Problem 270 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility product of a salt having general formula \(\mathrm{MX}_{2}\) in water is, \(4 \times 10^{-12}\) \([2005]\) The concentration of \(\mathrm{M}^{2+}\) ions in the aqueous solution of the salt is (a) \(1.6 \times 10^{-4} \mathrm{M}\) (b) \(2.0 \times 10^{-6} \mathrm{M}\) (c) \(1.0 \times 10^{-4} \mathrm{M}\) (c) \(4.0 \times 10^{-10} \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of \( \mathrm{M}^{2+} \) ions is \( 1.0 \times 10^{-4} \) M. (c)

Step 1: Understand the Dissolution Equation

The dissolution of the salt with the formula \( \mathrm{MX}_{2} \) in water can be represented as: \[ \mathrm{MX}_2 \rightarrow \mathrm{M}^{2+} + 2\mathrm{X}^- \]. In this equation, 1 mole of \( \mathrm{MX}_2 \) dissolves to produce 1 mole of \( \mathrm{M}^{2+} \) ions and 2 moles of \( \mathrm{X}^- \) ions.

Step 2: Express Solubility in Terms of \( s \)

Let \( s \) be the molar solubility of \( \mathrm{MX}_2 \). When \( \mathrm{MX}_2 \) dissolves, it produces \( s \) moles of \( \mathrm{M}^{2+} \) ions and \( 2s \) moles of \( \mathrm{X}^- \) ions per liter of solution.

Step 3: Write the Expression for the Solubility Product

The solubility product constant (\( K_{sp} \)) for \( \mathrm{MX}_2 \) is given as \( 4 \times 10^{-12} \). The expression for \( K_{sp} \) is \[ K_{sp} = [\mathrm{M}^{2+}][\mathrm{X}^-]^2 \]. Substituting \( s \) for \( [\mathrm{M}^{2+}] \) and \( 2s \) for \( [\mathrm{X}^-] \), we get: \[ K_{sp} = s(2s)^2 = 4s^3 \].

Step 4: Substitute and Solve for \( s \)

Substitute \( K_{sp} = 4 \times 10^{-12} \) into the equation: \[ 4s^3 = 4 \times 10^{-12} \]. Dividing both sides by 4, we get \[ s^3 = 10^{-12} \]. Solving for \( s \), we take the cube root of both sides \[ s = (10^{-12})^{1/3} = 10^{-4} \].

Step 5: Determine the Concentration of \( \mathrm{M}^{2+} \)

The concentration of \( \mathrm{M}^{2+} \) ions is the value of \( s \), which is \( 1.0 \times 10^{-4} \) M. Thus, the correct option is the concentration of \( \mathrm{M}^{2+} \) ions in the aqueous solution.

Key Concepts

Ionic Equilibrium

In the world of chemistry, ionic equilibrium deals with the balance between ions in a solution. When salts like \( \mathrm{MX}_2 \) dissolve in water, they dissociate into ions. In our example, \( \mathrm{MX}_2 \) breaks apart to form \( \mathrm{M}^{2+} \) and \( 2\mathrm{X}^- \) ions. Ionic equilibrium is all about understanding how these ions interact and exist in balance within the solution.
By knowing the amount of these ions, we can predict properties of the solution. The concentration and interaction of ions directly affect the solution's electrical conductivity and overall chemical behavior.

• The balance of ions is crucial for many natural processes.
• In the chemical world, ionic equilibrium helps us to understand reactions in aqueous solutions.
• Knowing ionic concentrations allows for predictions of reaction directions and outcomes.

Understanding ionic equilibrium helps chemists to tackle various real-world problems, from industrial processes to biological systems.

Molar Solubility

Molar solubility is a measure of how much of a compound can dissolve in a solvent to reach saturation. For a salt like \( \mathrm{MX}_2 \), the molar solubility, represented by \( s \), is the concentration of the dissolved \( \mathrm{M}^{2+} \) ions in a saturated solution.
This concept is pivotal when we want to know how much of a salt can be dissolved in a solution before it starts precipitating. Here, the molar solubility of \( \mathrm{MX}_2 \) is calculated based on the dissolution equation:

• One mole of \( \mathrm{MX}_2 \) results in one mole of \( \mathrm{M}^{2+} \) and two moles of \( \mathrm{X}^- \).
• The saturation point is when no more salt can dissolve, which is quantified with \( s \).
• In our specific example, the calculated molar solubility \( s \) turns out to be \( 1.0 \times 10^{-4} \) M.

The molar solubility is vital for predicting the point at which precipitation occurs, which is important for many scientific and industrial applications.

Chemical Equilibrium

Chemical equilibrium refers to a state in a chemical reaction where the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products over time. In equilibrium, the system is dynamic, i.e., reactants and products are continually being transformed into each other, even though there are no net changes.
For the dissolution of \( \mathrm{MX}_{2} \), equilibrium is reached when the rate of dissolving for \( \mathrm{MX}_2 \) equals the rate at which \( \mathrm{M}^{2+} \) and \( \mathrm{X}^- \) recombine to form undissolved \( \mathrm{MX}_{2} \).

• The balance in concentrations indicates chemical equilibrium.
• Equilibrium does not mean equal concentrations; it means constant concentrations over time.
• Knowledge of the equilibrium state helps in predicting the solubility of salts in different conditions.

Chemical equilibrium is a fundamental concept, as it helps chemists control and predict the outcomes of various reactions and conditions.

Ksp Calculation

The solubility product constant, \( K_{sp} \), is a key value that provides insights into the extent of a salt's solubility in water. For slightly soluble salts like \( \mathrm{MX}_2 \), \( K_{sp} \) quantifies the product of the molar concentrations of the ions produced by the salt:
\[ K_{sp} = [\mathrm{M}^{2+}]\cdot[\mathrm{X}^-]^2 \]
In our example with \( \mathrm{MX}_2 \), \( K_{sp} \) of \( 4 \times 10^{-12} \) can be used to find molar solubility, \( s \). Through the expression \( 4s^3 \), we solve for \( s \) to understand how much \( \mathrm{MX}_2 \) dissolves:

• The calculation starts with substituting the known \( K_{sp} \) value.
• Solving the resulting equation gives us the molar solubility.
• In this particular case, solving the equation shows that \( s = 10^{-4} \).

Calculating \( K_{sp} \) is invaluable for chemists who need to predict solubility and understand solution dynamics in various environments.