Problem 59 - Chemistry The Central Science In Si Units

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

Calculate the x value from Ka or Kb, and then use it to find pH.

Step 1: Identify the equilibrium expression and constants

For each substance, identify whether it acts as an acid or base in solution, and write the equilibrium expression using its ionization constant from Appendix D. (a) Propionic acid (\( C_2H_5COOH\)) is a weak acid with a given \( K_a\) value. It partially ionizes according to:\[C_2H_5COOH \rightleftharpoons C_2H_5COO^- + H^+\](b) Hydrogen chromate ion (\( HCrO_4^-\)) can donate a proton, thus behaves as a weak acid; use \( K_a\) value as provided.(c) Pyridine (\( C_5H_5N\)) acts as a weak base in solution, so we reference its \( K_b\) to form its equilibrium:\[C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^-\]

Step 2: Write the expression for ionization

(a) For the propionic acid, set up the ionization expression: \[K_a = \frac{{[H^+][C_2H_5COO^-]}}{{[C_2H_5COOH]}}\](b) For hydrogen chromate ion, it is:\[K_a = \frac{{[H^+][CrO_4^{2-}]}}{{[HCrO_4^-]}}\](c) For pyridine, convert \( K_b\) to \( K_a\) using \(K_a \cdot K_b = K_w\) (where \(K_w = 1.0 \times 10^{-14}\)).It becomes:\[K_a = \frac{{[H^+][C_5H_5N]}}{{[C_5H_5NH^+]}}\]

Step 3: Set up the equilibrium concentrations

Assume the initial concentration of the weak acid or base is the initial molarity given. Define 'x' as the change in concentration of products and write the equilibrium concentrations in terms of x:(a) Initial: \([C_2H_5COOH] = 0.150\) MAt equilibrium:\([C_2H_5COOH] = 0.150 - x\), \([C_2H_5COO^-] = x\), \([H^+] = x\)(b) Initial: \([HCrO_4^-] = 0.250\) MAt equilibrium:\([HCrO_4^-] = 0.250 - x\), \([CrO_4^{2-}] = x\), \([H^+] = x\)(c) Initial: \([C_5H_5N] = 0.750\) MAt equilibrium:\([C_5H_5NH^+] = x\), \([OH^-] = x\), \([C_5H_5N] = 0.750 - x\)

Step 4: Solve for x (Hydrogen ion or hydroxide ion concentration)

Insert the equilibrium concentrations into the expression for each case and solve for x, which gives the concentration of H+ or OH-:For an acid like propionic acid and hydrogen chromate, solve:(a)\[K_a = \frac{x^2}{0.150 - x} \approx \frac{x^2}{0.150}\](b)\[K_a = \frac{x^2}{0.250 - x} \approx \frac{x^2}{0.250}\]For a base like pyridine, solve for x using \(K_b\):(c)\[K_b = \frac{x^2}{0.750 - x} \approx \frac{x^2}{0.750}\]

Step 5: Calculate pH

(a) Find \( x\) and calculate pH:\[\text{pH} = -\log([H^+]) = -\log(x)\](b) Similarly calculate \( x\) and find the pH:\[\text{pH} = -\log(x)\](c) For base pyridine, first find \([OH^-] = x\), then:\[\text{pOH} = -\log([OH^-]) = -\log(x)\]Finally, calculate pH:\[\text{pH} = 14 - \text{pOH}\]

Key Concepts

Weak Acids and Bases

In chemistry, weak acids and bases are substances that do not completely dissociate into their ions when dissolved in water. Unlike strong acids and bases, which fully ionize in solution, weak acids and weak bases only partially ionize. This partial ionization means that an equilibrium exists between the undissociated molecules and the ions present in the solution. For weak acids, this equilibrium is represented by the equation:

• Acid dissociation: \[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\]

For bases, the scenario is quite similar but involves hydroxide ions. Pyridine, for example, is a weak base that partially ionizes:

• Base ionization: \[\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-\]

The importance of weak acids and bases in pH calculation is due to their tendency to maintain a certain level of ionization, which is crucial for calculating the concentrations of ions in the solution and ultimately the pH.

Ionization Constant

The ionization constant, whether it be the acid dissociation constant \(K_a\) or the base dissociation constant \(K_b\), is a crucial factor when dealing with weak acids and bases. It reflects the extent of ionization of a weak acid or base and how far the equilibrium lies on the side of the ions.For weak acids, the acid dissociation constant \(K_a\) is expressed as:

• \[ K_a = \frac{[H^+][A^-]}{[HA]}\]

This expression shows the ratio of the concentration of the ionized forms (protons and the conjugate base) to the non-ionized acid concentration. A higher \(K_a\) value implies a stronger weak acid, meaning it ionizes more in solution.Similarly, weak bases have a base dissociation constant \(K_b\), given by:

• \[ K_b = \frac{[BH^+][OH^-]}{[B]}\]

For bases, a higher \(K_b\) value suggests stronger base properties, with more ionization in solution. The ionization constant helps predict the pH of the solution and is fundamental in establishing the equilibrium expressions used in pH calculations.

Equilibrium Expressions

Equilibrium expressions are vital when assessing the ionization of weak acids and bases. They describe the balance between reactants and products in a solution at equilibrium. To establish these expressions, it’s essential to write out the balanced equation for the dissociation and use the given initial concentrations as a reference point.Consider the general expression setup for an acid:

• \[ K_a = \frac{[H^+][A^-]}{[HA]} \]

For bases, the process is similar:

• \[ K_b = \frac{[BH^+][OH^-]}{[B]} \]

When solving the pH for a weak acid or base, we often assume that the change in concentration (denoted as \(x\)) is very small compared to the initial concentration. Therefore, it simplifies the math as we approximate \([HA]_0 - x \approx [HA]_0\), for acids, or similarly for bases.Understanding and applying equilibrium expressions allows for accurate calculation of ion concentrations, leading to a precise determination of pH, which is an essential concept particularly in chemistry.