Calculate the molar solubility of Cr(OH)3 in 0.50 M NaOH; Kf for Cr(OH)4- is 8 x 10^29.

Verified step by step guidance

1. Identify the relevant chemical equilibria: Cr(OH)_3(s) \rightleftharpoons Cr^{3+}(aq) + 3OH^-(aq) and Cr^{3+}(aq) + 4OH^-(aq) \rightleftharpoons Cr(OH)_4^-(aq).
2. Write the expression for the solubility product (K_{sp}) of Cr(OH)_3: K_{sp} = [Cr^{3+}][OH^-]^3.
3. Write the expression for the formation constant (K_f) of Cr(OH)_4^-: K_f = \frac{[Cr(OH)_4^-]}{[Cr^{3+}][OH^-]^4}.
4. Assume the initial concentration of OH^- from NaOH is 0.50 M and set up the equilibrium expressions considering the formation of Cr(OH)_4^-.
5. Combine the K_{sp} and K_f expressions to solve for the molar solubility of Cr(OH)_3 in the presence of 0.50 M NaOH.

Key Concepts

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at a specific temperature, expressed in moles per liter (M). It is a crucial concept in understanding how substances interact in solution and is often determined through equilibrium expressions involving solubility product constants (Ksp). In this case, we need to calculate the molar solubility of Cr(OH)3 in the presence of NaOH.

Common Ion Effect

The common ion effect describes the decrease in solubility of a salt when a common ion is added to the solution. In this scenario, the addition of NaOH introduces OH- ions, which shifts the equilibrium of the dissolution of Cr(OH)3, reducing its solubility. Understanding this effect is essential for calculating the new molar solubility in the presence of a strong base like NaOH.

Formation Constant (Kf)

The formation constant (Kf) quantifies the stability of a complex ion formed in solution. For Cr(OH)4-, the Kf value indicates how favorably Cr(OH)4- forms from Cr(OH)3 and OH- ions. This constant is critical for determining the extent of complexation in the solution, which influences the overall solubility of Cr(OH)3 when NaOH is present.