Question

. A 0.100 M solution of the weak acid HA was titrated with 0.100 M NaOH. The pH measured when Vb = ½Ve was 4.62. Using activity coefficients, calculate pKa. The size of the A- anion is 450 pm.

Answer 1

`pk_a = 4.69`

Step-by-step explanation:

`V_b = (1)/(2) V_e`

`pk_a(c) = 4.62\\pk_a(c) = - logk_a(c)\\4.62 = - logk_a(c)\\k_a(c) = antilog (-4.62)\\k_a(c) = 2.4 * 10^(-65)`

Activity coefficients:

`f_A = 0.854\\f_H = 1.00\\f_(HA) = 1.00`

`k_a = k_a(c) *f_A* (f_H)/(f_(HA)) \\k_a = 2.4 * 10^(-5) *0.854* (1.0)/(1.0)\\k_a = 2.05 * 10^(-5)`

`pk_a = -log(k_a)\\pk_a = -log(2.05*10^(-5))\\pk_a = 4.69`