The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14} .\) (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?\) \((\mathbf{b} ) \)The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

First, let's write the solubility equilibrium equation for Cd(OH)₂: Cd(OH)₂(s) ⇌ Cd²⁺(aq) + 2OH⁻(aq) Now, let's denote the molar solubility of Cd(OH)₂ as 's'. Therefore, the concentration of Cd²⁺ will be 's' and the concentration of OH⁻ will be '2s' in the solution. The solubility product constant (Kₛₚ) is given by: Kₛₚ = [Cd²⁺][OH⁻]² We have the value of Kₛₚ as 2.5 × 10⁻¹⁴. Substituting the values, we get: \(2.5 \times 10^{-14} = s \times (2s)^2\) Now, solve this equation for 's' to find the molar solubility of Cd(OH)₂.

To find the initial concentration of NaBr, we need to use the complex formation constant (Kf) for the complex ion CdBr₄²⁻. The complex formation reaction can be written as: Cd²⁺(aq) + 4Br⁻(aq) ⇌ CdBr₄²⁻(aq) Now, let's denote the initial concentration of NaBr required as 'x'. Since NaBr is a strong electrolyte, it will completely dissociate in water and will provide 'x' moles of Br⁻ per liter. The final concentration of Br⁻ should be 4 times the increased molar solubility. Now, we can write the expression for the complex formation constant (Kf) as: Kf = [CdBr₄²⁻] / ([Cd²⁺][Br⁻]⁴) We have the value of Kf as 5 × 10³ and are given that the increased molar solubility of Cd(OH)₂ (concentration of Cd²⁺) is 1.0 × 10⁻³ mol/L, which can be written in terms of 'x': Kf = (1.0 × 10⁻³) / ((1.0 × 10⁻³)(x⁴)) Now, solve this equation for 'x' to find the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)₂.