Answer: Dissociation constant of the acid is [tex]10^{-11}[/tex].
Explanation: Assuming the acid to be monoprotic, the reaction follows:
[tex]HA\rightleftharpoons H^++A^-[/tex]
pH of the solution = 6
and we know that
[tex]pH=-log([H^+])[/tex]
[tex][H^+]=antilog(-pH)[/tex]
[tex][H^+]=antilog(-6)=10^{-6}M[/tex]
As HA ionizes into its ions in 1 : 1 ratio, hence
[tex][H^+]=[A^-]=10^{-6}M[/tex]
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:
[tex][HA]=[HA]-[H^+][/tex]
[tex][HA]=0.1M-10^{-6}M[/tex]
[tex][HA]=0.09999M[/tex]
Dissociation Constant of acid, [tex]K_a[/tex] is given as:
[tex]K_a=\frac{[A^-][H^+]}{HA}[/tex]
Putting values of [tex][H^+],[A^-]\text{ and }[HA][/tex] in the above equation, we get
[tex]K_a=\frac{(10^{-6}M).(10^{-6}M)}{0.09999M}[/tex]
[tex]K_a=1.0001\times 10^{-11}M[/tex]
Rounding it of to one significant figure, we get
[tex]K_a=1.0\times 10^{-11}M\approx 10^{-11}M[/tex]
