Problem 87 - Chemistry: The Central Science in SI Units, Global Edition

Question

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid 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 HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{b}\) for \(B\) ?

Step-by-Step Solution

Answer

The chemical equation for the reaction between the acid HA and the base B is \(HA + B^- \rightleftharpoons A^- + HB\), with the equilibrium constant expression \(K = \frac{[A^-][HB]}{[HA][B^-]}\). The equilibrium constant K for the reaction between HA and B is 12.65. The value of Kb for B is 1.25 x 10^(-10).

Step 1: Write the chemical equation and equilibrium constant expression

First, let's write the chemical equation for the reaction between the acid HA and the base B. Since we are dealing with an acid-base reaction, we can write the equation as follows: \(HA + B^- \rightleftharpoons A^- + HB\) Now, let's write the equilibrium constant expression for this reaction: \(K = \frac{[A^-][HB]}{[HA][B^-]}\)

Step 2: Calculate the concentration of H+ ions

The given pH of the resulting solution is 9.2. We can use the pH to calculate the concentration of H+ ions using the formula: \([H^+] = 10^{-pH}\) \([H^+] = 10^{-9.2} = 6.31 \times 10^{-10} M\)

Step 3: Calculate the concentration of A- ions

Now, let's use the given Ka value for HA to find the concentration of A- ions. The equation for Ka is: \(K_a = \frac{[A^-][H^+]}{[HA]}\) Given the Ka value as 8.0 x 10^(-5), we can rearrange the equation and solve for the concentration of A- ions: \([A^-] = \frac{K_a[HA]}{[H^+]}\) \([A^-] = \frac{(8.0 \times 10^{-5})(0.010)}{6.31 \times 10^{-10}}\) \([A^-] = 1.27 \times10^{-4} M\)

Step 4: Calculate the equilibrium constant for the reaction between HA and B

Now, let's find the equilibrium constant for the reaction between HA and B using the equilibrium constant expression and the concentrations of the reacting species: \(K = \frac{[A^-][HB]}{[HA][B^-]}\) Given that the reaction was between equal quantities of HA and B, their concentrations will be equal at equilibrium: \([HB] = [B^-] = [HA] - [A^-]\) Using the known values, \(K = \frac{[A^-][HA] - [A^-]}{[HA]^2 - [A^-][HA]}\) \(K = \frac{(1.27 \times 10^{-4})(0.010 - 1.27 \times 10^{-4})}{(0.010)^2 - (1.27 \times 10^{-4})(0.010)}\) \(K = 12.65\)

Step 5: Calculate the value of Kb for B

Lastly, we will find the value of Kb for B. We can use the relationship between Ka and Kb for conjugate acid-base pairs, which is: \(K_a \times K_b = K_w\) Where Kw is the ion product of water (1.0 x 10^(-14) at 25°C). Rearranging the equation, we can solve for Kb: \(K_b = \frac{K_w}{K_a}\) \(K_b = \frac{1.0 \times 10^{-14}}{8.0 \times 10^{-5}}\) \(K_b = 1.25 \times 10^{-10}\) So, the value of Kb for B is 1.25 x 10^(-10).

Key Concepts

Equilibrium Constant

The equilibrium constant (K) is a value that expresses the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. It gives insight into the position of equilibrium, indicating whether the products or reactants are favored. When you write an equilibrium constant expression, you use the concentrations of the involved substances, raised to the power of their coefficients in the balanced equation. For instance, in the chemical reaction between an acid HA and base B as presented: \[HA + B^- \rightleftharpoons A^- + HB\] The equilibrium constant expression based on this reaction is: \[K = \frac{[A^-][HB]}{[HA][B^-]}\] Here, K reflects the concentrations of the species at equilibrium. A larger value of K suggests a greater concentration of products compared to reactants at equilibrium.

Chemical Equation

A chemical equation represents the substances involved in a chemical reaction and their relationships. It clearly indicates the reactants and products, utilizing chemical formulas to express each species involved. In the given problem, we are dealing with an acid-base reaction between HA and B, forming A- and HB. - **Reactants:** HA + B^- - **Products:** A^- + HB Writing a chemical equation accurately is crucial as it forms the basis for deriving other calculations such as equilibrium constants. It serves as a blueprint for understanding the transformations that happen during the reaction.

pH Calculation

pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration. Mathematically, it is represented as: \[pH = -\log_{10}([H^+])\] In this exercise, we are given a pH value of 9.2 for the solution. To find the concentration of hydrogen ions ([H^+]), you can rearrange the formula: \[[H^+] = 10^{-pH}\] Substituting the known pH: \[[H^+] = 10^{-9.2} = 6.31 \times 10^{-10} \ M\] This method allows you to determine how acidic or basic a solution is based on the given pH, which is instrumental in various equilibrium and concentration calculations.

Ion Product of Water

The ion product of water (K_w) is a fundamental constant in acid-base chemistry, describing the equilibrium between water molecules and their ionization into hydrogen (H^+) and hydroxide (OH^-) ions:\[H_2O \rightleftharpoons H^+ + OH^-\] At 25°C, K_w is 1.0 \times 10^{-14}. This constant is used extensively when dealing with the acidity or basicity of solutions and exploring the relationship between Ka and Kb for conjugate acid-base pairs. The relationship is expressed by: \[K_a \times K_b = K_w\] Where K_a and K_b represent the acid and base dissociation constants, respectively. Kw helps in determining one dissociation constant if the other is known, offering a powerful tool for solving equilibrium problems, as illustrated by calculating Kb in this exercise.