To find the molar solubility of a salt in water, we need to use the solubility product constant (Ksp) and the stoichiometry of the salt.
For the general formula MX2, let's assume that the salt dissociates into M^2+ cations and 2X^- anions. Thus, the solubility product expression for this salt is:
Ksp = [M^2+][X^-]^2
Since the stoichiometry of the salt is 1:2 (1 M^2+ cation and 2 X^- anions), the molar solubility of the salt can be represented as "s" in terms of moles per liter:
[M^2+] = s [X^-] = 2s
Substituting these values into the solubility product expression, we have:
Ksp = (s)(2s)^2 Ksp = 4s^3
Now, we can solve for "s":
4s^3 = 5.56 x 10^(-10)
Dividing both sides of the equation by 4:
s^3 = (5.56 x 10^(-10)) / 4 s^3 = 1.39 x 10^(-10)
Taking the cube root of both sides:
s = (1.39 x 10^(-10))^(1/3) s = 1.38 x 10^(-3)
Therefore, the molar solubility of the salt MX2 in pure water is approximately 1.38 x 10^(-3) mol L^(-1).