Question

Calculate the pH and the equilibrium concentrations of H₂AsO₄^-, HAsO₄^2- and AsO₄^3- in a 0.2620 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

Given that conc. of Arsenic acid 0.2620 MM The Arsenic acid has the following ionizations HAS04-+ H2O 근 HAsO4-2 + H30+ HAsO4-2 + Нг。근 As04-3 + H30+ Ka2= 5.6 x 10-8 Ka3-3.OX 10-13

The dissociation of Arsenic acid (triprotic acid) is given by Initial 0.2620 0 0 Change + X Equilibrium 0.2620-х a1 [H,Aso0.2620-x X2 +2.5× 10-4 X-6.55x 10-5-0 By solving, x_-0.00822 and 0.007969 r cannot be negative, Sox 0.007969 H 0.007969M [H.Aso,-] = 0.007969M [H3Asaj-0.2620-x-0.2620-0.00796): 0.25403 IM 0.2540M

H2AsO will further dissociates into H+ and HAsO42 Let y be the change in the conc. of all species at equilibrium HAsO,-2 4 Initial 0.007969 0.007969 0 Change Equilibrium 0.007969 - y 0.007969+ y -21 (0.007969 + y)·y =5.6 x 10 0.007969-v Since y is very small, 0.007969-y-0.007969 and neglet y 0.007969 5.6x10 0.007969 :. y 5.6x10-5 (insignificant) [H30+]-0.0079694 y-0.007969 +(5.6x10-8-0.007969

HAsO2 will further dissociates into H+ and AsO43 Let z be the change in the conc. of all species at equilibrium Ka3 HAsO,-2 5.6 x10-8 Aso,-3 4 4 Initial 0.00796:9 0 Change Equilibrium 5.6 x108-z 0.007969 + z 0HAso, 5.6x10 Negletz and assume ( 5.6 10% )-~ ( 5 .бы 10 0.007969z 5.6x10 :. - 2.108x1018 (insignificant) [H3O+] 0.0079694 = 0.0079694 ( 2.108x 10-18 )-0.007969 pHlog (0.007969) 2.099 2.10 ") 3.0x10-13 pH -2.10 H,Aso,10.007969M [HAso,] 5.6x10- M [AsO,2.108x10"M