Problem 74 - Chemistry and Chemical Reactivity

Question

Calculate the hydronium ion concentration and the \(\mathrm{pH}\) of the solution that results when \(20.0 \mathrm{mL}\) of \(0.15 \mathrm{M}\) acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H},\) is mixed with \(5.0 \mathrm{mL}\) of \(0.17 \mathrm{M} \mathrm{NaOH}\).

Step-by-Step Solution

Answer

The solution has a hydronium ion concentration of approximately 1.24 x 10^-3 M and a pH of 2.91.

Step 1: Write the chemical equation

The reaction between acetic acid (\( \text{CH}_3\text{CO}_2\text{H} \)) and sodium hydroxide (\( \text{NaOH} \)) is a neutralization reaction. The equation is:\[ \text{CH}_3\text{CO}_2\text{H} + \text{NaOH} \rightarrow \text{CH}_3\text{CO}_2\text{Na} + \text{H}_2\text{O} \]

Step 2: Calculate initial moles of reactants

To find the moles of each reactant, use the formula: \( \text{Moles} = \text{Concentration (M)} \times \text{Volume (L)} \).For \( \text{CH}_3\text{CO}_2\text{H} \):\[ 20.0 \text{ mL} = 0.0200 \text{ L} \quad \text{and} \quad \text{Moles} = 0.15 \times 0.0200 = 0.0030 \]For \( \text{NaOH} \):\[ 5.0 \text{ mL} = 0.0050 \text{ L} \quad \text{and} \quad \text{Moles} = 0.17 \times 0.0050 = 0.00085 \]

Step 3: Determine limiting reactant and excess amount

Since \( \text{NaOH} \) is less in moles than \( \text{CH}_3\text{CO}_2\text{H} \), it is the limiting reactant.After the reaction, all \( \text{NaOH} \) will react fully, leaving:\[ 0.0030 - 0.00085 = 0.00215 \text{ moles of } \text{CH}_3\text{CO}_2\text{H} \text{ remaining} \]

Step 4: Calculate total volume and concentration of acetic acid

The total volume after mixing is:\[ 20.0 \text{ mL} + 5.0 \text{ mL} = 25.0 \text{ mL} = 0.0250 \text{ L} \]The concentration of remaining \( \text{CH}_3\text{CO}_2\text{H} \) is:\[ \text{Concentration} = \frac{0.00215}{0.0250} = 0.086 \text{ M} \]

Step 5: Calculate pH using equilibrium expression for weak acid

Acetic acid dissociates in water according to the equation:\[ \text{CH}_3\text{CO}_2\text{H} \rightleftharpoons \text{CH}_3\text{CO}_2^- + \text{H}_3\text{O}^+ \]We use the expression for the acid dissociation constant \( K_a \) of acetic acid, which is \( 1.8 \times 10^{-5} \):\[ K_a = \frac{[\text{CH}_3\text{CO}_2^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{CO}_2\text{H}]} \approx \frac{x^2}{0.086 - x} \approx \frac{x^2}{0.086} \]Assuming \( x \ll 0.086 \), solve for \( x \):\[ x = \sqrt{(1.8 \times 10^{-5}) \times 0.086} = 1.24 \times 10^{-3} \text{ M} \]This value of \( x \) is the concentration of \( \text{H}_3\text{O}^+ \).

Step 6: Calculate pH of the solution

The pH formula is:\[ \text{pH} = -\log[\text{H}_3\text{O}^+] \]Substitute the value of \( [\text{H}_3\text{O}^+] \):\[ \text{pH} = -\log(1.24 \times 10^{-3}) \approx 2.91 \]

Key Concepts

Neutralization Reaction

A neutralization reaction occurs when an acid reacts with a base to produce a salt and water. In the exercise, acetic acid \( \text{CH}_3\text{CO}_2\text{H} \) reacts with sodium hydroxide \( \text{NaOH} \). The chemical equation for this reaction is: \[ \text{CH}_3\text{CO}_2\text{H} + \text{NaOH} \rightarrow \text{CH}_3\text{CO}_2\text{Na} + \text{H}_2\text{O} \] This is a straightforward example of a neutralization reaction. The acid, acetic acid, donates a proton \((\text{H}^+)\) to the base, sodium hydroxide. The result is the formation of water and a salt, sodium acetate \((\text{CH}_3\text{CO}_2\text{Na})\).

