The solubility of product constant (ksp) for the dissolution of Pb3(Aso4)2 as represented by the chemical equation is 4.0 x 10^-36
Pb3(AsO4)2 (s)--->3Pb 2+ (aq) + 2 AsO4^3 (aq)
Calculate the molar solubility (mol/L) of Pb3(AsO4)2 in water
how do you do this
nevermind i got it :)
To calculate the molar solubility (mol/L) of Pb3(AsO4)2 in water, we need to set up an ICE table and use the given Ksp value.
Let's assume the molar solubility of Pb3(AsO4)2 is represented by "x".
The balanced chemical equation is:
Pb3(AsO4)2 (s) ⇌ 3Pb2+ (aq) + 2AsO4^3- (aq)
Using the stoichiometry of the balanced equation, we can see that:
1 mol of Pb3(AsO4)2 → 3 mol of Pb2+ ions
1 mol of Pb3(AsO4)2 → 2 mol of AsO4^3- ions
Therefore, the equilibrium concentrations in terms of "x" are:
[Pb2+] = 3x
[AsO4^3-] = 2x
The expression for the solubility product constant (Ksp) is:
Ksp = [Pb2+]^3 * [AsO4^3-]^2
Substituting the equilibrium concentrations, we have:
Ksp = (3x)^3 * (2x)^2
Ksp = 108x^5
Given that the Ksp value is 4.0 x 10^-36, we can set up the equation:
4.0 x 10^-36 = 108x^5
To solve for "x", let's rearrange the equation and take the fifth root of both sides:
x = (4.0 x 10^-36 / 108)^(1/5)
x ≈ 2.06 x 10^-8 mol/L
Hence, the molar solubility of Pb3(AsO4)2 in water is approximately 2.06 x 10^-8 mol/L.
To find the molar solubility of Pb3(AsO4)2 in water, we need to use the Ksp expression and solve for the concentration of Pb2+ ions. Let's break down the steps:
1. Write the balanced equation: Pb3(AsO4)2 (s) ⇌ 3Pb2+ (aq) + 2AsO4^3- (aq)
2. Write the Ksp expression:
Ksp = [Pb2+]^3[AsO4^3-]^2
3. Substitute the given value of Ksp:
4.0 x 10^-36 = [Pb2+]^3 x [AsO4^3-]^2
4. Since we are trying to find the molar solubility of Pb3(AsO4)2, let x be the concentration of Pb2+ ions in mol/L when the compound dissolves.
5. The concentration of Pb2+ ions is then 3x, as indicated by the stoichiometry of the balanced equation.
6. Substitute these values into the Ksp expression:
4.0 x 10^-36 = (3x)^3 x [AsO4^3-]^2
7. Simplify the equation:
4.0 x 10^-36 = 27x^3 x [AsO4^3-]^2
8. Since the compound is assumed to be soluble, the concentration of the anion (AsO4^3-) can be considered constant and equal to the initial concentration of Pb2+ ions.
9. Therefore, [AsO4^3-] = 2x.
10. Substitute this value back into the equation:
4.0 x 10^-36 = 27x^3 x (2x)^2
11. Simplify further:
4.0 x 10^-36 = 108x^5
12. Rearrange the equation and solve for x:
x^5 = (4.0 x 10^-36) / 108
x^5 ≈ 3.7 x 10^-38
13. Take the fifth root of both sides to solve for x:
x ≈ (3.7 x 10^-38)^(1/5)
x ≈ 1.15 x 10^-7 mol/L
Therefore, the molar solubility of Pb3(AsO4)2 in water is approximately 1.15 x 10^-7 mol/L.