Problem 135 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the \(\mathrm{pH}\) value at which \(\mathrm{Mg}(\mathrm{OH})_{2}\) begins to precipitate from a solution containing \(0.10 \mathrm{M} \mathrm{Mg}^{+2}\) ion? Ksp of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1 \times 10^{-11}\) (a) 3 (b) 6 (c) 9 (d) 11

Step-by-Step Solution

Answer

The \( \text{pH} \) value at which \( \text{Mg(OH)}_2 \) begins to precipitate is 9. Answer: (c) 9.

Step 1: Understand the chemical equilibrium

The compound \( \text{Mg(OH)}_2 \) dissociates as follows: \[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{OH}^- (aq) \] The solubility product constant (\( K_{sp} \)) expression for this equilibrium is \[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \] where \( K_{sp} = 1 \times 10^{-11} \).

Step 2: Calculate the concentration of hydroxide ions

Given that \( [\text{Mg}^{2+}] = 0.10 \text{ M} \), plug this value into the \( K_{sp} \) expression: \[ 1 \times 10^{-11} = (0.10)[\text{OH}^-]^2 \] Isolate \([\text{OH}^-]\): \[ [\text{OH}^-]^2 = \frac{1 \times 10^{-11}}{0.10} = 1 \times 10^{-10} \] Therefore, \([\text{OH}^-] = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \text{ M}\).

Step 3: Calculate the pOH from [OH⁻]

The \( \text{pOH} \) of a solution is calculated using: \[ \text{pOH} = -\log[\text{OH}^-] \] Substitute the concentration of \([\text{OH}^-]\): \[ \text{pOH} = -\log(1 \times 10^{-5}) = 5 \].

Step 4: Convert pOH to pH

\( \text{pH} \) and \( \text{pOH} \) are related by the equation: \[ \text{pH} + \text{pOH} = 14 \] Given \( \text{pOH} = 5 \), calculate \( \text{pH} \): \[ \text{pH} = 14 - 5 = 9 \].

Key Concepts

Solubility Product Constant

The solubility product constant, or \(K_{sp}\), is crucial when examining sparingly soluble salts. It represents the maximum capacity of ions in a saturated solution before precipitation occurs. For a salt like \(\text{Mg(OH)}_2\), which dissociates to \(\text{Mg}^{2+}\) and \(\text{OH}^-\) ions, the \(K_{sp}\) expression comes into play as:

• \[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]

This expression helps predict whether a precipitate will form in the solution. If the product of the ion concentrations exceeds the \(K_{sp}\), precipitation occurs. If it is less, the solution remains homogeneously mixed. Understandably, knowing the \(K_{sp}\) allows chemists to manipulate systems to precipitate specific ions by adjusting concentrations accordingly.
In this exercise, we used the \(K_{sp}\) value of \(1 \times 10^{-11}\) to ascertain the needed concentration of hydroxide ions at which magnesium hydroxide begins to precipitate. This ensures that calculations for systems involving sparingly soluble compounds are accurate and predictable.

Chemical Equilibrium

Chemical equilibrium is a fundamental principle in chemistry where the rate of the forward and reverse reactions become equal, leading to a stable state of reactants and products. For \(\text{Mg(OH)}_2\), we observe:

• \[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{OH}^- (aq) \]

At equilibrium, the rate of dissolution of \(\text{Mg(OH)}_2\) is equal to the rate of precipitation of its ions back into solid form. The system can be described by an equilibrium constant, \(K_{sp}\), for the reaction, representing the product of the ionic concentrations.
Maintaining chemical equilibrium is essential for controlling reactions and predicting outcomes in various chemical processes. It ensures that reactions do not proceed infinitely in one direction but balance out over time. Comprehension of equilibrium concepts allows chemists to manipulate reaction conditions to favor the formation of desired products, such as in synthesis and purification processes.

Precipitation Reaction

A precipitation reaction is when two solutions containing soluble salts combine to form an insoluble salt that precipitates out as a solid. In this exercise, precipitation occurs once the product of the ion concentrations in solution reaches the value of the solubility product constant, \(K_{sp}\). When \(\text{Mg}^{2+}\) and \(\text{OH}^-\) ions exceed the \(K_{sp}\) threshold, \(\text{Mg(OH)}_2\) precipitates from the solution.

• Precipitation is useful for removing certain ions from solutions, purifying substances, and analytical applications.
• By controlling concentrations, we can selectively precipitate compounds from mixtures, aiding in separation techniques.

In the exercise, by calculating when the concentration of \(\text{OH}^-\) reaches \(1 \times 10^{-5} \text{ M}\), we predict the pH at which \(\text{Mg(OH)}_2\) begins to form. The understanding of such reactions allows chemists to develop protocols for processes like water treatment and mineral extraction efficiently.