How do you do this problem?

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)