Question

A buffer solution is made by adding 10 mL of 2M NaOH to 50mL of a weak acid HA. The pKa of HA is 3.42, and the activity coefficient of the weak acid is 1.6 and the activity coefficient of the weak base is 0.75. If the pH of the resulting solution is 4.59, what was the concentration of the original HA solution?

Answer 1

[NaOH] = 2 M x 0.01 L = 0.02 mol

let us assume , molarity of HA = C

[HA] = C x 0.05 L = 0.05C mol

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

0.05C                  0.02                        0

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

0.05C -0.02           0                           0.02 mol  

Hence,

      [HA] = (0.05C -0.02 ) mol

    [NaA] = 0.02 mol

Given that

       activity coefficient of the weak acid HA, Y_HA = 1.6     

      activity coefficient of the weak base NaA, Y_NaA =  0.75

                         pKa = 3.42

                        pH = 4.59

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

4.59 = 3.42 + log { 0.02 x 0.75 / (0.05C -0.02 ) x 1.6 }

4.59 = 3.42 + log { 0.009375 /(0.05C -0.02 )}

0.17 = log { 0.009375 /(0.05C -0.02 )}

0.009375 /(0.05C -0.02 ) = 1.48

  0.009375 = 0.074 C - 0.0296

0.074 C = 0.038975

    C = 0.526 M

Therefore,

concentration of the original HA solution = 0.526 M