Question

A 5.171 g sample of a solid, weak, monoprotic acid is used to make a 100.0 mL solution. 26.00 mL of the resulting acid solution is then titrated with 0.09671 MNaOH. The pH after the addition of15.00 mL of the base is 5.51, and the endpoint is reached after the addition of 47.85 mL of the base.

(c) What is the pK_a of the acid?

Answer 1

The Ka is calculated as = [H+] [A-] / [HA]

let us find the moles of NaOH used

For 47.85 mL 0.09671 M of NaOH, moles = M x L

                                                                   = 0.09671 x 0.048

                                                                   = 0.0046 moles

thus, the moles of acid in 26 mL of acid solution will also be = 0.0046 moles

Now at pH of 5.51

thus, [H+] = 0.0031 x 10^-3 M

moles of NaOH = 0.09671 x 0.015

                         = 0.0015 moles

Thus, at pH 5.5 remaining acid in solution would be = 0.00463 - 0.0015

                                                                                   = 0.0031 moles

Total volume of solution = 15 + 26 = 41 mL = 0.041 L

concentration of acid = 0.0031 / 0.041

                                    = 0.076 M

Feeding the values in the above Ka equation,

Ka = (0.0031 x 10^-3) (0.0031 x 10^-3) / 0.076

     = 1.26 x 10^-10

pKa = -log[Ka]

        = -log(1.26 x 10^-10)

        = 9.90

thus, the pKa is found to be 9.90