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

Question

What is the \(\mathrm{pH}\) value at which \(\mathrm{Mg}(\mathrm{OH})\), 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 pH at which \(\mathrm{Mg}( ext{OH})_2\) begins to precipitate is 9.

Step 1: Understanding the Precipitation Condition

To determine when \(\mathrm{Mg}( ext{OH})_2\) begins to precipitate, we need to know the concentration of \(\text{OH}^-\) ions necessary to reach the solubility product constant (\(K_{sp}\)). The formula for \( \mathrm{Mg}( ext{OH})_2 \) dissociation is: \[ \mathrm{Mg}( ext{OH})_2(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \] This suggests the equation for the \(K_{sp}\) is: \[ K_{sp} = [\mathrm{Mg}^{2+}][\text{OH}^-]^2 \] We have \([\mathrm{Mg}^{2+}] = 0.10 \, \text{M}\) and \(K_{sp} = 1 \times 10^{-11}\).

Step 2: Solve for Hydroxide Ion Concentration [OH-]

To find \([\text{OH}^-]\), rearrange the \(K_{sp}\) equation: \[ 1 \times 10^{-11} = 0.10 \times [\text{OH}^-]^2 \] Solving for \([\text{OH}^-]\), we have: \[ [\text{OH}^-]^2 = \frac{1 \times 10^{-11}}{0.10} = 1 \times 10^{-10} \] Taking the square root: \[ [\text{OH}^-] = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \text{ M} \]

Step 3: Convert [OH-] to pH

We know \([\text{OH}^-]\) and want to find the pH. First, find pOH: \[ \text{pOH} = -\log([\text{OH}^-]) = -\log(1 \times 10^{-5}) = 5 \] Then, use the relation between pH and pOH: \[ \text{pH} = 14 - \text{pOH} = 14 - 5 = 9 \]

Step 4: Conclusion and Answer Selection

Based on the calculation, \(\text{pH} = 9\) is the point at which \(\mathrm{Mg}( ext{OH})_2\) begins to precipitate. Thus, the corresponding answer is option (c) 9.

Key Concepts

Precipitation

In chemistry, precipitation occurs when ions in a solution combine to form an insoluble compound, which then separates from the solution as a solid. This process is significant when studying solubility and reactions in aqueous solutions. Precipitation happens when the product of the reactant ion concentrations exceeds the solubility product constant ( K_{sp} ) for the compound.

• Consider the reaction: ext{Mg}(OH)_2(s) ightleftharpoons ext{Mg}^{2+}(aq) + 2 ext{OH}^-(aq)
• Precipitation begins when the concentration of [ ext{Mg}^{2+}] and [ ext{OH}^-] in solution reach a level where their product equals or exceeds K_{sp} .

Understanding precipitation helps predict whether a reaction will proceed by forming a solid or staying dissolved.

Solubility Product Constant (Ksp)

The solubility product constant, or K_{sp} , is an equilibrium constant for the dissolution of a sparingly soluble compound. It represents the maximum product of the ion concentrations before the compound begins to precipitate. K_{sp} provides insight into the solubility of different compounds.

• For Mg(OH)_2 , the dissociation is representable as: Mg(OH)_2(s) ightleftharpoons ext{Mg}^{2+}(aq) + 2 ext{OH}^-(aq) .
• Hence, K_{sp} = [ ext{Mg}^{2+}][ ext{OH}^-]^2

The K_{sp} value for Mg(OH)_2 is 1 imes 10^{-11} . A lower K_{sp} signifies lower solubility. Understanding this value helps predict when precipitation will start in varying concentrations of ions.

Hydroxide Ion Concentration (OH-)

Hydroxide ions ( ext{OH}^- ) greatly influence the solubility of metal hydroxides. By calculating the concentration of ext{OH}^- ions, you uncover information about when a compound will precipitate. The equilibrium expression for Mg(OH)_2 solubility is crucial.

• Starting from K_{sp} = [ ext{Mg}^{2+}][ ext{OH}^-]^2 , plug in the known [ ext{Mg}^{2+}] and K_{sp} to find [ ext{OH}^-].
• We solve for [ ext{OH}^-] : [ ext{OH}^-]^2 = rac{1 imes 10^{-11}}{0.10} = 1 imes 10^{-10} .

After taking the square root: [ ext{OH}^-] = 1 imes 10^{-5} ext{ M} . This value reveals the point at which hydroxide concentration will trigger precipitation.

pH and pOH Relationship

Understanding pH and ext{pOH} is essential in the realm of chemical equilibria. Since pH measures hydrogen ion concentration, while ext{pOH} measures hydroxide ion concentration, their relationship is inversely connected through water's constant ion product.

• The equation for their relation is: ext{pH} + ext{pOH} = 14 .
• When [ ext{OH}^-] = 1 imes 10^{-5} ext{ M} , calculate ext{pOH} as 5, using - ext{log}([ ext{OH}^-]) .
• To find ext{pH} : ext{pH} = 14 - ext{pOH} = 9 .

Identifying the ext{pH} allows you to predict at what point a compound like ext{Mg}(OH)_2 starts to precipitate, offering a clear understanding of acid-base equilibria in the solution.