Problem 276 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(25^{\circ} \mathrm{C}\), the solubility product of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1.0 \times 10^{-11} .\) At which \(\mathrm{pH}\), will \(\mathrm{Mg}^{2+}\) ions start precipitating in the form of \(\mathrm{Mg}(\mathrm{OH})_{2}\) from a solution of \(0.001\) M \(\mathrm{Mg}^{2+}\) ions? \([2010]\) (a) 9 (b) 10 (c) 11 (d) 8

Step-by-Step Solution

Answer

The pH at which \(\text{Mg}^{2+}\) ions start precipitating is 10.

Step 1: Understanding the Solubility Product

The problem gives the solubility product (Ksp) of \(\text{Mg(OH)}_2\) as \(1.0 \times 10^{-11}\). This means that at equilibrium, the ion product of \(\text{Mg}^{2+}\) and \(\text{OH}^-\)^2 must equal \(1.0 \times 10^{-11}\).

Step 2: Writing the Ksp Expression

The dissociation of \(\text{Mg(OH)}_2\) in water is given by: \[\text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq)\] The solubility product expression is: \[K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\].

Step 3: Substitute Known Values

We know the concentration of \([\text{Mg}^{2+}] = 0.001\, \text{M}\), so the Ksp expression becomes: \[1.0 \times 10^{-11} = 0.001 \times [\text{OH}^-]^2\].

Step 4: Solve for [OH]-

Rearrange the expression to solve for \([\text{OH}^-]^2\): \[[\text{OH}^-]^2 = \frac{1.0 \times 10^{-11}}{0.001} = 1.0 \times 10^{-8}\]Solving this gives \([\text{OH}^-] = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4}\, \text{M}\).

Step 5: Calculate pOH

The \(\text{pOH}\) is calculated using the formula: \[\text{pOH} = -\log( [\text{OH}^-]) = -\log(1.0 \times 10^{-4})\]Thus, \(\text{pOH} = 4\).

Step 6: Convert pOH to pH

Use the relation \( \text{pH} + \text{pOH} = 14 \), so: \[\text{pH} = 14 - \text{pOH} = 14 - 4 = 10\].

Step 7: Conclusion

Therefore, at \(\text{pH} = 10\), \(\text{Mg}^{2+}\) ions will start precipitating as \(\text{Mg(OH)}_2\).

Key Concepts

Precipitation

Precipitation occurs when a solid forms and separates out from a solution. This happens when the concentration of ions in the solution exceeds a certain level, known as the solubility product ( K_{sp} ). For example, in a solution containing magnesium ions ( Mg^{2+} ) and hydroxide ions ( OH^- ), precipitation of magnesium hydroxide ( Mg(OH)_2 ) occurs if the product of the ion concentrations exceeds its K_{sp} value.

Precipitation is crucial in chemical processes. In water treatment, it's used to remove contaminants by converting dissolved substances into solids, which can be easily filtered out. Similarly, precipitation reactions are used in laboratories to separate different ions in a mixture by selectively precipitating specific compounds.

Dissociation

Dissociation is the process by which a compound breaks down into its constituent ions in a solution. For example, magnesium hydroxide, when dissolved in water, dissociates into magnesium ions (Mg^{2+}) and hydroxide ions (OH^-). The chemical equation for this process is:

\[\text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq)\]The extent of dissociation impacts the solution's conductivity, acidity, and reactivity. Factors such as temperature, solvent nature, and the original concentration of the solute influence dissociation. Understanding dissociation is important for manipulating conditions to favor the formation of desired products, such as in industrial chemical synthesis.

pH and pOH relationship

The pH and pOH scale measure hydrogen ion concentration ([H^+]) and hydroxide ion concentration ([OH^-]) in a solution, respectively. The relationship between pH and pOH is given by the equation:

• \[ \text{pH} + \text{pOH} = 14 \]

At 25°C, this equation holds due to the ion product of water being constant. The equation allows conversion between pH and pOH, helping to determine the acidity or basicity of a solution.

A solution with a pH less than 7 is acidic, while a pH greater than 7 is basic. If the pH equals 7, it indicates a neutral solution. This understanding is pivotal for chemical reactions, environmental science, and even everyday processes like cooking and cleaning.

Ionic equilibrium

Ionic equilibrium refers to the state in a solution where the rate of dissociation of ions equals the rate of their recombination. In this state, the concentration of ions remains constant over time. This concept is integral in reactions involving weak acids or bases and slightly soluble salts.

In the case of Mg(OH)_2 in water, equilibrium is reached as it partially dissociates into Mg^{2+} and OH^- ions. The concentration of these ions determines the solvolytic equilibrium. Meanwhile, the solubility product constant ( K_{sp} ) provides a measure of how far the dissociation proceeds.

• Equilibrium helps predict the direction of a reaction.
• Considerations of K_{sp} can determine when precipitation occurs.

Understanding ionic equilibrium is key for controlling reactions in various scientific and industrial processes.