What is the molar solubility of barium phosphate in a 0.282M barium acetate solution?

If I set up and ice table I would have 0.282M - X = ? Would I use the molar mass of barium acetate to complete the equation to find X?

Ba3(PO4)2 ==> 3Ba^2+ + 2PO4^3-

Ba(C2H3O2)2 ==> Ba^2+ + 2C2H3O2^-
(Ba^2+) = 3x from Ba3(PO4)2 + 0.282 from Ba(C2H3O2)
(PO4^3-) = 2x from Ba3(PO4)2
Substitute into Ksp and solve for x = molar solubility Ba3(PO4)2.

To determine the molar solubility of barium phosphate in a 0.282M barium acetate solution, you will need to consider the solubility product constant (Ksp) for barium phosphate.

The solubility product constant expression for barium phosphate (Ba3(PO4)2) is:

Ksp = [Ba2+][PO42-]^2

First, write the equation for the dissociation of barium acetate in water:

Ba(C2H3O2)2 -> Ba2+ + 2C2H3O2-

Since the barium acetate is dissociated in water, the concentration of barium ion ([Ba2+]) in the solution is equal to the initial concentration of barium acetate, which is 0.282M.

Now, let's assume the molar solubility of barium phosphate is "x" mol/L. The concentration of phosphate ion ([PO42-]) is then 2x mol/L, based on the stoichiometry of the balanced equation.

Therefore, the solubility product expression can be rewritten as:

Ksp = (0.282)(x)(2x)^2

Now, you can solve for the molar solubility (x) by rearranging the equation and solving for x.

To determine the molar solubility of barium phosphate in a 0.282M barium acetate solution, you need to consider the relevant equilibrium equation and the solubility product constant (Ksp) for barium phosphate.

The solubility equilibrium equation for barium phosphate is as follows:
Ba3(PO4)2(s) ⇌ 3Ba2+(aq) + 2PO42-(aq)

Let's assume that the molar solubility of barium phosphate is represented by 'x'. Thus, the concentrations of Ba2+ and PO42- ions in the equilibrium state will both be 3x and 2x, respectively.

Considering the 0.282M barium acetate solution, you can use an ICE (Initial, Change, Equilibrium) table to set up the equation correctly:

Ba(C2H3O2)2(aq) ⇌ Ba2+(aq) + 2C2H3O2-(aq)

Here's the ICE table to represent the dissociation of barium acetate:

Ba(C2H3O2)2(aq) ⇌ Ba2+(aq) + 2C2H3O2-(aq)
Initial: 0.282M 0 0
Change: -x +x +2x
Equilibrium: 0.282M - x x 2x

To find the value of 'x' (molar solubility of barium phosphate), you need to consider the relationship between the concentrations of barium ions (Ba2+) and phosphate ions (PO42-) indicated by the balanced equation of barium phosphate. In this case, the ratio is 3:2.

Since the concentration of Ba2+ is 3 times the concentration of barium phosphate (3x), and the concentration of PO42- is twice the concentration of barium phosphate (2x), you can write the following equation:

(3x)^3 * (2x)^2 = Ksp

Now, you can insert the value of the solubility product constant (Ksp) for barium phosphate to solve for 'x' and determine the molar solubility of barium phosphate in the 0.282M barium acetate solution.