Problem 126 - 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

The concentration of hydroxyl ions is \(1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\).

Step 1: Understanding the Dissociation Constant

The dissociation constant of a base, denoted \(K_b\), is a measure of the extent to which a base dissociates into its ions in solution. For the base \(BOH\), it dissociates as follows: \[ BOH \rightleftharpoons B^+ + OH^- \] The given dissociation constant is \(K_b = 1.0 \times 10^{-12}\).

Step 2: Identifying the Equation for Equilibrium Concentration

For the dissociation of \(BOH\), the expression for the equilibrium constant is \[ K_b = \frac{[B^+][OH^-]}{[BOH]} \] For a weak base, we can assume \(x \) as the concentration of \(OH^-\) ions formed. Then, the expression simplifies to \[ K_b \approx x^2/c \] where \(c\) is the initial concentration of \(BOH\) and \(x\) is the concentration of hydroxyl ions at equilibrium.

Step 3: Applying the Initial Conditions

The initial concentration of the base, \(c\), is given as \(0.01 \, \text{M}\). Inserting the known values into the equation \( K_b = \frac{x^2}{c} \), we have: \[ 1.0 \times 10^{-12} = \frac{x^2}{0.01} \] which simplifies to \[ x^2 = 1.0 \times 10^{-14} \]

Step 4: Solving for Hydroxyl Ion Concentration

We solve the equation \( x^2 = 1.0 \times 10^{-14} \) for \(x\): \[ x = \sqrt{1.0 \times 10^{-14}} \] \[ x = 1.0 \times 10^{-7} \] This result gives us the concentration of hydroxyl ions in the solution.

Step 5: Verifying and Concluding

The calculated hydroxyl ion concentration is \(1.0 \times 10^{-7} \, \text{mol L}^{-1}\). Therefore, the correct answer to the problem, based on our calculations, is (d) \(1.0 \times 10^{-7} \, \text{mol L}^{-1}\). This verifies our solution as we followed the steps methodically, confirming the hydroxide concentration.

Key Concepts

Equilibrium Constant Expression

When studying chemical reactions, particularly those involving weak bases, we use the equilibrium constant expression to describe the balance between reactants and products. For a weak base like \(BOH\), the dissociation in water is represented as follows: \[BOH \rightleftharpoons B^+ + OH^-\].

The equilibrium constant for this reaction is the dissociation constant, \(K_b\). It reflects how much the base will dissociate in a solution. Mathematically, it's expressed as:

• \( K_b = \frac{[B^+][OH^-]}{[BOH]} \)

This expression tells us that the product of the molar concentrations of \(B^+\) and \(OH^-\) ions divided by the molar concentration of the undissociated base \(BOH\) gives the constant \(K_b\).

In simpler terms, if the equilibrium favors the products (more ions formed), the \(K_b\) will be larger. If more of the base remains undissociated, \(K_b\) is smaller, indicating a weaker base.

Concentration of Hydroxyl Ions

The concentration of hydroxyl ions, \(OH^-\), is crucial in understanding the chemical behavior of bases in solution. When a weak base such as \(BOH\) dissociates in water, it releases \(OH^-\) ions. This concentration determines the solution's basicity.

In our specific problem, calculating the concentration of \(OH^-\) requires inserting given values into the rearranged equilibrium expression. Initially, with \(x\) representing the concentration of \(OH^-\) formed, we insert it into our equation:

• \(K_b \approx \frac{x^2}{c} \)

where \(c\) is the initial concentration of the base. Solving for \(x\), we find the equilibrium concentration. In this exercise:

• \(x^2 = 1.0 \times 10^{-14}\)
• Solving gives \(x = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}\)

This solution displays the low concentration of \(OH^-\) ions, showcasing the weak base's minimal ionization.

Weak Base Dissociation

Weak base dissociation is a fundamental concept in chemistry. It refers to the ability of a base to partially dissociate into its ions in a solution. Using \(BOH\) as an example, it dissociates into \(B^+\) and \(OH^-\) ions but only partially, as indicated by its low \(K_b\) value, \(1.0 \times 10^{-12}\).

Here's what happens in weak base dissociation:

• Only a small fraction of \(BOH\) molecules dissociate.
• The majority of \(BOH\) remains in its original form.
• This leads to a relatively low concentration of \(OH^-\) ions.
• As a result, the overall \(pH\) of the solution will be higher than that of neutral water, but not as high as strong bases.

A low \(K_b\) signifies a weak base, meaning it doesn't completely ionize in water, which is why solutions of weak bases generally have less drastic pH changes compared to strong bases.