Problem 87 - Chemistry The Central Science

Question

Equal quantities of 0.010\(M\) 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 \(\mathrm{HA}\) and \(\mathrm{B}\) . (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 \(K_{b}\) for \(\mathrm{B}\) ?

Step-by-Step Solution

Answer

(a) The chemical equation for the reaction between HA and B is: HA + B → A⁻ + HB. The equilibrium-constant expression is: K = [A⁻][HB] / [HA][B] (b) The value of the equilibrium constant, K, is: \(K = \frac{(0.010 - 10^{-4.8})^2}{(8.0 \times 10^{-5})}\) (c) The value of Kb for B is: \(K_b = \frac{1.0 \times 10^{-14}}{8.0 \times 10^{-5}}\)

Step 1: (a) Chemical equation and equilibrium-constant expression

First, let's write the reaction between the acid HA and the base B (assuming B is a generic hydroxide ion donor): HA + B → A⁻ + HB Next, write the equilibrium-constant expression for this reaction: K = [A⁻][HB] / [HA][B]

Step 2: (b) Value of equilibrium constant

We're given the pH of the resulting solution which is 9.2. To find the pOH of the solution, use the relationship between pH and pOH: pH + pOH = 14 9.2 + pOH = 14 pOH = 4.8 Now, we can find the OH⁻ concentration in the solution from the pOH value: \[pOH = - log [OH^-]\] \[4.8 = - log [OH^-]\] \[[OH^-] = 10^{-4.8}\] Since we're given the Ka value for HA as \(8.0 × 10^{-5}\), we can write the equilibrium-constant expression for HA as: \[K_a = [\mathrm{H^+}][\mathrm{A^-}]/[\mathrm{HA}]\] We're also given that equal quantities of 0.010 M solutions of HA and B are mixed. When HA reacts with B, it donates a proton to B. So the amount of H⁺ ions from HA is: \[[\mathrm{H^+}] = 0.010 - [OH^-]\] Now we can plug in all the values in the Ka expression: \[8.0 × 10^{-5} = (0.010 - [OH^-])[\mathrm{A^-}]/[\mathrm{HA}]\] We know that the concentration of A⁻ is equal to the concentration of HB, and the same for HA and B, we can find the value of the equilibrium constant K as follows: \[K = \frac{[\mathrm{A^-}][\mathrm{HB}]}{[\mathrm{HA}][\mathrm{B}]} = \frac{[\mathrm{A^-^2}]}{[\mathrm{HA}]^2} = \frac{(0.010 - [OH^-])^2 }{(8.0 × 10^{-5})}\] Now substitute the value of [OH⁻] in the above equation and calculate the value of K: \[K = \frac{(0.010 - 10^{-4.8})^2 }{(8.0 × 10^{-5})}\]

Step 3: (c) Value of Kb for B

The relation between Ka, Kb, and Kw for an acid-base conjugate pair is given by: \[K_a × K_b = K_w\] Where Kw is the ion-product constant for water: \(K_w = 1.0 × 10^{-14}\) Using the given Ka value and the calculated K value, we can find the Kb value for the base B: \[K_b = \frac{K_w}{K_a} = \frac{1.0 × 10^{-14}}{8.0 × 10^{-5}}\] Calculate the final value for Kb by dividing the constants.

Key Concepts

Equilibrium Constant

In any chemical reaction at equilibrium, the quantities of reactants and products reach a balance. The equilibrium constant, often represented as \(K\), helps in understanding this balance. For the reaction between an acid \(HA\) and a base \(B\), the equilibrium constant expression is as follows: - \(K = \frac{[A^-][HB]}{[HA][B]}\) Here, \([A^-]\) and \([HB]\) are the concentrations of the reaction products, and \([HA]\) and \([B]\) are the concentrations of the reactants at equilibrium. This expression shows the ratio of the product of the concentrations of the products to that of the reactants. Understanding the equilibrium constant is crucial for predicting the direction of the reaction. A large \(K\) value indicates that the products are favored, whereas a small \(K\) suggests that reactants are favored. It's a fundamental concept in assessing the extent of the reaction's progression.

pH and pOH Relationship

The pH and pOH relationship is a vital concept in understanding the acidity and basicity of a solution. They are related by the simple equation: - \(\text{pH} + \text{pOH} = 14\) This relationship holds true at 25°C, representing the equilibrium between hydrogen ions \((H^+)\) and hydroxide ions \((OH^-)\) in water. When given the pH or pOH of a solution, we can quickly determine the other value using this relationship. For example, a pH of 9.2 implies: - \(9.2 + \text{pOH} = 14\) - \(\text{pOH} = 4.8\)The pOH value then helps determine the concentration of \([OH^-]\) in the solution. This understanding is crucial in acid-base reactions, helping us evaluate the ionic composition of solutions.

Ka and Kb Calculations

Acid dissociation constant \((K_a)\) and base dissociation constant \((K_b)\) are fundamental in understanding the strength of acids and bases. For an acid \(HA\), the \(K_a\) represents how well it dissociates into its ions: - \(K_a = \frac{[H^+][A^-]}{[HA]}\)In reverse, the \(K_b\) is used for bases like \(B\), and it represents the base's ability to dissociate: - \(K_b = \frac{[HB^+][OH^-]}{[B]}\)Since \(K_a\), \(K_b\), and the ion-product constant for water \(K_w\) are connected, we can use the relationship: - \(K_a \times K_b = K_w\)This link allows us to calculate \(K_b\) when \(K_a\) and \(K_w\) are known, providing a deeper insight into the base's properties. Understanding these calculations aids in comparing different acids and bases.

Chemical Equilibrium Expressions

Chemical equilibrium expressions provide a mathematical way to represent chemical reactions at balance. In acid-base reactions, the expression is crucial to analyze the behavior of reactants and products over time.For the equilibrium involving \(HA\) and \(B\): - \(HA + B \leftrightarrow A^- + HB\)The corresponding equilibrium expression is: - \(K = \frac{[A^-][HB]}{[HA][B]}\)This equation represents how the reactants convert to products and vice versa, guided by the equilibrium constant \(K\). The expression helps determine concentrations at equilibrium, showing symmetry between forward and reverse reactions.Building these expressions involves understanding and using concentrations of species in the equilibrium mixture. They are pivotal in practical applications such as solving real-world problems related to acids and bases.