Calculate the concentration of bicarbonate ion,HCO₃^-,in a 0.010 M H₂CO₃ solution that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11. A)6.6 × 10^-5 M B)4.3 × 10^-7 M C)4.3 × 10^-9 M D)5.6 × 10^-11 M

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Quizplus · Accepted Answer

The equilibrium reactions and their corresponding equilibrium constants are: H₂CO3- ⇌ H+ + HCO3-, Ka1 = 4.3 × 10^-7 HCO3- ⇌ H+ + CO32-, Ka2 = 5.6 × 10^-11 Let's assume x is the concentration of HCO3- initially. Then, the concentration of H₂CO3- is (0.010 - x) because the two concentrations add up to the total bicarbonate concentration of 0.010 M. In the first equilibrium, we can write: Ka1 = [H+][HCO3-]/[H₂CO3-] = (x)(0.010 - x) / (0.010 - x) Simplifying yields: 4.3 × 10^-7 = x(0.010 - x) / (0.010 - x) 4.3 × 10^-7(0.010 - x) = x(0.010 - x) 0.000043 - 4.3 × 10^-7x = x^2 - 0.010x x^2 - 0.010x - 4.3 × 10^-7x + 0.000043 = 0 x^2 - 4.3 × 10^-7x - 0.01x + 0.000043 = 0 Solving this quadratic equation yields: x = 6.6 × 10^-5 M Therefore, the answer is (A), 6.6 × 10^-5 M.

Calculate the Concentration of Bicarbonate Ion,HCO3-,in a 0

Calculate the Concentration of Bicarbonate Ion,HCO₃^-,in a 0