what would be the concentration of chromate ion, CrO4^-2, in a saturated solution of CaCrO4. (Ksp= 7.1x10^-4)

To find the concentration of chromate ion, you need to first determine the solubility of CaCrO4 and then use the stoichiometry of the dissociation reaction to calculate the concentration of CrO4^2-.

The solubility product constant (Ksp) of CaCrO4 is given as 7.1x10^-4. The dissociation reaction for the dissolution of CaCrO4 in water is as follows:
CaCrO4(s) ⇌ Ca^2+(aq) + CrO4^2-(aq)

Let's assume that 'x' is the molar solubility of CaCrO4 in moles per liter (mol/L).

Using the stoichiometry of the reaction, the concentration of CrO4^2- in terms of 'x' can be expressed as 2x, as there is a 1:1 ratio between CaCrO4 and CrO4^2- ions.

Therefore, the expression for the solubility product (Ksp) can be written as:
Ksp = [Ca^2+][CrO4^2-] = (x)(2x)^2 = 4x^3

Substituting the given Ksp value:
7.1x10^-4 = 4x^3

To solve this equation, we can take the cube root of both sides:
∛(7.1x10^-4) = ∛(4x^3)

0.091 = 2x

To find the concentration of CrO4^2-, multiply 'x' by 2:
Concentration of CrO4^2- = 2x = 2(0.091) = 0.182 M

Therefore, the concentration of chromate ion, CrO4^2-, in a saturated solution of CaCrO4 would be 0.182 M.

To find the concentration of chromate ion, CrO4^-2, in a saturated solution of CaCrO4, you'll need to use the solubility product constant (Ksp) and the stoichiometry of the balanced equation. Here are the steps to calculate the concentration:

Step 1: Write the balanced equation for the dissolution of CaCrO4.
The balanced equation is: CaCrO4(s) ⇌ Ca^2+(aq) + CrO4^-2(aq)

Step 2: Express the Ksp expression for CaCrO4.
Ksp = [Ca^2+][CrO4^-2]

Step 3: Set up the solubility equilibrium expression and define the unknowns.
Let x be the concentration of Ca^2+ and CrO4^-2 in mol/L.
Ksp = x · x = x^2

Step 4: Substitute the Ksp value into the equation.
7.1x10^-4 = x^2

Step 5: Solve for x.
Take the square root of both sides of the equation.
√(7.1x10^-4) = x

Step 6: Calculate the concentration of chromate ion.
Since the concentration of chromate ion is equal to x, substituting the value we calculated in step 5:
[CrO4^-2] = √(7.1x10^-4)

So, the concentration of chromate ion, CrO4^-2, in a saturated solution of CaCrO4 is √(7.1x10^-4).

See my responses to your other posts.