The solubility of sparingly soluble salt A3 B2 (molar mass = 'M' g / mol ) in water is 'x' g/L.

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The solubility of sparingly soluble salt A3 B2 (molar mass...

The solubility of sparingly soluble salt $\mathrm{A}_{3} \mathrm{~B}_{2}$ (molar mass $=$ 'M' $\mathrm{g} / \mathrm{mol}$ ) in water is 'x' g/L. The ratio of molar concentration of $\mathrm{B}^{3}$ to the solubility product of the salt is (a) $\frac{108 x^{5}}{M^{5}}$ (b) $\frac{x^{4}}{108 M^{4}}$ (c) $\frac{x^{4}}{54 M^{4}}$ (d) $\frac{x^{3}}{27 M^{3}}$

The solubility of sparingly soluble salt $\mathrm{A}_{3} \mathrm{~B}_{2}$ (molar mass $=$ 'M' $\mathrm{g} / \mathrm{mol}$ ) in water is 'x' g/L. The ratio of molar concentration of $\mathrm{B}^{3}$ to the solubility product of the salt is
(a) $\frac{108 x^{5}}{M^{5}}$
(b) $\frac{x^{4}}{108 M^{4}}$
(c) $\frac{x^{4}}{54 M^{4}}$
(d) $\frac{x^{3}}{27 M^{3}}$

Key Concepts

Sparingly Soluble Salts

These salts have very limited solubility in water, and they establish a dynamic equilibrium between the undissolved solid phase and the ions dissolved in solution. Understanding sparingly soluble salts involves analyzing their dissolution equilibria which often require careful application of the solubility product constant (K??).

Dissolution Stoichiometry

The stoichiometry of the dissolution process describes how each formula unit of the salt dissociates into ions in solution. For a salt with the formula A?B?, the molar ratios of the resulting ions are determined by the coefficients in the formula. This stoichiometric relationship is crucial for relating the solubility in moles per liter to the concentrations of individual ions.

Solubility Product (K??)

The solubility product, K??, is an equilibrium constant that quantifies the extent of dissolution for sparingly soluble salts. It is defined as the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient. This constant is key in determining whether precipitation or complete dissolution occurs in a solution.

Conversion from Mass Solubility to Molar Solubility

Converting solubility expressed in grams per liter (g/L) to molar solubility (mol/L) requires dividing the mass solubility by the molar mass of the salt. This conversion is essential in problems involving dissolution equilibria, as it allows the use of molar ratios and the solubility product concept for quantitative analysis.

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