Question

At 298K, the equilibrium constant for the following reaction is 4.20×10-7: H2CO3(aq) + H2O H3O+(aq) + HCO3-(aq) The equilibrium constant for a second reaction is 4.80×10-11: HCO3-(aq) + H2O H3O+(aq) + CO32-(aq) Use this information to determine the equilibrium constant for the reaction: H2CO3(aq) + 2H2O 2H3O+(aq) + CO32-(aq)

Answer 1

Answer:

The correct answer is 2.016 x 10⁻¹⁷

Explanation:

We have the following chemical reactions and their equilibrium constants (K):

(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq)  K₁= 4.20×10⁻⁷

(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq)    K₂= 4.80×10⁻¹¹

And we have to obtain K for the following reaction:

H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)

If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.

H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq)     K₁= 4.20×10⁻⁷

+

HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq)      K₂= 4.80×10⁻¹¹

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H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)    K= K₁ x K₂

K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