Question

Use the Solver function in Excel to determine [Sr₂+], [F−], [HF], [OH−], and [H+]for a saturated solution of SrF₂ The pKsppKsp for SrF₂ is 8.58 The pKa for HF is 3.17 Ignore activity coefficients.

Answer 1

In a saturated solution of SrF₂, the concentrations of [Sr²+] and [F-] are approximately 2 M and 4 M, respectively. The concentrations of [HF], [OH-], and [H+] cannot be determined without additional information.

Step-by-step explanation:

The solubility product constant, Ksp, is given as 8.58. In a saturated solution of SrF₂, the equilibrium expression can be written as [Sr²+][F-]² = Ksp. Since SrF₂ dissociates into Sr²+ and 2F-, we can substitute [Sr²+] with x and [F-] with 2x. Therefore, the equilibrium expression becomes x(2x)² = Ksp.

Simplifying the equation further, we have 4x³ = Ksp. Using the given Ksp value of 8.58, we can solve for x. Taking the cube root of 8.58 gives us x ≈ 2. In a saturated solution of SrF₂, the concentration of [Sr²+] would be approximately 2 M, and the concentration of [F-] would be 2(2) = 4 M.

Since [HF] and [OH-] are not directly involved in the dissolution of SrF₂, their concentrations in the saturated solution will depend on other factors.

Without additional information, we cannot determine the exact concentrations of [HF] and [OH-]. The concentration of [H+] can be calculated using the pH scale and the pKa value of HF. However, since the pKa value is not provided in the question, we cannot determine the exact concentration of [H+].