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 \(9.2 .(\mathrm{a})\) Write the equilibrium equation and equilibrium-constant expression for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\). (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?

Short answer

The equilibrium constant for the reaction between HA and B is approximately \(1.56 \times 10^{6}\), and the Kb value for base B is approximately \(1.25 \times 10^{-10}\).

Step-by-step solution

Write the reaction and equilibrium equation

The reaction between acid HA and base B can be written as follows: \[ \mathrm{HA} + \mathrm{B}^{-} \rightleftharpoons \mathrm{A}^{-} + \mathrm{HB} \] The equilibrium constant expression for this reaction can be written as: \[ K = \frac{ [\mathrm{A}^{-}][\mathrm{HB}]} { [\mathrm{HA}][\mathrm{B}^{-}]} \]

Note the given values

We have the following given values: - The pH of the resulting solution is 9.2 - The Ka value for HA is \( 8.0 \times 10^{-5} \)

Calculate the concentration of \(\mathrm{A}^{-}\) and \(\mathrm{HB}\) in the resulting solution

We know that pH is related to the concentration of hydrogen ions as: \[ \mathrm{pH} = - \log[H^{+}] \] We are given that the pH of the resulting solution is 9.2. Hence, we can calculate the concentration of hydrogen ions: \[ H^{+} = 10^{-\mathrm{pH}} = 10^{-9.2} \] Since the resulting solution has negligible hydrogen ions, we can assume that most of the acid HA has dissociated into A- and HB. So, the concentrations of A- and HB in the resulting solution will be approximately equal to the initial concentration of HA: \[ [\mathrm{A}^{-}] \approx [\mathrm{HB}] \approx 0.010 \mathrm{M} \]

Calculate the equilibrium constant for the reaction between HA and B

We can now use the equilibrium constant expression: \[ K = \frac{ [\mathrm{A}^{-}][\mathrm{HB}]} { [\mathrm{HA}][\mathrm{B}^{-}]} \] We know that: \[ [\mathrm{A}^{-}] \approx [\mathrm{HB}] \approx 0.010 \mathrm{M} \] As equal quantities of HA and B are mixed, their initial concentrations can be considered equal: \[ [\mathrm{HA}] \approx [\mathrm{B}^{-}] \] So, the expression for K can be simplified to: \[ K = \frac{ (0.010)^{2}} { [\mathrm{HA}]^{2}} \] Substituting Ka for HA, we have: \[ K = \frac{(0.010)^{2}}{(8.0 \times 10^{-5})^{2}} \] \[ K \approx 1.56 \times 10^{6} \] So, the equilibrium constant for the reaction between HA and B is approximately \(1.56 \times 10^{6}\).

Calculate the Kb value for B

We know that for a conjugate acid-base pair, the following relationship holds true: \[ K_{a} \cdot K_{b} = K_{w} \] where \(K_{w}\) is the ion product of water, which is \(1.0 \times 10^{-14}\) at 25°C. We are given the Ka value for HA and we just found the equilibrium constant for the reaction between HA and B. Now, we need to find Kb for B. Re-write the above equation as: \[ K_{b} = \frac{K_{w}}{ K_{a} } \] Substitute the given values and the calculated constant: \[ K_{b} = \frac{(1.0 \times 10^{-14})}{(8.0 \times 10^{-5})} \] \[ K_{b} \approx 1.25 \times 10^{-10} \] So, the Kb value for base B is approximately \(1.25 \times 10^{-10}\).

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, denoted as \( K \), is a crucial factor that describes the relationship between the concentrations of reactants and products at equilibrium. It gives insight into the position of equilibrium and the extent to which a reaction proceeds. The equilibrium constant for a reaction involving an acid \( \mathrm{HA} \) and a base \( \mathrm{B}^{-} \) is given by:\[ K = \frac{ [\mathrm{A}^{-}][\mathrm{HB}]}{ [\mathrm{HA}][\mathrm{B}^{-}]} \]Here, \( [\mathrm{A}^{-}] \) and \( [\mathrm{HB}] \) are the concentrations of the products while \( [\mathrm{HA}] \) and \( [\mathrm{B}^{-}] \) are the concentrations of the reactants at equilibrium. The value of \( K \) tells us whether products or reactants are favored under the conditions of the reaction:

• If \( K \) is much greater than 1, the reaction goes almost to completion, favoring the formation of products.
• If \( K \) is much less than 1, the reaction favors the reactants, indicating only a small extent of conversion to products.

Understanding \( K \) helps chemists anticipate how changes in conditions, like concentration, temperature, or pressure, will affect the equilibrium position.

pH Calculation

Calculating pH is vital in determining the acidity or basicity of a solution. pH is calculated as the negative logarithm of the hydrogen ion concentration: \[ \mathrm{pH} = -\log[H^{+}] \] A pH value below 7 indicates an acidic solution, whereas a value above 7 suggests a basic solution. For neutral solutions, the pH is exactly 7. In our problem, the pH of the solution was given as 9.2, highlighting its basic nature. The hydrogen ion concentration \( [H^{+}] \) can be determined by: \[ H^{+} = 10^{-\mathrm{pH}} = 10^{-9.2} \]Here are a few guidelines for interpreting pH:

• A small change in pH represents a large change in [H+] due to its logarithmic nature.
• Higher pH corresponds to lower hydrogen ion concentration.

Understanding pH values is essential not just in laboratory settings but also in biological and environmental sciences, influencing processes like enzyme activity and pollution levels.

Conjugate Acid-Base Pair

In acid-base equilibrium, a conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton (\( H^+ \)). An acid donates a proton to become its conjugate base, while a base accepts a proton to transform into its conjugate acid. For the given exercise, the pair is \( \mathrm{HA} \) (acid) and \( \mathrm{A}^{-} \) (conjugate base), and \( \mathrm{B} \) (base) and \( \mathrm{HB} \) (conjugate acid). The relationship between the acid and base in a conjugate pair has significant implications:

• The strength of an acid is inversely related to the strength of its conjugate base. A strong acid has a weak conjugate base and vice versa.
• This concept is crucial in buffer solutions, where weak acids and their conjugate bases help maintain constant pH levels despite the addition of acids or bases.

By understanding conjugate acid-base pairs, we see how equilibrium is maintained in solutions and how they help buffer and resist drastic pH changes.