Question

The value of K_sp (solubility product constant) for silver carbonate, Ag_ 2 CO_ 3 , is 8.10×10^−12

. Calculate the solubility of Ag_ 2 CO_ 3 in grams per liter.

a) 9.25 x 10^3 g/L
b) 1.25 x 10^-4 g/L
c) 7.55 x 10^-6 g/L
d) 4.15 x 10^-9 g/L

Answer 1

The solubility of silver carbonate in water is calculated using its K_sp value and its dissociation in water. After determining the molarity 's' by solving for 's' in the K_sp expression, it's converted to grams per liter using the molar mass of Ag_2CO_3.

Step-by-step explanation:

The Ksp (solubility product constant) for silver carbonate, Ag2CO3, allows us to calculate the solubility of the substance in water. To do this, we let the solubility of Ag2CO3 be 's' where s represents the concentration of Ag2CO3 that dissolves to form ions in solution. The dissociation of Ag2CO3 in water is:

Ag2CO3(s) → 2Ag+(aq) + CO32-(aq)

According to the stoichiometry of the reaction, the concentration of Ag+ ions will be 2s and that of CO32- ions will be s. We can write the Ksp expression as:

Ksp = [Ag+]2[CO32-] = (2s)2(s) = 4s3

Given that Ksp = 8.1 × 10-12, we can set up the equation:

4s3 = 8.1 × 10-12

By solving for 's', we can then find the molarity of the saturated solution. After finding the molarity, we can convert it into grams per liter using the molar mass of Ag2CO3, which is 275.75 g/mol:

s(mol/L) × 275.75(g/mol) = solubility in g/L

After completing the calculations, the correct answer will be found to determine the solubility of Ag2CO3 in grams per liter.