Problem 189 - A Complete Resource Book in Chemistry for JEE Main

Question

The concentration of hydroxyl ion in a solution left after mixing \(100 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MgCl}_{2}\) and \(100 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{NaOH}\left[\mathrm{K}_{5}\right.\) of \(\left.\mathrm{Mg}(\mathrm{OH})_{2}=1.2 \times 10^{-11}\right]\) is (a) \(2.8 \times 10^{-3}\) (b) \(2.8 \times 10^{-2}\) (c) \(2.8 \times 10^{-4}\) (d) \(2.8 \times 10^{-5}\)

Step-by-Step Solution

Answer

Concentration of \(\text{OH}^-\) is approximately \(2.8 \times 10^{-3}\).

Step 1: Calculate Moles of Reactants

First, calculate the moles of each reactant. For \(\text{MgCl}_2\), moles \( = 100 \text{ mL} \times 0.1 \text{ M} = 0.01 \text{ mol}\). For \(\text{NaOH}\), moles \( = 100 \text{ mL} \times 0.2 \text{ M} = 0.02 \text{ mol}\).

Step 2: Determine Limiting Reactant and Reaction

The reaction is \(\text{Mg}^{2+} + 2\text{OH}^- \rightarrow \text{Mg(OH)}_2\). You have 0.01 mol of \(\text{Mg}^{2+}\), which can completely react with 0.02 mol of \(\text{OH}^-\). Thus, all \(\text{Mg}^{2+}\) and 0.02 mol of \(\text{OH}^-\) are used, leaving no \(\text{Mg}^{2+}\).

Step 3: Calculate Excess Hydroxide Ions

Initially, there were 0.02 mol \(\text{OH}^-\). Since it reacts with 0.01 mol \(\text{Mg}^{2+}\), consuming 0.02 mol of \(\text{OH}^-\), there is no excess \(\text{OH}^-\).

Step 4: Calculate Equilibrium Concentration of \(\text{OH}^-\)

The \(\text{Mg(OH)}_2\) partially dissociates, governed by \(K_{sp}\). \(K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = 1.2 \times 10^{-11}\).

Key Concepts

Hydroxyl Ion Concentration

Hydroxyl ions, represented as \( \text{OH}^- \), are basic ions present in aqueous solutions resulting from the dissociation of bases like NaOH. Calculating the concentration of hydroxyl ions involves understanding both initial concentrations and chemical reactions that might remove or produce these ions.

In the problem, NaOH contributes \( \text{OH}^- \) ions to the solution. Initially, the system had 0.02 moles of \( \text{OH}^- \) from 0.2 M NaOH in 100 mL. During a reaction with \( \text{MgCl}_2 \), \( \text{OH}^- \) ions react with \( \text{Mg}^{2+} \) to form \( \text{Mg(OH)}_2 \), consuming some \( \text{OH}^- \) ions.

Monitoring hydroxyl ion concentration is crucial, particularly when computing whether more base or metallic ions will remain in solution post reaction. Eventually, the dynamic equilibrium will determine the final concentration of hydroxyl ions, which could be crucial for further calculations.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, thus determining the amount of product formed. Identifying the limiting reactant involves calculating and comparing the moles of each reactant based on their stoichiometric ratios.

In our exercise, the reaction \( \text{Mg}^{2+} + 2\text{OH}^- \rightarrow \text{Mg(OH)}_2 \) indicates that two moles of hydroxide ions are needed for every mole of magnesium ions. Starting with 0.01 mol of \( \text{Mg}^{2+} \) and 0.02 mol of \( \text{OH}^- \), you would completely use all of the magnesium ions first to form the insoluble \( \text{Mg(OH)}_2 \). Thus, \( \text{MgCl}_2 \) is the limiting reactant here.

Understanding the limiting reactant helps in predicting the amounts of products formed and ensuring calculations involving chemical equations are accurate.

Solubility Product Constant (Ksp)

The Solubility Product Constant, or \( K_{sp} \), is a numerical value that represents the solubility of a sparingly soluble ionic compound. It is a product of the concentrations of the ions each raised to the power of their stoichiometric coefficients.

For \( \text{Mg(OH)}_2 \), which barely dissolves in water, the \( K_{sp} \) equation is \[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = 1.2 \times 10^{-11} \].

For this particular problem, calculating the remaining \( \text{OH}^- \) concentration involves setting the concentration of \( \text{Mg(OH)}_2 \) equal to its \( K_{sp} \) value under equilibrium conditions. As \( \text{Mg(OH)}_2 \) forms in the solution, it reaches a balance between dissolved ions and the solid, affecting the \( \text{OH}^- \) concentration in the solution.

Solution Stoichiometry

Solution stoichiometry is a quantitative relationship between the amounts of reactants and products in a chemical reaction. It allows you to calculate the quantities needed or produced in a reaction by applying known concentrations of reactants.

When solving stoichiometry problems, it's essential to convert volumes and molarities into moles. In this exercise, you start by determining the moles of \( \text{MgCl}_2 \) and \( \text{NaOH} \) based on their given concentrations and volumes. Then, using the balanced chemical equation, you can determine how these moles interact.

• Calculate moles from volume and molarity: \( \text{moles} = \text{Volume (L)} \times \text{Molarity (M)} \).
• Use the reaction stoichiometry to pinpoint reactants and products.

Understanding solution stoichiometry allows you to accurately predict the outcome of dissolutions and reactions in an aqueous environment.