Question

Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 \(\mathrm{L}\) of each of the following solutions: a. 0.050\(M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)\) and 0.080\(M\) sodium propanoate b. 0.50\(M\) propanoic acid and 0.80\(M\) sodium propanoate c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

Short answer

a) The pH of solution a after adding NaOH is \(4.34\). b) The pH of solution b after adding NaOH is \(4.87\). c) Solution a does not act as a buffer after adding NaOH.

Step-by-step solution

Calculate the initial moles and concentration of each species

Before adding NaOH, we can calculate the initial moles and concentration of propanoic acid (HC3H5O2) and sodium propanoate (C3H5O2- Na+) in each solution. For solution a: Initial [HC3H5O2] = 0.050 M Initial [C3H5O2-] = 0.080 M For solution b: Initial [HC3H5O2] = 0.50 M Initial [C3H5O2-] = 0.80 M

Calculate the moles and concentration of species after NaOH is added

When NaOH is added, it reacts with propanoic acid as follows: HC3H5O2 + OH- -> C3H5O2- + H2O Since 0.15 mole of NaOH is added to 1.00 L of solution, the concentration of NaOH is 0.15 M. Now we will calculate the new concentration of propanoic acid and its conjugate base. For solution a: New [HC3H5O2] = 0.050 M - 0.15 M New [C3H5O2-] = 0.080 M + 0.15 M For solution b: New [HC3H5O2] = 0.50 M - 0.15 M New [C3H5O2-] = 0.80 M + 0.15 M

Calculate the pH using the Henderson-Hasselbalch equation

Now we can use the Henderson-Hasselbalch equation to find the pH of each solution: pH = pKa + log([C3H5O2-]/[HC3H5O2]) First, we need to determine pKa from the given Ka value: pKa = -log(Ka) = -log(1.3 × 10^{-5}) ≈ 4.89 Now we can find the pH for each solution: For solution a: pH = 4.89 + log((0.080 + 0.15)/(0.050 - 0.15)) ≈ 4.34 For solution b: pH = 4.89 + log((0.80 + 0.15)/(0.50 - 0.15)) ≈ 4.87

Determine if the solution in part a is still a buffer after adding NaOH

To check if solution a still acts as a buffer solution, we can look at the ratio of [C3H5O2-] to [HC3H5O2] after adding NaOH. The new concentrations are: [C3H5O2-] = 0.080 M + 0.15 M = 0.23 M [HC3H5O2] = 0.050 M - 0.15 M = -0.10 M Since the concentration of HC3H5O2 is negative after adding NaOH, it means that all of the weak acid has reacted with NaOH, and there is no weak acid left to maintain a buffer system. Therefore, solution a does not act as a buffer after adding NaOH. Final answer: a) The pH of solution a after adding NaOH is 4.34. b) The pH of solution b after adding NaOH is 4.87. c) Solution a does not act as a buffer after adding NaOH.

Key concepts

Buffer Solution

A buffer solution is a special kind of chemical solution that resists changes in pH when small amounts of an acid or base are added. Buffer solutions are typically made from a weak acid and its conjugate base, or a weak base and its conjugate acid.

In this context, a buffer works by neutralizing added acid or base to maintain pH. For instance, in our exercise, when NaOH, a strong base, is added to the buffer solution of propanoic acid (a weak acid) and sodium propanoate (its conjugate base), the NaOH reacts with the remaining propanoic acid.

However, once the acid is entirely consumed, as in part a of the problem, the solution can no longer resist pH changes effectively. This means that if the weak acid or weak base gets entirely consumed, the "buffering" ability of the solution is lost.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a crucial tool for understanding buffer solutions. This equation relates the pH of a buffer solution to the concentration of its acid and base components. It is expressed as:
\[\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]
where

• \( \text{pKa} \) is the negative logarithm of the acid dissociation constant, \( \text{Ka} \).
• \([\text{A}^-]\) represents the concentration of the conjugate base.
• \([\text{HA}]\) is the concentration of the weak acid.

This equation assumes that the change in volume when adding NaOH is negligible. It beautifully illustrates how a buffer's capacity to resist pH change depends on the ratio of the concentrations of its weak acid and conjugate base. In our problem, after calculating the new concentrations, the Henderson-Hasselbalch equation was applied to determine the new pH. An important takeaway is that the buffer is most effective when the concentrations of acid and base are similar and not depleted completely.

Weak Acid-Strong Base Reaction

When a weak acid like propanoic acid reacts with a strong base like NaOH, the reaction leads to the formation of the conjugate base and water. The general reaction can be expressed as:
\[\text{Weak Acid (HA)} + \text{OH}^- \rightarrow \text{Conjugate Base (A}^-\text{) + H}_2\text{O}\]
Here, the hydroxide ions \( \text{OH}^- \) from NaOH react with the weak acid, propanoic acid, to convert it into its conjugate base, sodium propanoate, while producing water.

This is an essential reaction in buffer systems. It’s crucial to understand that as NaOH is added, it continually reacts with the weak acid until the acid is fully neutralized. This point was exemplified in part a of the solution, where the NaOH addition exceeded the available weak acid. This resulted in the exhaustion of the buffer system, demonstrating the buffer's range and its limitations.