Calculate the pH of a 0.025 mol dm-​3​ solution of methanoic acid. For HCOOH, Ka = 1.58 x 10-​4​ mol dm-​3

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