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

Step 1: Write the dissociation equation

The dissociation of the base \( \mathrm{BOH} \) in water can be represented by the equation: \( \mathrm{BOH} \rightarrow \mathrm{B^+} + \mathrm{OH^-} \).

Step 2: Define the dissociation constant

The dissociation constant \( K_b \) for the base is given by:\[ K_b = \frac{[\mathrm{B^+}][\mathrm{OH^-}]}{[\mathrm{BOH}]} \]

Step 3: Establish initial concentrations

Initially, the concentration of \( \mathrm{BOH} \) is given as \( 0.01 \mathrm{M} \). The concentrations of \( \mathrm{B^+} \) and \( \mathrm{OH^-} \) are both 0 at the start.

Step 4: Apply the change in concentrations

As \( \mathrm{BOH} \) dissociates, the concentration of \( \mathrm{OH^-} \) increases by \( x \) and the concentration of \( \mathrm{B^+} \) also increases by \( x \). The concentration of \( \mathrm{BOH} \) decreases by \( x \), leading to:- \( [\mathrm{B^+}] = x \)- \( [\mathrm{OH^-}] = x \)- \( [\mathrm{BOH}] = 0.01 - x \)

Step 5: Substitute into the dissociation constant formula

Substitute the concentrations from the previous step into the formula for \( K_b \):\[ 1.0 \times 10^{-12} = \frac{x^2}{0.01 - x} \]

Step 6: Approximate and solve for \( x \)

Given that \( K_b \) is very small, \( x \) will also be very small, so \( (0.01 - x) \approx 0.01 \). Thus, the equation simplifies to:\[ 1.0 \times 10^{-12} \approx \frac{x^2}{0.01} \]\[ x^2 = 1.0 \times 10^{-14} \]\[ x = \sqrt{1.0 \times 10^{-14}} \approx 1.0 \times 10^{-7} \]

Step 7: Conclusion

The concentration of hydroxyl ions \( [\mathrm{OH^-}] \) in the solution is \( 1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1} \).

Key Concepts

Base dissociation

Base dissociation is an important concept in chemistry, especially when dealing with aqueous solutions of weak bases. When a base like \( \mathrm{BOH} \) is dissolved in water, it dissociates, albeit partially, as it releases ions into the solution. In the case of \( \mathrm{BOH} \), the dissociation can be represented by the chemical equation:

• \( \mathrm{BOH} \rightarrow \mathrm{B^+} + \mathrm{OH^-} \)

This equation tells us that when \( \mathrm{BOH} \) dissolves, it breaks down into positively charged \( \mathrm{B^+} \) ions and negatively charged \( \mathrm{OH^-} \) ions.
Understanding base dissociation also involves knowing how much the base actually dissociates. This is where the dissociation constant, \( K_b \), is instrumental. It gives you a measure of the extent of dissociation. The lower the \( K_b \) value, the weaker the base, meaning it dissociates less in solution.
The dissociation constant is a fundamental indicator used by chemists to predict the behavior of bases in solution, such as how much of the base is converted into ions. This level of detail is crucial for calculations of equilibrium concentrations and for understanding the nature of acids and bases.

Hydroxyl ion concentration

The concentration of hydroxyl ions \( [\mathrm{OH^-}] \) in a solution is a pivotal factor in determining the solution's basicity. In the dissociation of \( \mathrm{BOH} \), the formation of \( \mathrm{OH^-} \) ion concentration is a direct result.
In our example, the dissociation constant \( K_b = 1.0 \times 10^{-12} \) provides the framework to calculate \( [\mathrm{OH^-}] \). When a weak base like \( \mathrm{BOH} \) dissolves in water, it produces \( \mathrm{OH^-} \) ions. Understanding the concentration of these ions is key to determining the pH level of the solution, as they significantly impact the basic nature.

• To find this concentration, you examine the equilibrium established as the base dissociates. In this case, initially, \( \mathrm{OH^-} \) starts at 0, but increases by \( x \), which represents the change in concentration.

Through approximation and solving the equilibrium equation, you identify \( [\mathrm{OH^-}] = x \), giving the final concentration.
This value is crucial not only for determining pH but also helps in further chemical calculations and reactions, particularly those involving titrations or buffer solutions.

Equilibrium concentrations

When a chemical reaction reaches a state where its forward and backward reactions occur at the same rate, it achieves equilibrium. The concept of equilibrium concentrations is about this balance in a dissociation reaction.
In the given example, establishing the equilibrium concentrations involves using the dissociation constant \( K_b \) and initial concentrations.

• Start with the initial concentration of the base \( \mathrm{BOH} = 0.01 \mathrm{M} \).
• As \( \mathrm{BOH} \) starts dissociating, the concentrations of \( \mathrm{B^+} \) and \( \mathrm{OH^-} \) increase by \( x \), while \( \mathrm{BOH} \) decreases by \( x \).

At equilibrium, the concentrations are expressed as:

• \( [\mathrm{B^+}] = x \)
• \( [\mathrm{OH^-}] = x \)
• \( [\mathrm{BOH}] = 0.01 - x \)

Substitute these values into the \( K_b \) expression to solve for \( x \). Given the very small \( K_b \), the approximation \( 0.01 - x \approx 0.01 \) simplifies calculations, making it easier to determine the small value of \( x \), and thereby \( [\mathrm{OH^-}] \).
These calculations ensure a clear understanding of how different concentration levels relate at equilibrium, helping with predictions of solution behavior and reactions.