Chemistry! Help!!

Question

Consider the following diprotic acid.
H2A+H2O⇀↽H3O+ +HA−

Ka1 =5.8×10^-3

HA− +H2O⇀↽H3O+ +A2−

Ka1 =8.5×10^-4
The initial concentration of H2A = 0.0010 M.
1. Determine the [H3O+] at equilibrium.
Since the Ka1 and Ka2 values are very close, treat both reactions simultaneously.

Answer 1

Since both K1 & K2 are so close, we can treat the two reactions simultaneously.  Adding the two equations gives you:

H₂A + 2 H₂O  ↔ 2 H₃O⁺ + A^2-

 

K₁K₂ = [H₃O⁺]² [A^2-] / [H₂A]

 

Let x = the change in the initial concentration of H₂A.

 

This results in the equation K₁K₂ = (5.8 X 10^-3)(8.5 x 10^-4) = (2x)²(x)/(0.0010 - x)

 

4.93 x 10^-9 = 4x³/(0.0010 - x)

 

The x in the denominator is negligable (because 4.93 x 10^-9 is <<< 3 powers of 10 of the initial concentration.  So, when solving fo x we ge 1.07 x 10^-2 and [H₃O⁺] = 2x, so [H₃O⁺] = 2.14 X 10^-2 m