Answer : The correct option is, [tex]3.5\times 10^{-4}[/tex]
Solution : Given,
pH = 2.03
Concentration of HF = 0.25 M
First we have to calculate the concentration of [tex]H^+[/tex] ion.
[tex]pH=-\log [H^+][/tex]
[tex]2.03=-\log [H^+][/tex]
[tex][H^+]=9.3\times 10^{-3}M[/tex]
Now we have to calculate the value of [tex]K_a[/tex] for HF.
The equilibrium reaction will be
[tex]HF\rightleftharpoons H^++F^-[/tex]
Concentration of [tex]H^+[/tex] = Concentration of [tex]F^-[/tex] = [tex]9.3\times 10^{-3}M[/tex]
The expression for [tex]K_a[/tex] for HF will be,
[tex]K_a=\frac{[H^+][F^-]}{[HF]}=\frac{(9.3\times 10^{-3})\times (9.3\times 10^{-3})}{0.25}=3.5\times 10^{-4}[/tex]
Therefore, the value of [tex]K_a[/tex] for HF is, [tex]3.5\times 10^{-4}[/tex]
