Determine the pH of a 0.10 M Na2CO3 solution using carbonate hydrolysis with Kb1 = Kw/Ka2 and Ka2 ≈ 4.7×10^-11.

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Multiple Choice

Determine the pH of a 0.10 M Na2CO3 solution using carbonate hydrolysis with Kb1 = Kw/Ka2 and Ka2 ≈ 4.7×10^-11.

Explanation:
This question tests recognizing that carbonate is a weak base in water and its basicity is described by a base hydrolysis constant. For CO3^2- the hydrolysis reaction is CO3^2- + H2O ⇌ HCO3^- + OH-, and its Kb is related to Ka2 of HCO3^- by Kb1 = Kw/Ka2. With Ka2 ≈ 4.7×10^-11 and Kw ≈ 1.0×10^-14, Kb1 ≈ 2.1×10^-4. For a 0.10 M solution of a weak base, the hydroxide concentration is approximated by [OH-] ≈ sqrt(Kb1 × C). So [OH-] ≈ sqrt(2.1×10^-4 × 0.10) = sqrt(2.1×10^-5) ≈ 4.6×10^-3 M. Then pOH ≈ −log(4.6×10^-3) ≈ 2.34, and pH ≈ 14 − 2.34 ≈ 11.66. This simple calculation captures the basic nature of the solution and aligns with the given result.

This question tests recognizing that carbonate is a weak base in water and its basicity is described by a base hydrolysis constant. For CO3^2- the hydrolysis reaction is CO3^2- + H2O ⇌ HCO3^- + OH-, and its Kb is related to Ka2 of HCO3^- by Kb1 = Kw/Ka2. With Ka2 ≈ 4.7×10^-11 and Kw ≈ 1.0×10^-14, Kb1 ≈ 2.1×10^-4. For a 0.10 M solution of a weak base, the hydroxide concentration is approximated by [OH-] ≈ sqrt(Kb1 × C). So [OH-] ≈ sqrt(2.1×10^-4 × 0.10) = sqrt(2.1×10^-5) ≈ 4.6×10^-3 M. Then pOH ≈ −log(4.6×10^-3) ≈ 2.34, and pH ≈ 14 − 2.34 ≈ 11.66. This simple calculation captures the basic nature of the solution and aligns with the given result.

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