This question is an acid calculation question so it requires you to recall key formulas relating to weak acids. As the Ka of HCOOH has been given, I would start by writing an expression for Ka. Ka= [HCOO- ] [H+]/[HCOOH] . As we want to calculate the pH, we recall the pH formula is pH=-log[H+]. We need to work out the concentration of [H+] so we rearrange the Ka expression to give [H+]. With the assumption that [H+]=[HCOO- ] , rearranging gives [H+] = √(Ka x [HCOOH])The last step is to input the values given in the question into the equation for [H+]. [H+] = √(1.58 x 10-4 x 0.025) = 1.98746 x 10-3. We then use the [H+] concentration to calculate pH, pH=-log(1.98746 x 10-3) therefore pH= 2.70