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

To determine the concentration of chromate ion in a saturated solution of CaCrO4, we need to use the solubility product constant, Ksp.

The balanced equation for the dissociation of CaCrO4 in water is:
CaCrO4 (s) ⇌ Ca2+ (aq) + CrO4^2- (aq)

The Ksp expression for the dissociation of CaCrO4 is:
Ksp = [Ca2+][CrO4^2-]

Since CaCrO4 is a 1:1 electrolyte, the concentration of Ca2+ and CrO4^2- ions will be the same when the compound is completely dissociated.

Therefore, let's assume that the concentration of Ca2+ and CrO4^2- ions in the saturated solution is x.

Thus, the Ksp expression becomes:
Ksp = x * x

Now, substitute the Ksp value (7.1 × 10^-4) into the equation:
7.1 × 10^-4 = x^2

Taking the square root of both sides:
x = √(7.1 × 10^-4)

Calculating this value gives:
x = 0.0267 M (rounded to four decimal places)

Therefore, the concentration of chromate ion (CrO4^2-) in a saturated solution of CaCrO4 is approximately 0.0267 M.

To determine the concentration of chromate ion (CrO4^-2) in a saturated solution of CaCrO4, you need to use the solubility product constant (Ksp) and the stoichiometry of the balanced equation.

The balanced equation for the dissociation of CaCrO4 is:
CaCrO4(s) ↔ Ca^2+(aq) + CrO4^2-(aq)

The Ksp expression for this equation is:
Ksp = [Ca^2+][CrO4^2-]

Since CaCrO4 is a sparingly soluble salt, it will dissociate very poorly into Ca^2+ and CrO4^2- ions. Thus, we can assume that the solubility of CaCrO4 is equal to the concentration of Ca^2+ and CrO4^2- ions in the saturated solution.

Let's assume that x represents the concentration of CrO4^2- ions in mol/L. Since the stoichiometry of the balanced equation is 1:1 for CaCrO4 and CrO4^2-, the concentration of Ca^2+ ions will also be x mol/L.

Substituting these values into the Ksp expression:
Ksp = [Ca^2+][CrO4^2-]
7.1x10^-4 = (x)(x)
7.1x10^-4 = x^2

To solve for x, take the square root of both sides:
√(7.1x10^-4) = x

Calculating this, we get:
x ≈ 0.0267 mol/L

Therefore, the concentration of chromate ion (CrO4^2-) in a saturated solution of CaCrO4 is approximately 0.0267 mol/L.

This is done the same way as the other two.