Problem 72 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

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) \(\frac{x^{3}}{27 M^{3}}\)

Step-by-Step Solution

Answer

\(\frac{x^{4}}{54 M^{4}}\)

Step 1: Write the Dissolution Equation

Start by writing the balanced chemical equation for the dissolution of the salt \(\mathrm{A}_{3} \mathrm{B}_{2}\) in water: \(\mathrm{A}_{3} \mathrm{B}_{2}(s) \rightleftharpoons 3\mathrm{A}^{+}(aq) + 2\mathrm{B}^{3+}(aq)\).

Step 2: Express Molar Solubility

If the solubility of the salt is 'x' g/L, convert this to molar solubility. Molar solubility is the number of moles of the salt that can dissolve in 1 L of water, which is given by the formula \(s = \frac{x}{M}\) moles/L, where 'M' is the molar mass of the salt.

Step 3: Determine Molar Concentrations

Using the stoichiometry from the dissolution equation, the molar concentration of \(\mathrm{A}^{+}\) will be \(3s\) and the molar concentration of \(\mathrm{B}^{3+}\) will be \(2s\), where \(s\) is the molar solubility.

Step 4: Write the Expression for Solubility Product

The solubility product (Ksp) for the salt is the product of the concentrations of the ions each raised to the power of their coefficients in the balanced equation: \(K_{sp} = [\mathrm{A}^{+}]^{3} \times [\mathrm{B}^{3+}]^{2} = (3s)^{3} \times (2s)^{2}\).

Step 5: Calculate Ratio of Molar Concentration of \(\mathrm{B}^{3+}\) to Solubility Product

Now, calculate the ratio of molar concentration of \(\mathrm{B}^{3+}\) to the solubility product: \(\frac{[\mathrm{B}^{3+}]}{K_{sp}} = \frac{2s}{(3s)^{3} \times (2s)^{2}} = \frac{2s}{27s^{3} \times 4s^{2}} = \frac{2s}{108s^{5}} = \frac{1}{54s^{4}}\).

Step 6: Substitute Molar Solubility Value

Substitute \(s = \frac{x}{M}\) into the ratio to find the relationship with the solubility 'x' and molar mass 'M': \(\frac{1}{54\left(\frac{x}{M}\right)^{4}} = \frac{M^{4}}{54x^{4}}\).

Key Concepts

Chemical Equilibrium

In chemistry, chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. When a sparingly soluble salt like \( \mathrm{A}_{3} \mathrm{~B}_{2} \) dissolves in water, it establishes a dynamic equilibrium with the solid phase and its dissolved ions in the solution. At this stage, the rate at which the solid salt transforms into ions is equal to the rate at which the ions combine to form the solid salt. This point of balance is essential to determining the solubility product, which is a unique constant for a given temperature and reflects how much of the salt can dissolve to form a saturated solution.

Understanding chemical equilibrium in the context of solubility helps us to predict whether a precipitate will form when different solutions are mixed, as well as to calculate the extent of dissolution in sparingly soluble salts.

Molar Solubility

Molar solubility is a measure of the ability of a compound to dissolve in solution and is defined as the number of moles of the compound that can dissolve in one liter of solution to form a saturated solution. For \( \mathrm{A}_{3} \mathrm{~B}_{2} \), if 'x' grams of the salt dissolve in one liter of water, we convert this mass to moles by dividing by the molar mass 'M' of the salt, obtaining \( s = \frac{x}{M} \) moles/L.

The concept of molar solubility is vital in determining how much of a substance will dissolve before reaching a point where the additional substance will start to form a precipitate, and also in calculations involving the solubility product.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In the case of \( \mathrm{A}_{3} \mathrm{~B}_{2} \) dissolving in water, the stoichiometry of the dissolution equation tells us for every one formula unit of \( \mathrm{A}_{3} \mathrm{B}_{2} \), three \( \mathrm{A}^{+} \) ions and two \( \mathrm{B}^{3+} \) ions are produced in solution. This means for molar solubility 's' of the salt, the concentration of \( \mathrm{A}^{+} \) ions will be \( 3s \) mol/L, while the concentration of \( \mathrm{B}^{3+} \) ions will be \( 2s \) mol/L.

Stoichiometry is essential for understanding many aspects of chemistry, including the calculation of solubility product, molar solubility, and the subsequent reactions that take place in solution.

Dissolution Equations

Dissolution equations represent the process by which a solid dissolves into its constituent ions in a solvent like water. For our salt \( \mathrm{A}_{3} \mathrm{~B}_{2} \), the balanced dissolution equation is \( \mathrm{A}_{3} \mathrm{B}_{2}(s) \rightleftharpoons 3\mathrm{A}^{+}(aq) + 2\mathrm{B}^{3+}(aq) \). The double arrow signifies the establishment of chemical equilibrium, where the forward reaction is the dissolution of the solid into ions, and the reverse reaction is the reformation of the solid from the ions.

Such dissolution equations are the starting point for calculating various parameters like the solubility product, understanding the stoichiometry of the reaction, and determining how the ions interact with each other and with other substances in solution.