The fictitious compound, pandemonium carbonate Pn(CO3)2 hase a Ksp=3.091×10−9 M3 in water at room temperature. Calculate the solubility of Pn(CO3)2 in an aqueous solution of 1.11 M pandemonium sulfate Pn(SO4)2 (aq). Express your answer in units of molarity.
2.639 x 10^-5
To calculate the solubility of Pn(CO3)2 in an aqueous solution of 1.11 M pandemonium sulfate Pn(SO4)2, we need to first write the balanced chemical equation for the dissociation of Pn(CO3)2 in water:
Pn(CO3)2 (s) ⇌ Pn2+ (aq) + 2CO32- (aq)
According to the principle of solubility product (Ksp), the solubility of Pn(CO3)2 can be determined by comparing the values of Ksp with the concentrations of the ions in the solution. The Ksp expression for the dissociation of Pn(CO3)2 is as follows:
Ksp = [Pn2+] * [CO32-]^2
Since Pn(SO4)2 is a strong electrolyte, it will completely dissociate into its ions in solution:
Pn(SO4)2 (aq) → Pn2+ (aq) + 2SO42- (aq)
Given that the concentration of pandemonium sulfate (Pn(SO4)2) is 1.11 M, we can assume that the concentration of Pn2+ ions ([Pn2+]) is also 1.11 M.
Now, we need to find the concentration of carbonate ions ([CO32-]). Since the stoichiometry of the balanced chemical equation tells us that the molar ratio of Pn2+ ions to CO32- ions is 1:2, the concentration of carbonate ions will be twice the concentration of Pn2+ ions. Therefore, [CO32-] = 2 * [Pn2+] = 2 * 1.11 M = 2.22 M.
Substituting these values into the Ksp expression, we have:
3.091×10^-9 = (1.11 M) * (2.22 M)^2
Now, we can solve for the solubility of Pn(CO3)2. Rearranging the expression, we get:
Solubility of Pn(CO3)2 = [Pn2+] = (3.091×10^-9) / (2 * (2.22)^2) ≈ 3.47×10^-10 M
Therefore, the solubility of Pn(CO3)2 in the 1.11 M pandemonium sulfate solution is approximately 3.47×10^-10 M.