Question

At \(25^{\circ} \mathrm{C}, 10.24 \mathrm{mg}\) of \(\mathrm{Cr}(\mathrm{OH})_{2}\) are dissolved in enough water to make \(125 \mathrm{~mL}\) of solution. When equilibrium is established, the solution has a pH of 8.49. Estimate \(K_{\mathrm{sp}}\) for \(\mathrm{Cr}(\mathrm{OH})_{2}\)

Short answer

Question: Estimate the Ksp value for Cr(OH)₂ given that 10.24 mg of Cr(OH)₂ is dissolved in 0.125 L of water and the resulting solution has a pH of 8.49. Answer: The estimated Ksp value for Cr(OH)₂ is approximately 1.51 × 10⁻¹².

Step-by-step solution

Calculate the concentration of OH⁻ ions from the given pH

We are given the pH of the solution, which is 8.49. To find the concentration of OH⁻ ions, we first need to find the pOH by using the relation pH + pOH = 14. pOH = 14 - pH = 14 - 8.49 = 5.51 Now, we can calculate the concentration of OH⁻ ions using the definition of pOH: pOH = -log([OH⁻]) Rearranging the equation to solve for [OH⁻]: [OH⁻] = 10^(-pOH) = 10^(-5.51)

Calculate the concentration of Cr²⁺ ions

For each mole of Cr(OH)₂ that dissolves, one mole of Cr²⁺ ions and two moles of OH⁻ ions are formed as per the reaction: Cr(OH)₂(s) ⇌ Cr²⁺(aq) + 2OH⁻(aq) The initial moles of Cr(OH)₂ can be found by converting the given mass (10.24 mg) to moles: initial moles of Cr(OH)₂ = (10.24 * 10⁻³) g / (160.0 g/mol) = 6.4 * 10⁻⁵ mol Now, calculate the initial concentration of Cr(OH)₂ by dividing the initial moles by the volume of the solution (0.125 L): Initial [Cr(OH)₂] = (6.4 * 10⁻⁵) mol / 0.125 L = 5.12 * 10⁻⁴ M Since each mole of Cr(OH)₂ produces one mole of Cr²⁺ ions, the equilibrium concentration of Cr²⁺ ions will be the same: [Cr²⁺] = 5.12 * 10⁻⁴ M

Calculate the Ksp value

Now that we have the equilibrium concentrations of Cr²⁺ and OH⁻ ions, we can plug them into the solubility product expression for Cr(OH)₂: Ksp = [Cr²⁺][OH⁻]² Using the calculated concentrations: Ksp = (5.12 * 10⁻⁴ M)(10^(-5.51))^2 Ksp ≈ 1.51 * 10⁻¹² So, the estimated Ksp value for Cr(OH)₂ is approximately 1.51 × 10⁻¹².