Question

Calculate the number of moles of \(\mathrm{HCl}(g)\) that must be added to 1.0 \(\mathrm{L}\) of 1.0 \(\mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) to produce a solution buffered at each pH. $$ \text{(a)}\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} \quad \text { b. } \mathrm{pH}=4.20 \quad \text { c. } \mathrm{pH}=5.00 $$

Short answer

To achieve the desired pH values in the buffer solution, the following number of moles of \(HCl(g)\) must be added: (a) For pH = pKa, add 0.5 moles of \(HCl(g)\). (b) For pH = 4.20, add approximately 0.361 moles of \(HCl(g)\). (c) For pH = 5.00, add approximately 0.212 moles of \(HCl(g)\).

Step-by-step solution

Determine the pKa of the weak acid (HC_2H_3O_2)

The first thing we need to do is determine the pKa of the weak acid. The formula for acetic acid, a common weak acid, is \(HC_2H_3O_2\). The acid dissociation constant, \(K_a\), for acetic acid is approximately \(1.8 \times 10^{-5}\). Using the formula for pKa, we can calculate it as follows: \(pK_a = -\log{K_a}\) \(pK_a = -\log{(1.8 \times 10^{-5})} \) Using a calculator, we get: \(pK_a \approx 4.74\) Now, we will use the pKa value to determine the required moles of \(HCl(g)\) for each specified pH value.

Solve for the number of moles of HCl(g) at pH = pKa

At pH = pKa, the buffer capacity is at its maximum. In this case: \(pH = pK_a\) \(4.74 = 4.74 + \log{\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}}\) Since the left and right sides of the equation are equal, we get: \(0 = \log{\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}}\) Raising both sides as exponential, we have: \(1 = \frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}\) Because 1.0 L of 1.0 M NaC2H3O2 is used, there are initially 1 mole of \(C_2H_3O_2^-\) and 0 moles of \(HC_2H_3O_2\). For the solution to be buffered at pH = pKa, we need an equal number of moles of \(HC_2H_3O_2\) to \(C_2H_3O_2^-\). Since the reaction consumes one mole of \(C_2H_3O_2^-\) and produces one mole of \(HC_2H_3O_2\), adding x moles of \(HCl(g)\) gives: \(x = 1 - x\) \(2x = 1\) \(x = \frac{1}{2}\) Hence, at pH = pKa, we need to add 0.5 moles of \(HCl(g)\).

Solve for the number of moles of H_{2}SO_{4} at pH = 4.20

For pH = 4.20, we have: \(4.20 = 4.74 + \log{\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}}\) Rearranging the equation to solve for the ratio of acid/base: \(\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]} = 10^{-0.54}\) Now, let the number of moles of \(HCl(g)\) added be x. Then: \(\frac{1-x}{x} = 10^{-0.54}\) Rearranging and solving for x: \(x = \frac{1}{10^{-0.54} + 1}\) Using a calculator: \(x \approx 0.361\) Thus, to get a pH of 4.20, we need to add approximately 0.361 moles of \(HCl(g)\).

Solve for the number of moles of H_{2}SO_{4} at pH = 5.00

Finally, for pH = 5.00, we have: \(5.00 = 4.74 + \log{\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}}\) Rearranging the equation to solve for the ratio of acid/base: \(\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]} = 10^{0.26}\) Now, let the number of moles of \(HCl(g)\) added be x. Then: \(\frac{1-x}{x} = 10^{0.26}\) Rearranging and solving for x: \(x = \frac{1}{10^{0.26} + 1}\) Using a calculator: \(x \approx 0.212\) Thus, to get a pH of 5.00, we will need to add approximately 0.212 moles of \(HCl(g)\).

Key concepts

pKa calculation

The concept of pKa is fundamental when dealing with acids and bases, particularly in the context of buffer solutions. In simple terms, pKa is the negative logarithm of the acid dissociation constant \(K_a\). It provides an insight into the strength of an acid.
To calculate pKa, you use the formula:

• \(pK_a = -\log{K_a}\)

Acetic acid, a common weak acid in buffer solutions, has a \(K_a\) of approximately \(1.8 \times 10^{-5}\). By using the pKa formula, you can calculate:

• \(pK_a = -\log{(1.8 \times 10^{-5})} \approx 4.74\)

A lower pKa value indicates a stronger acid, meaning the acid donates protons more easily. Understanding this helps students grasp how different acids influence buffer systems and their equilibrium. The pKa tells you the pH at which the concentrations of the acid and its conjugate base are equal. This is crucial for constructing effective buffer solutions.

acetic acid equilibrium

Acetic acid equilibrium is an essential concept when working with buffer solutions. It involves understanding how acetic acid, \(HC_2H_3O_2\), dissociates in solution to release protons (\(H^+\)) and form acetate ions (\(C_2H_3O_2^-\)).
This equilibrium can be represented as:

• \(HC_2H_3O_2 \rightleftharpoons H^+ + C_2H_3O_2^-\)

At equilibrium, the rate of the forward reaction (acetic acid dissociating) equals the rate of the reverse reaction (acetate forming acetic acid).
The position of this equilibrium affects the pH of the solution and can be shifted by adding strong acids or bases. For example, adding \(HCl\) increases \([H^+]\), shifting the equilibrium to the left, forming more \(HC_2H_3O_2\).
Understanding these shifts allows students to calculate the molar amounts needed to achieve desired pH levels in buffer systems. In buffered solutions, acetic acid acts as a weak acid, while the acetate ion acts as its conjugate base, both contributing to the buffering capacity of the solution.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a powerful tool for calculating the pH of buffer solutions. It relates pKa and the ratio of the concentration of the acid and its conjugate base in the buffer.
The equation is:

• \( \mathrm{pH} = \mathrm{pK_a} + \log{\frac{[\text{Base}]}{[\text{Acid}]}} \)

In our exercise, this helps determine how much \(HCl(g)\) to add to achieve specific pH levels. By rearranging and applying the equation, students can find the exact amount of acid or base needed to maintain a stable pH in their buffer.
For instance, when \(\mathrm{pH} = \mathrm{pK_a}\), the concentrations of acetic acid and acetate are equal, demonstrating maximum buffer capacity. When the pH shifts, the ratio of base to acid becomes crucial in maintaining the desired pH.
Understanding the Henderson-Hasselbalch equation provides insight into the chemistry of buffers and helps with practical applications like biochemistry and environmental science.