Textbook Question
Calculate the [H3O+] and pH of each H2SO4 solution. At approximately what concentration does the x is small approximation break down?
a. 0.50 M b. 0.10 M c. 0.050 M
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17. Acid and Base Equilibrium
Diprotic Acids and Bases Calculations
Multiple Choice
Carbonic acid, H₂CO₃, is a diprotic acid with Kₐ₁ = 4.2 x 10⁻⁷ and Kₐ₂ = 4.7 x 10⁻¹¹. Calculate the concentration of carbonate ions (CO₃²⁻) in a 0.80 M solution of carbonic acid.
A
4.7 x 10⁻¹¹ M
B
1.0 x 10⁻⁷ M
C
0.80 M
D
4.2 x 10⁻⁷ M
Verified step by step guidance1
Identify that carbonic acid (H₂CO₃) is a diprotic acid, meaning it can donate two protons (H⁺) in two steps. The first ionization step is H₂CO₃ ⇌ H⁺ + HCO₃⁻ with Kₐ₁ = 4.2 x 10⁻⁷, and the second ionization step is HCO₃⁻ ⇌ H⁺ + CO₃²⁻ with Kₐ₂ = 4.7 x 10⁻¹¹.
Assume that the initial concentration of H₂CO₃ is 0.80 M. For the first ionization, set up the expression for the equilibrium constant Kₐ₁: Kₐ₁ = [H⁺][HCO₃⁻] / [H₂CO₃].
Assume that the change in concentration of H₂CO₃ due to ionization is x, so at equilibrium, [H₂CO₃] = 0.80 - x, [H⁺] = x, and [HCO₃⁻] = x. Substitute these into the Kₐ₁ expression: 4.2 x 10⁻⁷ = (x)(x) / (0.80 - x).
Solve the equation for x, which represents the concentration of H⁺ and HCO₃⁻ at equilibrium after the first ionization. Since Kₐ₁ is small, you can approximate 0.80 - x ≈ 0.80 for simplification.
For the second ionization, use the concentration of HCO₃⁻ (which is approximately x from the first ionization) to set up the Kₐ₂ expression: Kₐ₂ = [H⁺][CO₃²⁻] / [HCO₃⁻]. Substitute the known values and solve for [CO₃²⁻], the concentration of carbonate ions.
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