Question

Calculate the concentrations of all species in a 0.180-M solution of H₂X if K₁ = 5.01E-05 and K₂ = 2.82E-08

[H₂X] =

M

[H₃O¹⁺] =

M

[HX^1-] =

M

[X^2-] =

M

Answer 1

H2x -----------------> HX-   +    H+

0.180                           0            0

0.180 - x                      x             x

Ka1 = [HX-][H+] / [H2X]

5.01 x 10^-5 = x^2 / 0.180 - x

x = 2.98 x 10^-3

[H3O+] = 2.98 x 10^-3 M

[H2X] = 0.177 M

[HX-] =2.98 x 10^-3 M

[X2-] = Ka2 = 2.82 × 10^-8 M