Respuesta :
Answer : The dissociation constant (Ka) = 9.025 × [tex]10^{-4}[/tex]
Solution : Given,
Concentration HF solution = 0.100 M
% Dissociation = 9.5 %
The equation for dissociation of HF is :
[tex]HF \rightleftharpoons H^{+}+ F^{-}[/tex]
The Ka expression for HF is :
[tex]Ka=\frac{[H^{+}][F^{-}]}{[HF]}[/tex] ............. (1)
Step 1 : we find the [tex][H^{+}][/tex] by using the concentration and % dissociation.
[tex][H^{+}][/tex] = Concentration HF solution × % Dissociation
[tex][H^{+}][/tex] = 0.100 M × [tex]\frac{9.5}{100}[/tex] = 9.5 × [tex]10^{-3}[/tex] M
Step 2 : For [tex][F^{-}][/tex] , the concentration of [tex][F^{-}][/tex] is equal to the [tex][H^{+}][/tex]. From the above equation the stoichiometry of [tex][F^{-}][/tex] and [tex][H^{+}][/tex] is 1:1.
Therefore,
[tex][F^{-}][/tex] = [tex][H^{+}][/tex] = 0.100 M × [tex]\frac{9.5}{100}[/tex] = = 9.5 × [tex]10^{-3}[/tex] M
Now, put all the values in equation (1), we get
[tex]Ka=\frac{(9.5\times10^{-3})\times(9.5\times10^{-3})}{(0.1)}[/tex]
= 9.025 × [tex]10^{-4}[/tex]
