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

Answer:

$pk_a = 4.69$

Explanation:

$V_b = \frac{1}{2} V_e$

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

Activity coefficients:

$f_A = 0.854\\f_H = 1.00\\f_{HA} = 1.00$

$k_a = k_a(c) *f_A* \frac{f_H}{f_{HA}} \\k_a = 2.4 * 10^{-5} *0.854* \frac{1.0}{1.0}\\k_a = 2.05 * 10^{-5}$

$pk_a = -log(k_a)\\pk_a = -log(2.05*10^{-5})\\pk_a = 4.69$