Question

What [Ag⁺] is required to reduce [CO₃²⁻] to 8.2 × 10⁻⁴ M by precipitation of Ag₂CO₃?

a) 4.1 × 10⁻⁴ M
b) 8.2 × 10⁻⁴ M
c) 1.64 × 10⁻³ M
d) 3.28 × 10⁻⁴ M

Answer 1

To find the required concentration of Ag⁺ ions to reduce [CO₃²⁻] to 8.2 × 10⁻⁴ M by precipitation of Ag₂CO₃, we use the solubility product constant and determine that the answer is 1.64 × 10⁻³ M.

Step-by-step explanation:

The solubility product constant (Ksp) of Ag₂CO₃ is given as 8.46 × 10⁻¹². The formula for the solubility product constant is Ksp = [Ag⁺]²[CO₃²⁻]. Since Ag₂CO₃ dissociates into 2 Ag⁺ and 1 CO₃²⁻, we can express the concentrations of silver and carbonate ions at equilibrium in terms of a single variable s, which stands for the solubility of Ag₂CO₃. Therefore, Ksp = (2s)²(s) = 4s³.

To find the concentration of silver ions, we should first find the value of s that corresponds to the given carbonate concentration and then substitute it back into the relationship [Ag⁺] = 2s. We are given [CO₃²⁻] = 8.2 × 10⁻⁴ M, so s = 8.2 × 10⁻⁴ M. Substituting into Ksp: 4(8.2 × 10⁻⁴ M)³ = 8.46 × 10⁻¹². Solving for the concentration of silver ions, we find [Ag⁺] = 2(8.2 × 10⁻⁴ M) = 1.64 × 10⁻³ M. Therefore, the correct answer is c) 1.64 × 10⁻³ M.