We’re given a weak diprotic acid and we are asked to calculate the contributions to [H3O+] from each ionization step.
Ka1 = 1.0 x10-4
• 1st equilibrium reaction: H2A(aq) + H2O(l) → HA-(aq) + H3O+(aq
Ka2 = 1.0 x10-5
• 2nd equilibrium reaction: HA-(aq) + H2O(l) → A2-(aq) + H3O+(aq)
• H3O+(aq) is also produced from the 2nd equilibrium reaction
Consider a 0.10 M solution of a weak polyprotic acid (H2 A) with the possible values of Ka1 and Ka2 given below. Calculate the contributions to [H3O+] from each ionization step.
Ka1 = 1.0 10 - 4 ; Ka2 = 1.0 10 - 5
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