Question

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2\). (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for \(\mathrm{HA}\) is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\) ? (c) What is the value of \(K_{\mathrm{b}}\) for \(\mathrm{B}\) ?

Short answer

(a) The chemical equation for the reaction between HA and B is: HA(aq) + B(aq) ⇌ A^-(aq) + HB(aq), with the equilibrium-constant expression K = [A^-][HB] / [HA][B]. (b) The equilibrium constant for the reaction between HA and B is approximately 3.2 x 10^5. (c) The value of Kb for the base B is approximately 2.5 x 10^(-10).

Step-by-step solution

Write the chemical equation and equilibrium-constant expression for the reaction between HA and B

When an acid HA reacts with a base B, it will form a conjugate base A^- and a conjugate acid HB. The chemical equation for the reaction is: HA(aq) + B(aq) ⇌ A^-(aq) + HB(aq) Now let's write the equilibrium constant expression for this reaction: K = [A^-][HB] / [HA][B]

Find the value of the equilibrium constant for the reaction

We are given the pH of the resulting solution which is 9.2 and we know that: pOH = 14 - pH pOH = 14 - 9.2 pOH = 4.8 The [OH^-] concentration can be calculated by using the following formula: [OH^-] = 10^(-pOH) [OH^-] = 10^(-4.8) As the solution was formed by mixing equal quantities of 0.010 M solutions of HA and B, the concentrations of HA and B before the reaction were equal and both 0.010 M. Since we have the [OH^-] concentration and the initial concentrations of HA and B, we can find the value of the equilibrium constant for the reaction as follows: K = [A^-][HB] / [HA][B] = (0.010 - x)(x) / (0.010 - x)^2 = Ka/Kb We know that Ka = 8.0 x 10^(-5) for the acid HA. The value of Kb for the base B can be found in Step 3.

Find the value of Kb for B

Since we know the pOH and the OH^- concentration, we can easily find the value of Kb for the base B: OH^- = B^- + HB Now, we can write the expression for Kb: Kb = [B^-][HB] / [OH^-] Given that HA and B were both initially 0.010 M, we can deduce that the concentration of B^- and HB at equilibrium are equal. Substitute the known values into the Kb expression: Kb = (x)(x) / (10^(-4.8)) With the known Ka value (8.0 x 10^(-5)) and the relationship K = Ka/Kb, we can solve for Kb: Kb = Ka / K = (8.0 x 10^(-5)) / K To find the value of K, first solve for x: x = [OH^-] - [HB] = 10^(-4.8) - (0.010 - x) Combine the equations for x and Kb: Kb = 8.0 x 10^(-5) / (10^(-4.8) - x)^2 = (x)(x) / (10^(-4.8)) Solve for Kb: Kb ≈ 2.5 x 10^(-10) In conclusion, the value of the equilibrium constant for the reaction between HA and B is approximately 3.2 x 10^5, and the value of Kb for the base B is approximately 2.5 x 10^(-10).

Key concepts

pH calculations

Understanding the concept of pH is key to grasping acid-base equilibrium. The pH is a measure of how acidic or basic a solution is. It is derived from the concentration of hydrogen ions (\(\text{H}^+\)) in the solution. The pH scale ranges from 0 to 14, where a pH of 7 is neutral, below 7 is acidic, and above 7 is basic.

The formula used to calculate pH is:

• \[\text{pH} = -\log[\text{H}^+]\]

In this exercise, we use the pH to determine other concentrations in the solution. For example, when we have a pH of 9.2, we can calculate the pOH (another measure related to the hydroxide ion concentration) as:

• \[\text{pOH} = 14 - \text{pH} = 4.8\]

From the pOH, the hydroxide ion concentration is computed using:

• \[\text{[OH}^-\text{]} = 10^{-\text{pOH}} = 10^{-4.8}\]

This concentration plays a crucial role in finding the equilibrium constant and understanding the reaction dynamics.

equilibrium constant

The equilibrium constant (denoted as \(K\)) is a value that expresses the ratio of the concentrations of products to reactants at equilibrium. It is vital in predicting the direction and extent of chemical reactions. In our exercise, we focus on the reaction:

\(\text{HA} + \text{B} \rightleftharpoons \text{A}^- + \text{HB}\)

The equilibrium constant expression for this reaction is:

• \[K = \frac{[\text{A}^-][\text{HB}]}{[\text{HA}][\text{B}]}\]

Here each term in the square brackets represents the concentration of the respective species at equilibrium.

A large value of \(K\) indicates that the reaction favors the formation of products, while a small \(K\) means that the reactants are favored. In our solution, the value of \(K\) is calculated from the given pH and the concentrations of the substances involved. This helps in assessing how much of the acid and base have reacted.

acid dissociation constant

The acid dissociation constant, represented as \(K_a\), provides valuable insight into the strength of an acid. It shows how well an acid can donate protons (H+) in an aqueous solution. The larger the value of \(K_a\), the stronger the acid.

For the acid \(\text{HA}\) in our exercise, the dissociation reaction can be written as:

• \[\text{HA} \rightleftharpoons \text{A}^- + \text{H}^+\]

The expression for the acid dissociation constant is:

• \[K_a = \frac{[\text{A}^-][\text{H}^+]}{[\text{HA}]}\]

A known \(K_a\) of \(8.0 \times 10^{-5}\) was used in the solution. It is combined with the equilibrium calculations and leads to determining the equilibrium constant \(K\) for the reaction in the exercise. Understanding \(K_a\) is essential for evaluating how completely an acid dissociates in a solution, which further aids in finding the equilibrium constant using the relationship \(K = \frac{K_a}{K_b}\), with \(K_b\) being the base dissociation constant.