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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Our tutors have indicated that to solve this problem you will need to apply the Diprotic Acid concept. You can view video lessons to learn Diprotic Acid. Or if you need more Diprotic Acid practice, you can also practice Diprotic Acid practice problems.
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