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 \(0.2 \mathrm{M} \mathrm{NaOH}\left[\mathrm{K}_{\mathrm{sp}}\right.\) 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

The concentration of hydroxyl ion is \(2.8 \times 10^{-4} \text{ M}\).

Step 1: Determine Moles of Each Reactant

Calculate the number of moles of each reactant. For \( MgCl_2 \), we have: \( 0.1 \text{ L} \times 0.1 \text{ M} = 0.01 \text{ moles of } MgCl_2 \). For \( NaOH \), we have: \( 0.1 \text{ L} \times 0.2 \text{ M} = 0.02 \text{ moles of } NaOH \).

Step 2: Reaction Between MgCl2 and NaOH

Write the chemical reaction: \( MgCl_2 + 2NaOH \rightarrow Mg(OH)_2(s) + 2NaCl \). From the stoichiometry, 1 mole of \( MgCl_2 \) reacts with 2 moles of \( NaOH \). So, \( 0.01 \) moles of \( MgCl_2 \) would react with \( 0.02 \) moles of \( NaOH \) completely.

Step 3: Identify Remaining Hydroxide Ions

Since \( 0.01 \) moles of \( NaOH \) react (as 2:1 stoichiometry), no extra \( NaOH \) is left since both the reactants are limiting. However,\( Mg(OH)_2 \) can dissolve partially according to its solubility product \( K_{sp} \).

Step 4: Calculate Concentration of Hydroxide Ions

Write the \( K_{sp} \) expression: \( Mg(OH)_2 \leftrightarrow Mg^{2+} + 2OH^- \), with \( K_{sp} = [Mg^{2+}][OH^-]^2 \). Assume a small amount 'x' of \( Mg(OH)_2 \) dissolves: \( x \) moles \( Mg^{2+} \), \( 2x \) moles \( OH^- \). So, \( 1.2 \times 10^{-11} = (x)(2x)^2 = 4x^3 \). Solving for \( x = 1.4 \times 10^{-4} \), doubling gives \(OH^- = 2.8 \times 10^{-4} \text{M}.\)

Key Concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that involves calculating the relationships between reactants and products in chemical reactions. It's essentially the math behind chemistry that ensures that matter is neither created nor destroyed in a chemical process. In the exercise, stoichiometry helps us directly determine the quantities of substances needed or produced in a reaction.

• First, it aids in determining the exact number of moles of each reactant. For example, we calculated 0.01 moles of \( MgCl_2 \) and 0.02 moles of \( NaOH \). This initial step is crucial to understanding how much of each substance participates in the reaction.
• Then, using the chemical reaction provided, \( MgCl_2 + 2NaOH \rightarrow Mg(OH)_2(s) + 2NaCl \), stoichiometry tells us that one mole of magnesium chloride reacts with two moles of sodium hydroxide. This stoichiometric ratio allows us to confirm that the solution initially contains the exact amounts needed to fully react with each other. This gives us confidence that the resulting solution composition is accurately represented and leads to no leftover reactants.

By understanding stoichiometry, you can predict the outcomes of the reaction and manage the reactants efficiently.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. They are written using chemical equations that succinctly express changes from reactants to products with a balanced format. In our scenario, the reaction between \( MgCl_2 \) and \( NaOH \) is:

• \( MgCl_2 + 2NaOH \rightarrow Mg(OH)_2(s) + 2NaCl \): This balanced equation outlines the transformation that occurs when these chemicals are combined, resulting in the formation of magnesium hydroxide and sodium chloride.
• The reaction with \( NaOH \) clearly illustrates a double displacement reaction where ions between the reactants are exchanged.
• Understanding this type of reaction is pivotal because it affects the solubility and formation of the \( Mg(OH)_2 \), which then influences the subsequent concentration calculations for excess ions following the product formation.

Chemical reactions help us comprehend the transformations at a molecular level, predict the products formed, and provide insight into important aspects such as solubility.

Concentration Calculations

Concentration calculations help determine the amount of a given solute present in a solution and are vital in interpreting the final composition of the chemical system. Solving the concentration of ions post-reaction involves some critical steps:

• After the reaction reaches completion, we need to calculate remaining hydrogen ion concentrations, which inevitably impact solubility products. In the case of \( Mg(OH)_2 \), understanding its dissociation is necessary.
• Using the solubility product constant (\( K_{sp} \)), we identify the minimal concentration of ions in equilibrium. For example, the \( K_{sp} \) expression for \( Mg(OH)_2 \) is: \([Mg^{2+}][OH^-]^2\) and this helps predict how much \( Mg(OH)_2 \) will dissolve slightly, re-supplying the system with \( OH^- \) ions.
• This precise calculation leads us to derive the concentration of hydroxide ions: doubling the dissolved hydroxide amount gives the concentration \( 2.8 \times 10^{-4} \text{ M} \).

Through concentration calculations, we accurately assess the final ionic composition of a solution, crucial for predicting chemical behavior in practical applications.