First, calculate the concentration of hydronium.
[tex]K_a=\frac{\lbrack H_3O^+\rbrack\lbrack\text{HCOO}^-\text{\rbrack}}{\lbrack\text{HCOOH\rbrack}}[/tex]Where the concentration of HCOOH is 75 g/L, or 75 M.
[tex]\begin{gathered} K_a=\frac{x\cdot x}{75M}=1.8\times10^{-4} \\ x^2=135\times10^{-4} \\ x=\sqrt[]{135\times10^{-4}} \\ x\approx0.116 \end{gathered}[/tex]Therefore, the concentration of hydronium is 0.116 M.
Then, we find the pH.
[tex]pH=-\log \lbrack H_3O^+\rbrack[/tex][tex]pH=-\log \lbrack0.116\rbrack\approx0.94[/tex]Therefore, the pH is 0.94, which means is a highly acidic solution.