• These reactions typically involve equimolar amounts of acid and base reacting completely, resulting in their neutralization.

In the context of titration, neutralization is used to determine the concentration of an unknown solution through gradual reaction with a solution of known concentration.

Limiting Reactant

The concept of the limiting reactant is crucial for understanding how much product can be formed in a chemical reaction. In simple terms, the limiting reactant is the substance that is completely consumed first, stopping the reaction from proceeding further. Here, sodium hydroxide \((\text{NaOH})\) is the limiting reactant. To identify it, we calculate the moles of each reactant:

• For acetic acid \(20.0 \text{ mL} = 0.0200 \text{ L}\): \(0.15 \times 0.0200 = 0.0030 \text{ moles}\)
• For sodium hydroxide \(5.0 \text{ mL} = 0.0050 \text{ L}\): \(0.17 \times 0.0050 = 0.00085 \text{ moles}\)

Since \(0.00085 \text{ moles}\) of \(\text{NaOH}\) is less than \(0.0030 \text{ moles}\) of acetic acid, all \(\text{NaOH}\) is used up first.

• This limits the amount of sodium acetate \((\text{CH}_3\text{CO}_2\text{Na})\) produced and leaves \(0.00215 \text{ moles}\) of acetic acid remaining unreacted.

pH Calculation

pH is a scale used to specify the acidity or basicity of an aqueous solution. To find the pH after a titration involves several steps, starting with finding the hydronium ion \((\text{H}_3\text{O}^+)\) concentration. After the reaction, a portion of acetic acid is left. This contributes to the acidic environment in the solution. A simplified equilibrium approach is used because the remaining acetic acid is a weak acid. It does not completely dissociate in water.

• Using the equilibrium expression for acetic acid: \[ \text{CH}_3\text{CO}_2\text{H} \rightleftharpoons \text{CH}_3\text{CO}_2^- + \text{H}_3\text{O}^+ \] This allows for calculation of \([\text{H}_3\text{O}^+]\).

Insert the known values into the expression for the acid dissociation constant \( K_a \). For acetic acid, \( K_a = 1.8 \times 10^{-5} \):\[ K_a = \frac{x^2}{0.086} \] Solving for \( x \) gives \( x = 1.24 \times 10^{-3} \text{ M}\), which is \([\text{H}_3\text{O}^+]\).Finally, calculate pH using: \[ \text{pH} = -\log[\text{H}_3\text{O}^+] \] Thus, \( \text{pH} = -\log(1.24 \times 10^{-3}) \approx 2.91 \). This pH reflects the solution's acidic nature after neutralization.

Weak Acid Equilibrium

Acetic acid is a weak acid, meaning it only partially dissociates in water, establishing an equilibrium between the undissociated acid and its ions. This partial dissociation is represented by the equilibrium expression: \[ \text{CH}_3\text{CO}_2\text{H} \rightleftharpoons \text{CH}_3\text{CO}_2^- + \text{H}_3\text{O}^+ \] The extent of dissociation is quantified by the acid dissociation constant \( K_a \). For acetic acid, \( K_a = 1.8 \times 10^{-5} \), indicating a low level of dissociation.

• This equilibrium is affected by the presence of any bases, such as \(\text{NaOH}\), which react with \(\text{H}_3\text{O}^+\) to form water, shifting the equilibrium position.
• When \(\text{NaOH}\) is added, initially, it leads to the formation of \(\text{CH}_3\text{CO}_2\text{Na}\), reducing the amount of \(\text{H}_3\text{O}^+\).

The equilibrium shifts to replenish some of the \(\text{H}_3\text{O}^+\) ions, as dictated by Le Chatelier's principle. This results in a weakly acidic solution, with a specific \(\text{pH}\), which was previously calculated as 2.91 post-neutralization. Understanding weak acid equilibria is essential for accurately determining the behavior of solutions in acid-base titrations.