Question

can someone assist with this problem? a classmate and i reviewed our answers and we got different answers.

6. Arsenic Acid (H3AsO4) has the ability to act as a triprotic acid. Calculate the equilionidan concentration of all possible species in a 0.483M arsenic acid solution. The Kal = 6.0 x 10-3, K2=1.0 x 10-7, and K3 = 3.2 x 10-12 1.) H ₃ As Oy caq? - H + H₂ AsOy i 0.483 x2x 102 = 6.04 10 3 0.483 - + x + x assume 0.493 -X = 0.483 0.483-8 6.0 x103 x² As 04] -0.483 M 0.483 x= 6.0x103 (0.483) x² = 2.89 x 103 x= 2.89 x 10-3 6.38x102 M = [HD (5.38 x 10² M:[H₂ AsOy Kaz= 10 x 107 mim assume 5.20 x 102 3 PH= - 100 coluna 12 ASO4 I HI + HAS Oy x 102 5.38 x 102 e

Answer 1

------ H3AsO4(aq) -------------> H^+ (aq) + H2AsO4^-(aq)

I ----- 0.483                                0                  0

C----   -x     ----------------------    +x -------    +x

E ---- 0.483-x -------------------   +x   --------- + x

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

    6*10^-3   = x*x/(0.483-x)

   6*10^-3*(0.483-x) = x^2

       x   = 0.051

    [H^+]   = x   = 0.051M

    [H2ASO4^-] = x = 0.051M

H2AsO4^- (aq) ------------> H^+ (aq) + HAsO4^2-(aq)

   Ka2    = [H^+][HAsO4^2-]/[H2AsO4^-]

1*10^-7   = 0.051[H2AsO4^2-]/0.051

   [H2AsO4^2-]    = 1*10^-7 M

         HAsO4^2- (aq) ------------> H^+ (aq) + AsO4^3- (aq)

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

3.2*10^-12   = 0.051[AsO4^3-]/1*10^-7

           [AsO4^3-]    = 3.2*10^-12*1*10^-7/(0.051)   = 6.3*10^-18M

PH   = -log[H^+]

        = -log0.051

         = 1.2924