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 

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Answered by Radia H. Chemistry tutor

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