Chemistry HELP!

Question

Consider the following diprotic acid.
H2A+H2O⇀↽H3O+ +HA− Ka1 =5.0×10−5 HA− +H2O⇀↽H3O+ +A2− Ka1 =4.5×10−5
The initial concentration of H2A = 1.0 × 10−5 M.
1. Since the Ka1 and Ka2 values are very close, treat both reactions simultaneously. (a) Determine the [H3O+] at equilibrium.
2. Now, as a comparison, ignore the second reaction. (a) Determine the [H3O+].
3. Is the approximation of ignoring the second reaction valid?

Answer 1

Not entirely sure what is meant by "treat both reactions simultaneously", because while they may be happening at the same time (following initial dissociation of H₂A), they are separate and distinct in how they provide H₃O⁺.

H₂A + H₂O ==> H₃O⁺ + HA^-  Ka1 = 5x10^-5 = [H₃O⁺][HA^-]/[H₂A]

5x10^-5 = (x)(x)/1x10^-5 - x

x = 8.5x10^-6 M = [H₃O⁺] = [HA^-]

 

HA^- + H₂O ==> H₃O⁺ + A^2-  Ka2 = 4.5x10^-5 =  [H3O⁺][A^2-]/[HA^-]

4.5x10^-5 = (8.5x10^-6 + x)(x)/(8.5x10^-6 - x)

x = ~1.6x10^-5

[H3O⁺] = 1.6x10^-5 + 8.5x10^-6 = ~2.5x10^-5 M

 

It would not be prudent to ignore the second reaction.