What is the molar solubility of Sr3(AsO4)2 and the concentration of Sr^+2 and AsO4^-3 in a saturated solution at 25C? the Ksp is 1.3 E -18. Show the balanced reaction.

........Sr3(AsO4)2 ==> 3Sr^2+ + AsO4^2-

initial....0.............0........0
change..................3x.......2x
equil...................3x.......2x
Ksp = (Sr^2+)^3(AsO4^2-)^2
Substitute and solve for x, then 2x and 3x.

To find the molar solubility of Sr3(AsO4)2 and the concentration of Sr^+2 and AsO4^-3 in a saturated solution at 25°C, we first need to write the balanced reaction and then use the solubility product constant (Ksp) to calculate the concentrations.

The balanced reaction for the dissolution of Sr3(AsO4)2 in water is:

Sr3(AsO4)2(s) ⇌ 3Sr^2+(aq) + 2AsO4^3-(aq)

According to this reaction, one mole of Sr3(AsO4)2 solid dissociates to yield three moles of Sr^2+ ions and two moles of AsO4^3- ions.

Let's assume the molar solubility of Sr3(AsO4)2 is "x". Therefore, at equilibrium, the concentration of Sr^2+ ions will be 3x, and the concentration of AsO4^3- ions will be 2x.

The solubility product constant (Ksp) expression can be written as:

Ksp = [Sr^2+]^3 * [AsO4^3-]^2

We can substitute the concentrations into the Ksp expression:

Ksp = (3x)^3 * (2x)^2
1.3E-18 = 54x^3 * 4x^2

Simplifying the equation:

1.3E-18 = 216x^5

Now, we can solve for "x" by taking the fifth root of both sides and then calculating the value:

x = (1.3E-18)^(1/5)
x ≈ 3.04E-4

Therefore, the molar solubility of Sr3(AsO4)2 in a saturated solution at 25°C is approximately 3.04 × 10^-4 mol/L.

The concentration of Sr^2+ ions in the saturated solution is 3 times the molar solubility, so it is approximately 9.12 × 10^-4 mol/L (3 × 3.04 × 10^-4).

The concentration of AsO4^3- ions in the saturated solution is 2 times the molar solubility, so it is approximately 6.08 × 10^-4 mol/L (2 × 3.04 × 10^-4).

To find the molar solubility of Sr3(AsO4)2 and the concentration of Sr^+2 and AsO4^-3 in a saturated solution, we need to use the given Ksp (solubility product constant) and the balanced reaction.

The balanced reaction for the dissociation of Sr3(AsO4)2 in water is as follows:
3 Sr3(AsO4)2(s) ⇌ 6 Sr^2+(aq) + 2 AsO4^3-(aq)

From the balanced reaction, we can see that each formula unit of Sr3(AsO4)2 produces 6 moles of Sr^2+ ions and 2 moles of AsO4^3- ions.

Let's assume that the molar solubility of Sr3(AsO4)2 is represented by "x".

Therefore, the concentration of Sr^2+ ions in a saturated solution will be 6x (since there are 6 moles of Sr^2+ ions for every mole of Sr3(AsO4)2 that dissolves).

The concentration of AsO4^3- ions in a saturated solution will be 2x (since there are 2 moles of AsO4^3- ions for every mole of Sr3(AsO4)2 that dissolves).

Now, we can set up the equation for the solubility product expression using the given Ksp value:
Ksp = (Sr^2+)^6 * (AsO4^3-)^2

Substituting the concentrations, we have:
Ksp = (6x)^6 * (2x)^2

Simplifying the equation:
Ksp = 36x^6 * 4x^2
Ksp = 144x^8

Now, we can solve the equation to find the value of x:
x = √(Ksp/144)

Substituting the given Ksp value into the formula:
x = √(1.3E-18/144)

Evaluating the expression, we find:
x ≈ 3.29E-7

So, the molar solubility of Sr3(AsO4)2 is approximately 3.29E-7 M.

The concentration of Sr^2+ ions is 6 times the molar solubility:
Sr^2+ concentration = 6 * 3.29E-7 = 1.97E-6 M.

The concentration of AsO4^3- ions is 2 times the molar solubility:
AsO4^3- concentration = 2 * 3.29E-7 = 6.58E-7 M.

Therefore, in a saturated solution at 25°C, the molar solubility of Sr3(AsO4)2 is approximately 3.29E-7 M, the concentration of Sr^2+ ions is approximately 1.97E-6 M, and the concentration of AsO4^3- ions is approximately 6.58E-7 M.