The solubility product of BaCL2 is 3.2×10*-11. solubility in Mol L-1 is P×10*-4 . Find value of p

The solubility product (Ksp) of a compound is given by the product of the concentrations of its ions raised to their stoichiometric coefficients. For BaCl2, which dissociates as:

BaCl2 -> Ba2+ + 2Cl-

The solubility product expression is:

Ksp = [Ba2+][Cl-]^2

Given that the solubility of BaCl2 is P x 10^-4 mol/L, we can say that [Ba2+] = P x 10^-4 mol/L and [Cl-] = 2P x 10^-4 mol/L.

Substituting these values into the Ksp expression, we get:

Ksp = (P x 10^-4)(2P x 10^-4)^2 = 4P^3 x 10^-12

Given that Ksp = 3.2 x 10^-11, we can set these two expressions equal to each other and solve for P:

3.2 x 10^-11 = 4P^3 x 10^-12

Divide both sides by 4 x 10^-12 to isolate P^3:

P^3 = (3.2 x 10^-11) / (4 x 10^-12) = 8

Take the cube root of both sides to solve for P:

P = cube root of 8 = 2

So, the value of P is 2.