Problem 59 - Chemistry: The Central Science in SI Units, Global Edition

Question

Calculate the pH of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix D): (a) \(0.150 \mathrm{M}\) propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\) (b) \(0.250 \mathrm{M}\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}^{-}\right),(\mathbf{c}) 0.750 \mathrm{M}\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\)

Step-by-Step Solution

Answer

The pH of the given solutions can be calculated as follows: (a) Propionic acid: Using the given \(K_a\), set up an equilibrium expression and solve for \(\mathrm{H^+}\) ions concentration. Then, use the pH formula to find the pH: \(\mathrm{pH} = -\log_{10} [\mathrm{H^+}]\). (b) Hydrogen chromate ion: Repeat the process for the acidic solution using the appropriate \(K_a\), and find the pH using the \(\mathrm{H^+}\) ions concentration. (c) Pyridine: As a basic solution, use the given \(K_b\) and solve for \(\mathrm{OH^-}\) ions concentration. Find the \(\mathrm{H^+}\) ions concentration using the ion product of water \((K_w)\), and finally calculate the pH using the \(\mathrm{H^+}\) ions concentration. By following these steps, you can determine the pH of each solution.

Step 1: Identify the nature of each compound

(a) Propionic acid is an organic acid and will dissociate in the solution to form \(\mathrm{H^+}\) ions. Therefore, it is an acidic solution. We will use its \(K_a\) to find the concentration of \(\mathrm{H^+}\) ions. (b) Hydrogen chromate ion is an anion containing a hydrogen ion and will also dissociate in the solution to form \(\mathrm{H^+}\) ions. Therefore, it is also an acidic solution. We will use its \(K_a\) to find the concentration of \(\mathrm{H^+}\) ions. (c) Pyridine is a weak base and will accept \(\mathrm{H^+}\) ions from water. Therefore, it is a basic solution. We will use its \(K_b\) to find the concentration of \(\mathrm{OH^-}\) ions and then calculate the concentration of \(\mathrm{H^+}\) ions.

Step 2: Write the dissociation equilibrium equations

(a) For propionic acid: \[\mathrm{C_2H_5COOH} \rightleftharpoons \mathrm{C_2H_5COO^-} + \mathrm{H^+}\] (b) For hydrogen chromate ion: \[\mathrm{HCrO_4^-} \rightleftharpoons \mathrm{CrO_4^{2-}} + \mathrm{H^+}\] (c) For pyridine: \[\mathrm{C_5H_5N} + \mathrm{H_2O} \rightleftharpoons \mathrm{C_5H_5NH^+} + \mathrm{OH^-}\]

Step 3: Calculate the concentration of H+ or OH- ions

For each solution, we will use the \(K_a\) or \(K_b\) values and create an equilibrium table to find the concentrations of \(\mathrm{H^+}\) or \(\mathrm{OH^-}\) ions. Since the exercise provides the \(K_a\) and \(K_b\) values in Appendix D, we can use them for our calculations. (a) Let \(x\) be the concentration of \(\mathrm{C_2H_5COO^-}\) and \(\mathrm{H^+}\): \[K_{a}=\frac{[\mathrm{C_2H_5COO^-}][\mathrm{H^+}]}{[\mathrm{C_2H_5COOH}]}\] (b) Let \(x\) be the concentration of \(\mathrm{CrO_4^{2-}}\) and \(\mathrm{H^+}\): \[K_{a}=\frac{[\mathrm{CrO_4^{2-}}][\mathrm{H^+}]}{[\mathrm{HCrO_4^-}]}\] (c) Let \(x\) be the concentration of \(\mathrm{C_5H_5NH^+}\) and \(\mathrm{OH^-}\): \[K_{b}=\frac{[\mathrm{C_5H_5NH^+}][\mathrm{OH^-}]}{[\mathrm{C_5H_5N}]}\]

