Question

A buffer with a pH of 4.52 contains 0.33 m of sodium benzoate and 0.16 m of benzoic acid. What is the concentration of [H+] in the solution after the addition of 0.052 mol of HCL to a final volume of 1.3 L? Assume that any contribution of HCL is to the volume is negligible. [H+] =

Answer 1

Answer – We are given, [Benzoic acid] = 0.16 M, [Benzoate] = 0.33 M

Volume = 1.3 L

Moles of benzoic acid = 0.16 M * 1.3 L

                                     = 0.208 moles

Moles of benzoate = 0.33 M * 1.3 L

                                     = 0.429 moles

So after added HCl there is moles of benzoic acid increase and benzoate decrease

Moles of benzoic acid = 0.208 + 0.052 = 0.260 moles

Moles of benzoate = 0.429 moles – 0.052 = 0.377 moles

So, [Benzoic acid] = 0.260 moles / 1.3 L

                                = 0.2 M

[Benzoate] = 0.377 moles / 1.3 L

                  = 0.29 M

So, pKa for benzoic acid = 4.19

So, pH = pKa + log [Benzoate] / [Benzoic acid]

            = 4.19 + log 0.29 M / 0.20 M

            = 4.35

So, [H⁺] = 10^-pH

              = 10^-4.35

              = 4.45*10^-5 M