Question

How do you do this problem?

Answer 1

Answer:

Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)

Explanation:

Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq).

Silver chromate is the salt of a strong base (AgOH) and a weak acid (H₂CrO₄).

HCrO₄⁻ is an even weaker acid than H₂CrO₄, so CrO₄²⁻ is a strong base.

Any added H⁺ will immediately combine with the chromate ions according to the reaction

H⁺ + CrO₄²⁻ ⟶ HCrO₄⁻

thereby removing chromate ions from solution.

According to Le Châtelier's Principle, more silver chromate will dissolve to replace the chromate ions that the H⁺  removes.

The overall equation for the reaction is

Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)

H⁺(aq) + CrO₄²⁻(aq) ⟶ HCrO₄⁻(aq)

Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)