Multiple Choice
What is the pH of a 0.090 M H2CO3 solution, given that the first dissociation constant (Ka1) is 4.3×10^-7 and the second dissociation constant (Ka2) is 5.6×10^-11?
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a solution that is 8.0 × 10⁻² M in H₂CO₃, given that the first dissociation constant (Ka1) of H₂CO₃ is 4.3 × 10⁻⁷?
A
2.36
B
5.67
C
3.45
D
4.12
Verified step by step guidance1
Identify that H₂CO₃ is a weak acid and will partially dissociate in water. The dissociation can be represented as: H₂CO₃ ⇌ H⁺ + HCO₃⁻.
Use the given first dissociation constant (Ka1) for H₂CO₃, which is 4.3 × 10⁻⁷, to set up the expression for the equilibrium constant: Ka1 = [H⁺][HCO₃⁻]/[H₂CO₃].
Assume that the initial concentration of H₂CO₃ is 8.0 × 10⁻² M and that the change in concentration of H₂CO₃ due to dissociation is x. Therefore, at equilibrium, [H⁺] = x, [HCO₃⁻] = x, and [H₂CO₃] = 8.0 × 10⁻² - x.
Substitute these equilibrium concentrations into the Ka1 expression: 4.3 × 10⁻⁷ = (x)(x)/(8.0 × 10⁻² - x).
Assume that x is much smaller than 8.0 × 10⁻², allowing the simplification 8.0 × 10⁻² - x ≈ 8.0 × 10⁻². Solve for x, which represents [H⁺], and then calculate the pH using the formula pH = -log[H⁺].
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