the fictitious compound pandemonium carbonate has a ksg=3.091*10^-9 M 3 in water at poor temperature .calculate the solubility of PnCo3 in an aqueous solution of 1.11 M pandemonium sulfate.
express your answer in units of molarity.
.......PnCO3 ==> Pn^2+ + CO3^2-
I......solid.....1.11.....0
C....x dissolves..x.......x
E......solid......x.......x
Ksp = (Pn^2+)(CO3^2-)
3.091E-9 = (x+1.11)(x)
Solve for x = solubility of PnCO3 in mols/L.
To calculate the solubility of PnCo3 in an aqueous solution of pandemonium sulfate, we need to use the concept of solubility product constant (Ksp). The Ksp is an equilibrium constant that represents the degree to which a chemical compound is soluble in water.
The given Ksg value of 3.091*10^-9 M^3 represents the solubility product constant for the fictitious compound PnCo3 in pure water at poor temperature. However, we need to calculate the solubility in an aqueous solution of pandemonium sulfate, which has a concentration of 1.11 M.
To calculate the solubility, we need to consider the reaction that occurs when PnCo3 dissolves in water:
PnCo3(s) ⇌ Pn2+(aq) + Co3^2-(aq)
Let's assume that the solubility of PnCo3 in the presence of pandemonium sulfate is represented by "x" M.
The concentration of Pn2+ ions will also be "x" M, and the concentration of Co3^2- ions will also be "x" M.
Now, let's consider the pandemonium sulfate in the solution. It will dissociate into pandemonium ions (Pn+) and sulfate ions (SO4^2-). Since the concentration of pandemonium sulfate is given as 1.11 M, the concentration of Pn+ ions and SO4^2- ions will both be 1.11 M.
Therefore, the equilibrium expression for the dissolution reaction can be written as:
Ksp = [Pn2+][Co3^2-]
= (x)(x)
= x^2
Given that Ksp = 3.091*10^-9 M^3, we can now find the value of x, which represents the solubility of PnCo3 in the aqueous solution of pandemonium sulfate.
Solving for x:
x^2 = 3.091*10^-9 M^3
Taking the square root of both sides:
x = √(3.091*10^-9 M^3)
x ≈ 5.56*10^-5 M
Therefore, the solubility of PnCo3 in the aqueous solution of 1.11 M pandemonium sulfate is approximately 5.56*10^-5 M.