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We’re being asked to calculate the pH of a 0.080 M carbonic acid (H2CO3). Carbonic acid is a diprotic acid, it can donate two protons (H+) and it will have two equilibrium reactions.
To calculate for pH, we use the following equation:
Ka1 = 4.3x10-7
• 1st equilibrium reaction: H2CO3(aq) + H2O(l) → HCO3-(aq) + H3O+(aq
Ka2 = 5.6x10-11
• 2nd equilibrium reaction: HCO3-(aq) + H2O(l) → CO32-(aq) + H3O+(aq)
• H3O+(aq) is also produced from the 2nd equilibrium reaction
Calculate the pH of a 0.080 M carbonic acid solution, H 2CO3(aq), that has the stepwise dissociation constants Ka1 = 4.3 × 10 -7 and Ka2 = 5.6 × 10 -11.
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