Question

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(B\) are mixed. The \(p H\) of the resulting solution is 9.2. (a) Write the equilibrium equation and equilibriumconstant 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 HA and B? (c) What is the value of \(K_{b}\) for B?

Short answer

In summary, for the reaction between an acid HA and a base B, the equilibrium constant expression can be written as \(K_c = \frac{[A^-][HB^+]}{[HA][B]}\). Given the Ka for the acid HA is \(8.0\times10^{-5}\), we determined the equilibrium constant for the reaction between HA and B to be \(K_c = \frac{(8.0\times10^{-5})(1\times10^{-14})}{K_b}\). Finally, we calculated the Kb value for the base B to be \(1.25\times10^{-10}\).

Step-by-step solution

Write the equilibrium equation and equilibrium constant expression

For the reaction between an acid HA and a base B, we can write the following generic equilibrium equation: \[HA + B \rightleftharpoons A^- + HB^+\] Now we need to write the equilibrium constant expression for this reaction. We know that equilibrium constant Kc is defined as the ratio of the concentrations of the products to the concentration of the reactants, with each concentration raised to the power of their respective stoichiometric coefficients. So, for the given reaction, the expression for the equilibrium constant Kc can be written as: \[K_c = \frac{[A^-][HB^+]}{[HA][B]}\] Now let's move on to the next part.

Determine the equilibrium constant for the reaction

We can relate Ka, Kb, and Kc with the following equation: \[K_a\times K_b = K_c\times K_w\] or \[K_c = \frac{K_a\times K_w}{K_b}\] We are given that the Ka value for the acid HA is \(8.0\times10^{-5}\). We also know that the ion product of water, Kw, is \(1\times10^{-14}\) at 25°C. Now, we can substitute these values into the equation to find the Kc for the reaction: \[K_c = \frac{(8.0\times10^{-5})(1\times10^{-14})}{K_b}\] Now, let's move on to the next part of the question to find the Kb of the base B.

Determine the Kb value for the base B

We're given the pH of the resulting solution as 9.2. From this, we can find the pOH using the following equation: \[pH + pOH = 14\] \[pOH = 14 - pH = 14-9.2 = 4.8\] Now, we can find the hydroxide ion concentration, [OH-], using the pOH value: \[pOH = -\log{[OH^-]}\] \[4.8 = -\log{[OH^-]}\] \[[OH^-] = 10^{-4.8} = 1.58\times10^{-5}\] Since the concentrations of HA and B are equal and given as 0.010 M, we can now calculate the concentration of A- and B. Since we're given the Ka, let's find the hydronium ion concentration, [H3O+]: \[[H3O^+] = \frac{K_a}{[A^-]} = \frac{8.0\times10^{-5}}{0.010} = 8.0\times10^{-4}\] Now, since we know the concentration of hydroxide ions, we can find the concentration of the base B and therefore, the value of Kb: \[K_b = \frac{K_c}{K_a} \times K_w = \frac{(1\times10^{-14})}{(8.0\times10^{-5})} = 1.25 \times 10^{-10}\] So, the value of Kb for the base B is \(1.25\times10^{-10}\).

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant (\(K_c\)) is an important parameter that helps us understand how the reaction behaves at equilibrium. It is the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients in the balanced chemical equation. Consider the reaction between the acid \(\text{HA}\) and base \(\text{B}\):

• The reaction is \(\text{HA} + \text{B} \rightleftharpoons \text{A}^- + \text{HB}^+\).
• The equilibrium constant expression is \(K_c = \frac{[A^-][HB^+]}{[HA][B]}\).

The value of \(K_c\) gives us insights into the extent of the reaction:

• If \(K_c\) is large, the reaction favors the formation of products.
• If \(K_c\) is small, the reactants are favored.

This is crucial for understanding the balance within the system at a particular temperature.

pH Calculation

The pH of a solution is a measure of its acidity or basicity. It is calculated as the negative logarithm of the hydronium ion concentration \([H^+]\). The pH scale typically ranges from 0 to 14, with values under 7 indicating an acidic solution, and values over 7 indicating a basic solution. In this scenario:

• The pH of the resulting solution is given as 9.2.

To find the \(\text{pOH}\), which is the equivalent measure for hydroxide ions \([OH^-]\), use:

• \(\text{pH} + \text{pOH} = 14\).

Substituting the given pH:

• \(\text{pOH} = 14 - 9.2 = 4.8\).

The \(pH\) and \(\text{pOH}\) are crucial for determining the concentrations of hydrogen and hydroxide ions, which help us further understand the chemical equilibrium of an acid-base reaction.

Ionization Constant

The ionization constant of an acid (\(K_a\)) or a base (\(K_b\)) quantifies the strength of the acid or base by measuring its ability to donate or accept protons in solution. For the given acid \(\text{HA}\), \(K_a\) is provided as \(8.0 \times 10^{-5}\):

• This value indicates that \(\text{HA}\) is a weak acid.

Similarly, the base ionization constant \(K_b\) tells us about the strength of base \(\text{B}\). Using the relation:

• \(K_a \times K_b = K_c \times K_w\),
• Where \(K_w\) is the ion product constant of water, typically \(1 \times 10^{-14}\) at 25°C.

Knowing \(\text{pH} = 9.2\), we can calculate \([OH^-]\) and find \(K_b\) to further detail the characteristics of base \(\text{B}\). Understanding these constants is key to predicting the direction of the equilibrium shift and the extent of the reaction.

Chemical Equations

Chemical equations symbolize chemical reactions using chemical formulas. They provide information about the reactants and products and their respective quantities. For the reaction between \(\text{HA}\) and \(\text{B}\), the balanced equation is as follows:

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

This indicates that one molecule of \(\text{HA}\) reacts with one molecule of \(\text{B}\) to produce one molecule each of \(\text{A}^-\) and \(\text{HB}^+\). The purpose of writing chemical equations includes:

• Clarifying the stoichiometry of the reaction.
• Helping with the calculation of reactant/product concentrations.
• Providing a basis for deriving equilibrium expressions.

By analyzing this equation, one can derive the equilibrium constant expression, enabling further exploration of how the mixture of chemicals behaves at equilibrium.