Problem 85 - Chemistry The Central Science In Si Units

Question

Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4} \mathrm{at}\) \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathbf{a})\) a solution formed by adding \(30.0 \mathrm{~g}\) of furoic acid and \(25.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.300 \mathrm{~L}\) of solution, \((\mathbf{b})\) a solution formed by mixing \(20.0 \mathrm{~mL}\). of \(0.200 \mathrm{M}\) \(\mathrm{HC}_{\mathrm{s}} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL},(\mathbf{c})\) a solution prepared by adding \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) solution to \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{MHC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\)

Step-by-Step Solution

Answer

(a) pH ≈ 3.04, (b) pH ≈ 3.43, (c) pH ≈ 12.30.

Step 1: Calculate Molar Mass of Furoic Acid and Sodium Furoate

First, we need to calculate the molar mass of furoic acid (HC5H3O3) and sodium furoate (NaC5H3O3). Furoic acid has a molar mass of approximately 112 g/mol, while sodium furoate has a molar mass of approximately 134 g/mol.

Step 2: Compute Moles in (a)

For furoic acid: \[ \text{Moles of HC}_5\text{H}_3\text{O}_3 = \frac{30.0 \text{ g}}{112 \text{ g/mol}} \approx 0.268 \text{ mol} \] For sodium furoate: \[ \text{Moles of NaC}_5\text{H}_3\text{O}_3 = \frac{25.0 \text{ g}}{134 \text{ g/mol}} \approx 0.187 \text{ mol} \] The initial concentrations in 0.300 L solution are approximately 0.893 M for furoic acid and 0.623 M for sodium furoate.

Step 3: Apply Henderson-Hasselbalch Equation in (a)

The Henderson-Hasselbalch equation is given by: \[ \text{pH} = \text{p}K_a + \log{\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)} \] Here, \( \text{p}K_a = -\log(6.76 \times 10^{-4}) \approx 3.17 \). Substituting the values: \[ \text{pH} = 3.17 + \log{\left(\frac{0.623}{0.893} \right)} \approx 3.04 \]

Step 4: Calculate Moles in (b)

For HC5H3O3: \[ \text{Moles of HC}_5\text{H}_3\text{O}_3 = 20.0 \text{ mL} \times 0.200 \text{ M} = 0.004 \text{ mol} \] For NaC5H3O3: \[ \text{Moles of NaC}_5\text{H}_3\text{O}_3 = 30.0 \text{ mL} \times 0.250 \text{ M} = 0.0075 \text{ mol} \] The total solution volume is 125 mL or 0.125 L, yielding concentrations of 0.032 M for the acid and 0.06 M for the salt.

Step 5: Apply Henderson-Hasselbalch Equation in (b)

Using the same \( \text{p}K_a \) value, substitute the concentrations:\[ \text{pH} = 3.17 + \log{\left(\frac{0.06}{0.032} \right)} \approx 3.43 \]

Step 6: Calculate Moles in (c)

For NaOH: \[ \text{Moles of NaOH} = 25.0 \text{ mL} \times 1.00 \text{ M} = 0.025 \text{ mol} \] For HC5H3O3: \[ \text{Moles of HC}_5\text{H}_3\text{O}_3 = 100.0 \text{ mL} \times 0.100 \text{ M} = 0.010 \text{ mol} \] The NaOH will react completely with furoic acid, turning it into a salt.

Step 7: Determine pH After Reaction (c)

NaOH completely reacts with furoic acid, resulting in 0.015 mol of sodium furoate in 125 mL or 0.125 L solution, which translates to a concentration of 0.12 M. Since no weak acid remains, the pH is determined by the strong base NaOH:\[ \text{pOH} = -\log(0.02) \approx 1.70 \]\[ \text{pH} = 14.00 - 1.70 = 12.30 \]

Key Concepts

Henderson-Hasselbalch Equation

Calculating the pH of solutions, especially those involving weak acids and their conjugate bases, can be efficiently done using the Henderson-Hasselbalch equation. This equation helps us relate the pH of a buffer solution to its acid and conjugate base concentrations, using the formula:

• \( \text{pH} = \text{p}K_a + \log{\left( \frac{[\text{A}^-]}{[\text{HA}]} \right)} \)

Here, \([\text{A}^-]\) is the concentration of the base and \([\text{HA}]\) is the concentration of the acid. The \(\text{p}K_a\) is a measure of the acid strength, specifically the inverse logarithm of the acid dissociation constant \(K_a\). Using this equation, we can approximate the pH of buffer solutions by knowing the \(\text{p}K_a\) and the ratio of the concentrations of the acid and base components.
It provides a simple way to understand how changes in the ratio of acid and base in a solution affect its pH, making it essential for acid-base reaction calculations.

Acid-Base Reactions

Acid-base reactions are chemical reactions that involve the transfer of a proton from an acid to a base. In the context of buffer solutions, these reactions play a crucial role. When you add a strong acid to a buffer, the base component of the buffer neutralizes the added acid. Conversely, when a strong base is added, the acid component neutralizes the base. This results in minimal changes to the pH. In the exercise provided, sodium furoate acts as the conjugate base of furoic acid. When calculating the pH, the interaction between the weak acid (furoic acid) and its conjugate base (sodium furoate) in the solution exemplifies acid-base reaction principles. The ability to predict how an acid and a base will react based on their relative strengths is foundational in understanding buffer actions and pH calculations.
The importance of acid-base balance is highlighted in the scenario with NaOH reacting with furoic acid, where complete neutralization leads to remaining base calculation for the final pH.

Buffer Solution

Buffer solutions are crucial in maintaining a stable pH in chemical and biological systems. They consist of a weak acid and its conjugate base (or a weak base and its conjugate acid). The pivotal property of a buffer is its capacity to resist drastic pH changes upon the addition of small amounts of acids or bases.In our example, the furoic acid-sodium furoate solution acts as a buffer. The presence of both a weak acid and its conjugate base allows the solution to neutralize added acids or bases, maintaining the pH within a narrow range. This is due to the equilibrium between the weak acid and its conjugate base, which shifts in response to the added substances, addressing any increase or decrease in H⁺ ions.
Key points about buffer solutions are:

• They stabilize pH by neutralizing added acids/bases.
• They are essential in many laboratory and physiological systems.
• The effectiveness of a buffer depends on the \( [\text{A}^-]/[\text{HA}] \) ratio.

Ka Value

The \(K_a\) value, or acid dissociation constant, represents the strength of a weak acid in solution. It measures the degree to which an acid dissociates into its ions in water. A larger \(K_a\) value indicates a stronger acid, as it dissociates more completely.In this exercise, the \(K_a\) value of furoic acid (6.76 \times 10^{-4}) is used to calculate the \(\text{p}K_a\), which is essential for applying the Henderson-Hasselbalch equation. The \(\text{p}K_a\) is calculated as the negative logarithm of \(K_a\):

• \(\text{p}K_a = -\log(K_a)\)

For instance, a \(K_a\) value around \(10^{-4}\) indicates an acid weaker than a strong acid like hydrochloric acid but strong enough to partially dissociate in solution.
Understanding \(K_a\) allows chemists to predict the behavior of acid-base reactions and effectively apply buffer systems. It is integral in understanding the strength and reactivity of acids in various chemical contexts.