Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(25^{\circ} \mathrm{C}\) the dissociation constant of a base, \(\mathrm{BOH}\) is \(1.0 \times 10^{-12}\), the concentration of hydroxyl ions \(0.01 \mathrm{M}\) aqueous solution of the base would become (a) \(2.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

Step-by-Step Solution

Answer

(d) \(1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\).

Step 1: Understand the Dissociation of Base

The base in question, \(\text{BOH}\), dissociates in water according to the reaction: \[ \text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^- \]. The dissociation constant, \(K_b\), is given as \(1.0 \times 10^{-12}\). We need to calculate the concentration of \(\text{OH}^-\) ions when the base is in solution.

Step 2: Expression for Kb

For the dissociation equilibrium \(\text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^-\), the dissociation constant \(K_b\) is given by the expression:\[ K_b = \frac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]} \]. Here, \([\text{B}^+]=[\text{OH}^-]=x\) and \([\text{BOH}]=0.01-x \approx 0.01\, \text{M}\) since \(x\) will be very small compared to 0.01 M.

Step 3: Set Up the Equation for x

Substituting the concentrations in the \(K_b\) expression gives:\[ 1.0 \times 10^{-12} = \frac{x^2}{0.01} \].We're solving for \(x\), the concentration of \([\text{OH}^-]\) ions.

Step 4: Solve for x

Rearrange the equation to find \(x^2\):\[ x^2 = 1.0 \times 10^{-12} \times 0.01 = 1.0 \times 10^{-14} \].Now solve for \(x\):\[ x = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \].

Step 5: Choose the Correct Option

The concentration of \([\text{OH}^-]\) ions in the solution is \(1.0 \times 10^{-7} \) mol/L. Reviewing the given options, the correct answer is (d) \(1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\).

Key Concepts

bases

In chemistry, bases are substances that can accept protons or donate pairs of electrons. When a base dissolves in water, it dissociates to form ions. Typically, a base will dissociate to produce hydroxyl ions \(\text{OH}^-\). These hydroxyl ions are responsible for making the solution basic or alkaline.
To further illustrate:

• The base \(\text{BOH}\) can dissociate in water to form \(\text{B}^+\) and hydroxyl ions \(\text{OH}^-\).
• This reaction can be represented as: \[ \text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^- \]
• In the problem, a base is a compound whose dissociation leads to the increase of hydroxyl ion concentration in a solution.

A higher concentration of \(\text{OH}^-\) ions in a solution characteristically means that the solution is more basic. Understanding how bases dissociate and interact in a solution is crucial for calculating hydroxyl ion concentration.

hydroxyl ion concentration

The hydroxyl ion concentration, represented by \(\text{OH}^-\), determines the basicity of a solution. The higher the concentration of hydroxyl ions, the more alkaline the solution is. We often calculate this concentration to understand the strength of a base.
To determine the hydroxyl ion concentration in a given solution, we use the dissociation constant \(K_b\) of the base. Here's how it's typically done:

• For the base \(\text{BOH}\), which dissociates into \(\text{B}^+\) and \(\text{OH}^-\), the equation is expressed as: \[ K_b = \frac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]} \]
• Assume initial concentrations: \(\text{B}^+ = \text{OH}^- = x\), and \(\text{BOH} = 0.01 - x\) approximately \(0.01 \, \text{M}\).
• Substituting into the equation gives: \[ 1.0 \times 10^{-12} = \frac{x^2}{0.01} \]
• Solving for \(x\) allows us to find that \(x = 1.0 \times 10^{-7} \, \text{mol/L}\), representing the hydroxyl ion concentration.

By calculating the hydroxyl ion concentration, chemists can assess how strong a base is and how it will behave in various chemical reactions.

equilibrium expressions

Equilibrium expressions are mathematically formulated equations that describe the state of balance between products and reactants in a reversible chemical reaction.
For bases that dissociate in water, equilibrium plays a significant role in determining concentrations of various ions in the solution. Here is a brief insight into how this is utilized:

• In the reaction \(\text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^-\), the point at which the forward and reverse reactions occur at the same rate is known as equilibrium.
• The equilibrium expression for this reaction, based on the dissociation constant \(K_b\), is: \[ K_b = \frac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]} \]
• This formula helps calculate how much of the base dissociates into ions at equilibrium, which in turn determines the concentration of hydroxyl ions.

Understanding equilibrium expressions allows chemists to predict the behavior of bases in a solution. By amplifying our comprehension of equilibrium, we can better control and direct chemical reactions to achieve desired results. This principle is pivotal in various fields, such as pharmacology, environmental science, and more.