The fictitious compound, pandemonium carbonate Pn(CO3)2 hase a Ksp=3.091x10^-9M3 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.

please, i need the answer too!

2.63e-5

0.917e-3

I had 2.63e-5 as the right answer

qwerty and paul are right. Thank you both.

To calculate the solubility of Pn(CO3)2 in an aqueous solution of 1.11 M pandemonium sulfate, we need to consider the solubility product constant (Ksp) of Pn(CO3)2 and the common ion effect.

The solubility product constant (Ksp) describes the equilibrium between the solid (Pn(CO3)2) and its dissolved ions (Pn2+ and CO32-) in water:

Pn(CO3)2(s) ⇌ Pn2+(aq) + 2 CO32-(aq)

The Ksp expression for this equilibrium is:

Ksp = [Pn2+] * [CO32-]^2

Given that the Ksp value for Pn(CO3)2 is 3.091x10^-9 M^3, we can set up the following equation:

3.091x10^-9 = [Pn2+] * [CO32-]^2

To determine the solubility of Pn(CO3)2 in the presence of 1.11 M Pn(SO4)2, we need to account for the common ion effect. The common ion effect occurs when the solubility of a compound is reduced in the presence of a common ion (in this case, the pandemonium ion, Pn2+).

Assuming that the solubility of Pn(CO3)2 is represented by "x" M, the concentration of Pn2+ ions in the solution will increase by 2x M (since 1 molecule of Pn(CO3)2 produces 1 Pn2+ ion). Therefore, the concentration of Pn2+ ions will be:

[Pn2+] = 1.11 + 2x

Substituting this value into the Ksp expression, we get:

3.091x10^-9 = (1.11 + 2x) * (x)^2

Now, we can solve this quadratic equation to find the solubility of Pn(CO3)2 (x):

3.091x10^-9 = (1.11x + 2x^2)

Simplifying the equation, we have:

3.091x10^-9 = 2x^2 + 1.11x

Rearranging the equation:

2x^2 + 1.11x - 3.091x10^-9 = 0

Using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

where a = 2, b = 1.11, and c = -3.091x10^-9, we can calculate the solubility.

After substituting the values into the formula, solving for x, and taking the positive root, we can find the solubility of Pn(CO3)2 in an aqueous solution of 1.11 M Pn(SO4)2.