Find the solubility of Cr(OH)3 in a buffer of pH 13.0[Ksp. of Cr(OH)3=6.3 ×10^-31 ; Kf of Cr r O

step 1 This question is a solubility calculation for a particular salt that is insoluble found in a buf

Find the solubility of Cr(OH)3 in a buffer of pH 13.0[Ksp....

Find the solubility of $\mathrm{Cr}(\mathrm{OH})_{3}$ in a buffer of $\mathrm{pH} 13.0\left[K_{\mathrm{sp}}\right.$ of $\mathrm{Cr}(\mathrm{OH})_{3}=6.3 \times 10^{-31} ; K_{\mathrm{f}}$ of $\mathrm{Cr} \mathrm{r} \mathrm{OH} )_{4}^{-}=8.0 \times 10^{99} ]$

Key Concepts

Solubility Product (Ksp)

The solubility product constant is an equilibrium constant that applies to the dissolution of a sparingly soluble ionic compound. It is defined as the product of the concentrations of the constituent ions each raised to the power of their stoichiometric coefficients. In this context, Ksp for Cr(OH)3 represents the equilibrium concentration of Cr3+ and OH– produced upon its dissolution, and its very small value indicates extremely low solubility under normal conditions.

Complex Ion Formation (Kf)

The formation constant, or stability constant (Kf), quantifies the formation of a complex ion from a metal ion and one or more ligands. A high Kf value signifies that the reaction strongly favors the complex ion formation. In the given problem, the complex ion formation of Cr(OH)4– with hydroxide ions significantly increases the effective solubility of Cr(OH)3 since the metal ion is 'tied up' in the complex, driving the dissolution equilibrium further.

pH and Buffer Effects

pH, a measure of acidity or basicity of a solution, directly affects the concentration of hydroxide ions. In a buffer system at pH 13, the hydroxide ion concentration is high. This increased concentration plays two critical roles: it shifts the solubility equilibrium of Cr(OH)3 by the common ion effect, and it provides the ligand needed for the formation of the complex ion. Thus, the buffer conditions significantly modify the apparent solubility of the sparingly soluble salt.

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