Question

Determine the pH of an aqueous solution of 0.0028 M Arsenic Acid.

K_a1(H₃AsO₄) = 5.8x10^-3, K_a2(H₃AsO₄) = 1.1x10^-7, K_a3(H₃AsO₄) = 3.2x10^-17

Charge balance equation: [H⁺]=[H₂AsO₄^-]+2[HAsO₄^2-]+3[AsO₄^3-]+[OH^-]

Mass Balance equation: 0.0028 M = [H₃AsO₄]+[H₂AsO₄^-]+[HAsO₄^2-]+[AsO₄^3-]

Answer 1

H3AsO4 -----> H+ + H2AsO4 - ........Ka1 = 5.8 x 10^-3

H2AsO4- ------> H+ + HAsO4 2- ...........Ka2 = 1.1 x 10^-7

HAsO4 2- -------> H+ + AsO4 3- ............Ka3 = 3.2 x 10^-17

We see that Ka1 >> Ka2 >> Ka3

Therefore we can say that,

[H+] from Ist reaction >> [H+] from second reaction >> [H+] from 3rd reaction

Hence we can neglect the concentration of H+ from the 2nd and 3rd reaction in comparison to the 1st

H3AsO4 -----> H+ + H2AsO4 -

0.0028 - X........X.............X

Ka1 = [H+] [H2AsO4-] / [H3AsO4]

5.8 x 10^-3 = X^2 / (0.0028 - X)

=> X = 2.065 x 10^-3 M = [H+]

We know that,

pH = - log [H+] = - log (2.065 x 10^-3) = 2.685