Problem 254 - A Complete Resource Book in Chemistry for JEE Main

Question

If the \(\mathrm{K}_{\text {sp }}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1 \times 10^{-12}\) then \(0.01 \mathrm{M} \mathrm{MgCl}_{2}\) will precipitate at a limiting \(\mathrm{pH}\) of

Step-by-Step Solution

Answer

The limiting pH is 9.

Step 1: Understand the Problem

We need to determine the limiting pH at which \( \mathrm{Mg(OH)}_2 \) starts to precipitate from a solution containing \( 0.01 \mathrm{M} \mathrm{MgCl}_2 \). We are provided the solubility product constant \( \mathrm{K}_{\mathrm{sp}} = 1 \times 10^{-12} \).

Step 2: Establish the Equilibrium Expression

The dissolution of \( \mathrm{Mg(OH)}_2 \) in water is represented as:\[ \mathrm{Mg(OH)}_2 \rightleftharpoons \mathrm{Mg}^{2+} + 2 \mathrm{OH}^- \]The equilibrium expression for \( \mathrm{K}_{\mathrm{sp}} \) is:\[ \mathrm{K}_{\mathrm{sp}} = [\mathrm{Mg}^{2+}][\mathrm{OH}^-]^2 \]

Step 3: Substitute Known Values

From the problem, \( [\mathrm{Mg}^{2+}] \) is given as \( 0.01 \mathrm{M} \). We substitute this into the \( \mathrm{K}_{\mathrm{sp}} \) expression:\[ 1 \times 10^{-12} = 0.01 \cdot [\mathrm{OH}^-]^2 \]

Step 4: Solve for Hydroxide Ion Concentration \( [\mathrm{OH}^-] \)

Rearrange the equation to solve for \( [\mathrm{OH}^-] \):\[ [\mathrm{OH}^-]^2 = \frac{1 \times 10^{-12}}{0.01} = 1 \times 10^{-10} \]Taking the square root gives:\[ [\mathrm{OH}^-] = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \]

Step 5: Calculate pH from \( [\mathrm{OH}^-] \)

We convert \( [\mathrm{OH}^-] \) to pOH:\[ \text{pOH} = -\log(1 \times 10^{-5}) = 5 \]Finally, use \( \text{pH} + \text{pOH} = 14 \) to find the pH:\[ \text{pH} = 14 - 5 = 9 \]

Key Concepts

pH Calculation

Understanding how to calculate pH is crucial in chemistry because it allows you to determine how acidic or basic a solution is. This measure is important, especially when dealing with reactions that are influenced by these properties.
When you calculate pH, the first step is usually to find the concentration of hydroxide ions (H⁻) or hydrogen ions (H⁺). In our example, we had the concentration of hydroxide ions from the dissolution of Mg(OH)₂ calculated. Once we had [OH⁻] = 1 \times 10^{-5}, we could find pOH, which is the counterpart to pH in base-heavy solutions.
For the pH calculation:

• You calculate pOH using the formula: \[ ext{pOH} = -\log([OH⁻]) \]
• Given [OH⁻] = 1 \times 10^{-5}, \text{pOH} = 5.
• You then use the relation \text{pH} + \text{pOH} = 14 to find the pH.
• Thus, \text{pH} = 14 - 5\Rightarrow = 9.

This process shows how pH calculations blend both numerical and conceptual skills in chemistry.

Equilibrium Expression

Equilibrium expressions are fundamental in understanding how concentrations of reactants and products behave in a chemical reaction at “equilibrium.”
In chemical reactions that can go in both directions, like the dissolution of Mg(OH)₂, equilibrium represents a state where the rates of the forward and backward reactions are equal.
For our scenario:

• The chemical equation is Mg(OH)₂ \rightleftharpoons Mg^{2+} + 2 OH^{-}
• The equilibrium expression based on the solubility product constant (Ksp) is:
  \[ K_{ ext{sp}} = [Mg^{2+}][OH^-]^2 \]
• This expression helps predict if a precipitate will form at a given concentration of reactants or products.

Understanding equilibrium expressions involves knowing not just the form of the equation but also what the terms represent:

• The terms inside the square brackets [ ] indicate the molarity or concentration of the ions in the solution.
• Ksp is a constant at a specific temperature, indicative of the extent to which a compound can dissolve; the lower the Ksp, the less soluble the compound.

In many chemistry problems, this equilibrium concept helps determine other reaction dynamics and outcomes.

Precipitation Reaction

A precipitation reaction occurs when ions in solution form an insoluble compound, removing ions from the solution in the form of a solid. It’s important, especially when trying to isolate a substance or predicting the results in solution chemistry.
For magnesium hydroxide ( Mg(OH)₂ ) in our example, precipitation begins when the concentration of ions exceeds that predicted by the Ksp. This means the solution gets super-saturated, prompting the formation of a solid.
Understanding precipitation reactions allows you to predict:

• When a particular ion will come out of solution as a precipitate.
• The conditions that lead to such precipitation, such as changes in pH or temperature.

The calculations in the exercise involved finding the limiting pH that would cause Mg(OH)₂ to start precipitating:

• By solving the Ksp expression and determining the hydroxide concentration necessary to exceed solubility limits.
• Adjusting pH is a common method to influence these reactions, as seen here where a pH of 9 was the threshold for starting precipitation.

These reactions are crucial in many industrial and laboratory processes, affecting both the purity and yield of the desired product.