Problem 242 - A Complete Resource Book in Chemistry for JEE Main

Question

The \(\mathrm{K}_{\mathrm{sp}}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1 \times 10^{-12}\), A \(0.01 \mathrm{M} \mathrm{MgCl}_{2}\) solution will precipitate at what limiting \(\mathrm{pH}\) value?

Step-by-Step Solution

Answer

The limiting pH value for precipitation is 9.

Step 1: Analyze the given equilibrium

Consider the dissolution of Mg(OH)₂: \[ \mathrm{Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2OH^- (aq)} \] The solubility product constant (\(K_{sp}\)) for this reaction is given as \(1 \times 10^{-12}\). This means \([\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2 = 1 \times 10^{-12}\).

Step 2: Set up the expression for precipitation condition

For precipitation to occur, the ionic product \([\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2\) must reach or exceed \(K_{sp}\). Given \([\mathrm{Mg^{2+}}] = 0.01\,M\), we set up the inequality: \[ 0.01 \times [\mathrm{OH^-}]^2 \ge 1 \times 10^{-12} \]

Step 3: Solve for \([\mathrm{OH^-}]\) concentration

Rearrange the inequality to solve for \( [\mathrm{OH^-}] \): \[ [\mathrm{OH^-}]^2 \ge \frac{1 \times 10^{-12}}{0.01} \] Thus, \[ [\mathrm{OH^-}]^2 \ge 1 \times 10^{-10} \] Taking the square root gives \([\mathrm{OH^-}] \ge 1 \times 10^{-5}\).

Step 4: Convert \([\mathrm{OH^-}]\) to \(pH\)

Knowing \(pOH = -\log_{10}([\mathrm{OH^-}])\), calculate: \[ pOH = -\log_{10}(1 \times 10^{-5}) = 5 \] Convert \(pOH\) to \(pH\) using \( 14 = pH + pOH \): \[ pH = 14 - 5 = 9 \]

Step 5: Determine the limiting \(pH\)

Since a \(0.01 \mathrm{M} \mathrm{MgCl}_2\) solution will start precipitating \(\mathrm{Mg(OH)_2}\) at a \(pH\) of 9, pH 9 is the limiting pH value for precipitation.

Key Concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, abbreviated as \(K_{sp}\), reflects the extent to which a compound can dissolve in water. It is a fundamental concept in the study of precipitation equilibrium.
Essentially, \(K_{sp}\) is an expression of solubility in quantitative terms for sparingly soluble salts. In our given exercise, the compound in focus is \( \mathrm{Mg(OH)_2} \), and its \(K_{sp}\) is \(1 \times 10^{-12}\). This indicates that \([\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2 = 1 \times 10^{-12}\).
This constant helps predict whether a precipitate will form in a solution. If the ionic product of the ions in solution exceeds the \(K_{sp}\), precipitation occurs, forming a solid. On the contrary, if the ionic product is less than \(K_{sp}\), the compound remains dissolved.
It is important to understand how manipulating concentrations of ions in a solution affects whether or not a precipitate forms.

pH and pOH Relationship

The relationship between pH and pOH is crucial when dealing with acidic or basic solutions. These values are interconnected through the water dissociation constant \([Kw]\), given by \( Kw = [H^+][OH^-] = 1 \times 10^{-14} \) at \(25^\circ C\).
From this relationship, we derive that \( pH + pOH = 14 \). In our case, knowing the pOH allows us to calculate the pH. From the earlier step-by-step solution, we found that \([\mathrm{OH^-}] \ge 1 \times 10^{-5}\), which converts to \( pOH = 5 \).
Subsequently, the pH is calculated as \(9\) using the formula \( pH = 14 - pOH \).
This calculation shows us how acids and bases in a solution impact the potential for precipitation to occur by affecting the concentration of hydroxide ions.

Ionic Product

The concept of the Ionic Product is crucial when discussing the precipitation conditions of ionic compounds. Essentially, it is the product of the concentrations of the ionic species in a solution. In equilibrium or precipitation discussions, an understanding of the ionic product determines whether a precipitate forms.
For our exercise, the ionic product of \(\mathrm{Mg(OH)_2}\) is \([\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2\). The solution will only precipitate if this ionic product matches or exceeds the solubility product constant \(K_{sp}\).
Using the step-by-step solution, an initial \([\mathrm{Mg^{2+}}]\) of \(0.01\,M\) is given. To maintain balance and prevent excess dissolution or precipitation, the \([\mathrm{OH^-}]^2\) is adjusted such that the ionic product aligns with or exceeds the \(K_{sp}\) of \(1 \times 10^{-12}\). This insight enables us to determine the point at which the solution will start precipitating.