)

Reaction of strong base (BOH) with weak acid can be written as:

$HA+OH^-\rightleftharpoons H_2O+A^-$

Dissociation constant can be written as:

$K=\frac{[A^-][H_2O]}{[HA][OH^-]}$           .....(3)

Combining equations 1,2 and 3, we get

$K=\frac{K_a}{K_w}$

Putting values, we get

$K=\frac{10^{-4}}{10^{-14}}$

$K=1\times 10^{10}$

Answer 2

Answer:

10 to the power 10 is the right answer

Explanation:

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