Question

Consider the acid-base neutralization reaction H2S(aq) + CN‒(aq) ⇌ HS‒(aq) + HCN(aq) For H2S, pKa = 7.0, and for HCN, pKa = 9.4. Which of the following statements is FALSE? (a) H2S is a stronger acid than HCN. (b) CN− is a stronger base than HS−. (c) At equilibrium, the equilibrium position favors the reactants. (d) At equilibrium, the equilibrium constant, Kc, is larger than 1.

Answer 1

Answer:

(c) At equilibrium, the equilibrium position favors the reactants.

Explanation:

To assess the strength of an acid or a base we need to consider the pKa or pKb. For an acid, the lower the pKa, the stronger the acid. For a base, the lower the pKb, the stronger the base.

Which of the following statements is FALSE?

(a) H₂S is a stronger acid than HCN. TRUE. The pKa of H₂S (7.0) is lower than the pKa of HCN (9.4).

(b) CN⁻ is a stronger base than HS⁻. TRUE. We can calculate the pKb of a base when we know the pKa of the conjugate acid using the following expression:

pKa + pKb = 14 ⇒ pKb = 14 - pka

The pKb of CN⁻ (4.6) is lower than the pKb of HS⁻(7.0) so it is a stronger base.

(c) At equilibrium, the equilibrium position favors the reactants. FALSE. In the reaction H₂S(aq) + CN⁻(aq) ⇌ HS⁻(aq) + HCN(aq), H₂S is the stronger acid and CN⁻ is the stronger base, so at the equilibrium the products are favored.

(d) At equilibrium, the equilibrium constant, Kc, is larger than 1. TRUE. Given the products are favored at the equilibrium, the equilibrium constant Kc is expected to be larger than 1 (the numerator is higher than the denominator).