Problem 137 - General Chemistry

Question

A generic base, \(\mathrm{B}^{-},\) is added to \(2.25 \mathrm{~L}\) of water. The \(\mathrm{pH}\) of the solution is found to be 10.10 . What is the concentration of the base \(\mathrm{B}^{-}\) in this solution? \(K_{a}\) for the acid \(\mathrm{HB}\) at \(25^{\circ} \mathrm{C}\) is \(1.99 \times 10^{-9}\)

Step-by-Step Solution

Answer

The concentration of the base \(\text{B}^-\) is approximately \(3.16 \times 10^{-3}\) M.

Step 1: Calculate pOH from pH

Since the pH is 10.10, we can find the pOH using the relation \( \text{pH} + \text{pOH} = 14 \). Thus, \( \text{pOH} = 14 - 10.10 = 3.90 \).

Step 2: Find hydroxide ion concentration

The concentration of hydroxide ions, \([\text{OH}^-]\), can be calculated from the pOH as \([\text{OH}^-] = 10^{-\text{pOH}} = 10^{-3.90} \approx 1.26 \times 10^{-4} \text{ M} \).

Step 3: Use relation between \(K_a\) and \(K_b\)

First, calculate \(K_b\) for the base \(\text{B}^-\) using the relation \(K_w = K_a \times K_b\). With \(K_w = 1 \times 10^{-14}\), \(K_a = 1.99 \times 10^{-9}\). Therefore, \(K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{1.99 \times 10^{-9}} \approx 5.03 \times 10^{-6}\).

Step 4: Calculate the initial concentration of the base

Assume that \( \text{B}^- \) about equally contributes to \([\text{OH}^-]\) at equilibrium. We know \([\text{OH}^-] = \sqrt{C \times K_b} \), so \(1.26 \times 10^{-4} = \sqrt{C \times 5.03 \times 10^{-6}}\). Solve for \(C\): \((1.26 \times 10^{-4})^2 = C \times 5.03 \times 10^{-6}\) implies \(C = \frac{(1.26 \times 10^{-4})^2}{5.03 \times 10^{-6}} \approx 3.16 \times 10^{-3} \text{ M}\).

Key Concepts

pH and pOH calculation

The relationship between pH and pOH is a fundamental concept in acid-base equilibrium. It denotes how acidic or basic a solution is. The formula linking them is:

• \( \text{pH} + \text{pOH} = 14 \) at 25°C

7) show basic solutions. In this exercise, the solution has a pH of 10.10, indicating it's basic.

To find the pOH from pH, simply subtract 10.10 from 14, giving a pOH of 3.90. This step is crucial because pOH further helps us calculate the hydroxide ion concentration \([\text{OH}^-]\), which is essential for determining base concentration. Understanding and accurately calculating pH and pOH based on their interrelationship is vital for solving various chemistry problems involving acidity or alkalinity of a solution.

Base concentration

The concentration of a base in a solution indicates how much of that base is present. Knowing the concentration is necessary to understand the strength and behavior of a solution. Here, we have \([\text{OH}^-]\) calculated from the pOH value, which is \([\text{OH}^-] = 10^{-3.90} \), giving approximately \(1.26 \times 10^{-4} \text{ M}\). This means there are \(1.26 \times 10^{-4} \) moles of hydroxide ions per liter of solution.

To find the base concentration, assume the base \(\text{B}^-\) contributes equally to the hydroxide ions at equilibrium. Apply the formula \([\text{OH}^-] = \sqrt{C \times K_b}\). Here, \(C\) is the initial concentration of the base, and \(K_b\) is the base dissociation constant. Solving for \(C\) using the equation, \((1.26 \times 10^{-4})^2 = C \times 5.03 \times 10^{-6}\), gives \(C \approx 3.16 \times 10^{-3} \text{ M}\). This value represents the base concentration. Understanding the connection between hydroxide concentration and base concentration helps predict how the base will behave when dissolved.

Equilibrium constant calculation

Equilibrium constants are vital in determining how reactants and products are positioned in a chemical equilibrium. For a base such as \(\text{B}^-\), the base dissociation constant is \(K_b\). It shows the extent to which a base dissociates in water to produce hydroxide ions.

In cases where both acid and base constants are known, use the relationship:

• \(K_w = K_a \times K_b\)

where \(K_w = 1 \times 10^{-14}\) at 25°C. In this exercise, given \(K_a = 1.99 \times 10^{-9}\), we calculate \(K_b\) by rearranging the formula:

• \(K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{1.99 \times 10^{-9}} \approx 5.03 \times 10^{-6}\)

This calculation shows the ability of the base \(\text{B}^-\) to dissociate. Having the equilibrium constant allows chemists to predict the behavior of acids and bases in solution, which is crucial for designing reactions and understanding biochemical pathways.