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