First, find the molality of the solution.
[tex]i=\frac{T_f}{k_f\cdot m}[/tex]Where k is the molal freezing point, which is 1.86 C/m for water.
[tex]i=\frac{0.758}{1.86\cdot0.4}=\frac{0.758}{0.744}=1.02[/tex]The equilibrium constant Ka would be
[tex]K_a=\frac{\lbrack HCOO\rbrack\lbrack H\rbrack}{\lbrack HCOOH\rbrack}[/tex][tex]K_a=\frac{x\cdot x}{0.4-x}[/tex]Then, to find x.
[tex]\begin{gathered} i=\frac{0.4-x+x+x}{0.4}=1.02 \\ x=1.02\cdot0.4-0.4=0.008 \end{gathered}[/tex]Once we have x, we can obtain the constant Ka
[tex]K_a=\frac{(0.008)^2}{0.4-0.008}=\frac{6.4\times10^{-5}}{0.392}=1.6\times10^{-4}[/tex]Therefore, the constant Ka of the reaction is 1.6x10^-4.
