Question

The pH of 0.10 M solution of an acid is 6. What is the acid dissociation constant of the acid?

10-11
10-12
10-5
10-6

Answer 1

Answer: Dissociation constant of the acid is
`10^(-11)`.

Step-by-step explanation: Assuming the acid to be monoprotic, the reaction follows:

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

pH of the solution = 6

and we know that

`pH=-log([H^+])`

`[H^+]=antilog(-pH)`

`[H^+]=antilog(-6)=10^(-6)M`

As HA ionizes into its ions in 1 : 1 ratio, hence

`[H^+]=[A^-]=10^(-6)M`

As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:

`[HA]=[HA]-[H^+]`

`[HA]=0.1M-10^(-6)M`

`[HA]=0.09999M`

Dissociation Constant of acid,
`K_a` is given as:

`K_a=([A^-][H^+])/(HA)`

Putting values of
`[H^+],[A^-]\text{ and }[HA]` in the above equation, we get

`K_a=((10^(-6)M).(10^(-6)M))/(0.09999M)`

`K_a=1.0001* 10^(-11)M`

Rounding it of to one significant figure, we get

`K_a=1.0* 10^(-11)M\approx 10^(-11)M`