Final Answer:
The concentration of Ba²⁺ ions and SO₄²⁻ ions in the saturated solution is approximately 0.01 M at 25 degrees C.
Step-by-step explanation:
In a saturated solution of BaSO₄, the solubility product constant (K_sp) is given by the product of the concentrations of the ions formed in the solution. The balanced chemical equation for the dissociation of BaSO₄ is BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. Let x represent the molar solubility of BaSO₄. Thus, the concentrations of Ba²⁺ and SO₄²⁻ ions in the saturated solution are both x M.
The expression for the solubility product constant is given by K_sp = [Ba²⁺][SO₄²⁻]. Substituting the expressions for the concentrations of the ions, we get K_sp = x * x. Given that K_sp is 10⁻⁶, we can solve for x: x² = 10⁻⁶. Taking the square root of both sides, x = 10⁻³. Therefore, the molar solubility of BaSO₄ is 10⁻³ M.
Since BaSO₄ dissociates into one Ba²⁺ ion and one SO₄²⁻ ion, the concentrations of Ba²⁺ and SO₄²⁻ ions in the saturated solution are both 10⁻³ M. Thus, the final answer is that the concentration of Ba²⁺ ions and SO₄²⁻ ions in the saturated solution is approximately 0.01 M at 25 degrees C.
This means that in equilibrium, 0.01 moles of Ba²⁺ and 0.01 moles of SO₄²⁻ ions coexist per liter of the solution. The solution is saturated, and any further addition of BaSO₄ will lead to precipitation, as the product of the ion concentrations will exceed the solubility product constant.