Question

Calculate the pH and the equilibrium concentrations of H₂AsO₄^-, HAsO₄^2- and AsO₄^3- in a 0.2920 M aqueous arsenic acid solution.

For H₃AsO₄, K_a1 = 2.5×10^-4, K_a2 = 5.6×10^-8, and K_a3 = 3.0×10^-13

pH = _____
[H2AsO4-] =	_____ M
[HAsO42-] =	_____ M
[AsO43-] =	_____ M

Answer 1

Ka1 = 2.5 x 10^-4

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

0.292 0          0

0.292 - x                                    x           x

Ka1 = [x][x]/[0.292-x]

2.5 x 10^-4 = x^2/(0.292-x)

x^2 + 2.5 x 10^-4 x- 7.3 x 10^-5 =0

x = 8.42 x 10^-3

[H2AsO4-] = x

[H2AsO4-] =8.42 x 10^-3 M

[H+] = x = 8.43 x 10^-3 M

almost maximum H+ ions comes from first ionisation so

pH = -log [H+] = -log(8.42 x 10^-3)

pH= 2.08

[HAsO4^-2] = Ka2 = 5.6 x10^-8 M

  HAsO4^2 --------------------->AsO4^-3 + H+

5.6 x10^-8                            0            8.42x 10^-3

5.6 x10^-8- z                        z                 8.42x 10^-3 +z

Ka3 = [ AsO4^-3][H+]/[HAsO4^2]

3.0x 10^-13 = (z) x 8.42x 10^-3 +z)/ (5.6 x10^-8 z )

z= 2.00 x 10^-18 M

[AsO4^-3] = z= 2.00 x 10^-18 M