Step 4: Calculate the pH of each solution

For each solution, find the concentration of \(\mathrm{H^+}\) ions and use the pH formula: \[\mathrm{pH} = -\log_{10} [\mathrm{H^+}]\] For the basic solution (c), first find the concentration of \(\mathrm{H^+}\) ions from the \(\mathrm{OH^-}\) ion concentration using the relation: \[[\mathrm{H^+}][\mathrm{OH^-}] = K_w\] where \(K_w (1\times 10^{-14})\) is the ion product of water. Then, calculate the pH using the found concentration of \(\mathrm{H^+}\) ions. Finally, you will have the pH for each of the given solutions.

Key Concepts

Acid Dissociation Constant

The acid dissociation constant, commonly represented as \(K_a\), is vital in understanding how strong or weak an acid is when dissolved in water. It provides insight into the extent to which an acid can donate protons (\(H^+\) ions) to water. A smaller \(K_a\) value means the acid is weaker because it donates fewer protons. On the other hand, a larger \(K_a\) indicates a stronger acid. When calculating \(K_a\), we look at the equilibrium of the dissociation of the acid in solution. For example, propionic acid dissociates in water:

• \[\mathrm{C_2H_5COOH} \rightleftharpoons \mathrm{C_2H_5COO^-} + \mathrm{H^+}\]

In this equation, \(K_a\) quantifies the concentrations of each species when equilibrium is reached. To find pH from \(K_a\), we note the formula \(K_{a} = \frac{[\mathrm{C_2H_5COO^-}][\mathrm{H^+}]}{[\mathrm{C_2H_5COOH}]}\). Solving for \([\mathrm{H^+}]\) allows us to calculate the pH using \(pH = -\log_{10} [\mathrm{H^+}]\). This step is crucial for accurately determining the acidity of the solution.

Equilibrium Equations

Equilibrium equations are essential tools in chemistry, as they represent the state at which both the reactants and products of a chemical reaction exist in precise balance. In the context of acids and bases, they tell us how reactants like acids dissociate or how bases accept protons.For each of the solutions in the exercise:

• Propionic acid dissociation: \[\mathrm{C_2H_5COOH} \rightleftharpoons \mathrm{C_2H_5COO^-} + \mathrm{H^+}\]
• Hydrogen chromate dissociation: \[\mathrm{HCrO_4^-} \rightleftharpoons \mathrm{CrO_4^{2-}} + \mathrm{H^+}\]
• Pyridine protonation: \[\mathrm{C_5H_5N} + \mathrm{H_2O} \rightleftharpoons \mathrm{C_5H_5NH^+} + \mathrm{OH^-}\]

These equations underline how reactants form products and the dynamic steady state that equilibrium establishes in solutions. When calculating pH, these equations help predict the concentrations of ions, particularly \([\mathrm{H^+}]\) or \([\mathrm{OH^-}]\), which are pivotal for determining the acidity or basicity of a solution.

Weak Acids and Bases

Understanding weak acids and bases is key to grasping many chemical reactions. A weak acid does not fully ionize in solution; instead, it establishes an equilibrium between its un-ionized form and the ions it produces. This partial ionization means that weak acids have a higher pH compared to strong acids, because they release fewer \(H^+\) ions.Similarly, weak bases do not completely dissociate to yield \(OH^-\) ions. An example of a weak base is pyridine, which accepts \(H^+\) ions from water. Unlike strong bases that almost fully dissociate in solution, weak bases like pyridine only partially engage in this exchange with water.When working with solutions of weak acids or bases, it's crucial to use the equilibrium constant \(K_a\) or \(K_b\) to define their behavior in water:- For weak acids, use \(K_a\) to determine how efficiently the acid donates \(H^+\) ions.- For weak bases, use \(K_b\) to understand the base's ability to accept \(H^+\) ions.Being familiar with these concepts helps students predict solution behaviors, understand their pH, and solve chemistry problems with precision.