Question

A buffer solution is made by adding 15mL of 2.4M NaOH to 55.0mL of a weak acid HA. The pKa of HA is 3.89. The activity coefficient of the weak acid is 1.21, and the activity coefficient of the weak base is 0.87. If the pH of the resulting solution is 4.75, what was the concentration of the original HA solution? Thanks!

Answer 1

[NaOH] = 2.4 M x 0.015 L = 0.036mol

let us assume , molarity of HA = C

[HA] = C x 0.055 L = 0.055C mol

HA                  +   NaOH ------------> NaA + H2O

0.055C                  0.024 0

---------------------------------------------------------------

0.055C -0.024 0                           0.024 mol  

Hence,

      [HA] = (0.055C -0.024 ) mol

    [NaA] = 0.024mol

Given that

       activity coefficient of the weak acid HA, Y_HA = 1.21

      activity coefficient of the weak base NaA, Y_NaA =  0.87

                         pKa = 3.89

                        pH = 4.75

pH = pKa + log { [NaA] Y_NaA / [HA] Y_HA}

4.75 = 3.89 + log { 0.024 x 0.87 / (0.055C -0.024 ) x 1.21}

4.75 = 3.89 + log  { 0.02088/(0.055C -0.024 ) x 1.21}

0.86= log  { 0.02088/(0.055C -0.024 ) x 1.21}

0.02088/(0.055C -0.024 ) x 1.21=7.24

0.02088/(0.055C -0.024 )=8.7604

0.02088=0.481822C-0.21024

0.481822C=0.23112

C=0.4796M

Therefore,concentration of the original HA solution = 0.4796 M