The solubility of silver(I)carbonate is 3.6 × 10^-2 g/L.What is the solubility product constant for silver(I)carbonate?

Question

Quizplus · Accepted Answer

The solubility product constant for a salt AB is given by the expression: 
Ksp= [A+]m[B-]n 
where m and n are the coefficients of the ions in the dissociation equation of the salt AB.

For silver(I)carbonate, the dissociation equation is: 
Ag2CO3 (s) ↔ 2Ag+ (aq) + CO3^2- (aq)

Therefore, 
Ksp = [Ag+]^2[CO3^2-]

The solubility of silver(I)carbonate is given as 3.6 × 10^-2 g/L.

The molar mass of Ag2CO3 = (2×107.87) + 12.01 + (3×16.00) = 331.75 g/mol

Therefore, the concentration of Ag2CO3 in mol/L is: 
(3.6 × 10^-2 g/L) / (331.75 g/mol) = 1.09 × 10^-4 mol/L

Since 1 molecule of Ag2CO3 gives 2 Ag+ ions and 1 CO3^2- ion, 
[Ag+] = 2 × (1.09 × 10^-4) = 2.18 × 10^-4 mol/L 
[CO3^2-] = 1.09 × 10^-4 mol/L

Substituting these values in the expression for Ksp, we get: 
Ksp = (2.18 × 10^-4)^2 × (1.09 × 10^-4) = 5.4 × 10^-12

Rounded to two significant figures, the answer is 11 × 10^-12 or 1.1 × 10^-11. 
Therefore, the best choice is B.

The Solubility of Silver(I)carbonate Is 3