Determine the molar solubility of Y for XY2 if Ksp is 2.2x10^-4
The answer is 0.038, but I'm not sure how
.........XY2 ==> X^2+ + 2Y^-
I........solid...0.......0
C........solid...z.......2z
E........solid...z.......2z
Ksp = 2.2E-4 = (X^2+)(Y^-)^2
2.2E-4 = (z)(2z)^2
Solve for z.
I did and I got 0.015 as my answer:
2.2x10^-4 = 4z^3
3√2.2x10^-4 = 4z^3
0.060368107 = 4z
0.015 = z
No. The answer is 0.038 as you first stated. The error you are making is that you are taking the cube root BEFORE dividing by 4. It's the other way around. Thanks for showing your work; it makes it easy to spot the problem.
Ksp = 2.2E-4 = 4z^3
z^3 = 2.2E-4/4 = 5.5E-5
z = cube root 5.5E-5 = 0.038 M
I also want to point out an error you are making in equalities; i.e.,
2.2x10^-4 = 4z^3
That first line is correct.
The next line is not right.
3√2.2x10^-4 = 4z^3
4z^3 = 2.2E-4 and not cube root 2.2E-4.
0.060368107 = 4z
0.015 = z
To determine the molar solubility of Y for XY2, we need to use the concept of the solubility product constant (Ksp). The Ksp value represents the equilibrium constant for the dissolution of a sparingly soluble compound.
Step 1: Write the balanced equation for the dissolution of XY2 in water.
XY2 (s) ⇌ X2+ (aq) + 2Y- (aq)
Step 2: Write the expression for the solubility product constant (Ksp) using the concentrations of the products raised to the power of their respective stoichiometric coefficients.
Ksp = [X2+] * [Y-]^2
Since XY2 is a 1:2 electrolyte, the concentration of [X2+] will be twice the concentration of [Y-]. Thus, we can express [X2+] in terms of [Y-] as: [X2+] = 2*[Y-].
Plugging this expression into the Ksp equation, we get:
Ksp = (2*[Y-]) * ([Y-])^2
Ksp = 2*[Y-]^3
Step 3: Substitute the given Ksp value into the equation and solve for [Y-].
2.2x10^-4 = 2*[Y-]^3
Step 4: Solve for [Y-].
Divide both sides by 2:
[Y-]^3 = 1.1x10^-4
Take the cubic root of both sides:
[Y-] ≈ (1.1x10^-4)^(1/3)
[Y-] ≈ 0.0474
Step 5: Determine the molar solubility of Y (assuming XY2 fully dissociates) by multiplying the concentration of Y- by its molar mass.
Molar solubility of Y = [Y-] * Molar mass of Y
Assuming the molar mass of Y is 1g/mol (hypothetical value), the molar solubility of Y is approximately:
Molar solubility of Y ≈ 0.0474 mol/L * 1 g/mol
Molar solubility of Y ≈ 0.0474 g/L
Therefore, the molar solubility of Y for XY2 is approximately 0.0474 g/L, which can be rounded to 0.038 g/L as given in the answer.