Problem 29 - Chemistry The Molecular Science

Question

Calculate the \(\mathrm{pH}\) change when \(1.0 \mathrm{~mL}\) of \(1.0-\mathrm{M} \mathrm{NaOH}\) is added to \(0.100 \mathrm{~L}\) of a solution of (a) 0.10 -M acetic acid and 0.10-M sodium acetate. (b) 0.010 -M acetic acid and 0.010 -M sodium acetate. (c) 0.0010 -M acetic acid and 0.0010 -M sodium acetate.

Step-by-Step Solution

Answer

(a) ΔpH = 0.02, (b) ΔpH = 0.12, (c) ΔpH = 6.24.

Step 1: Understanding the Reaction

When NaOH is added to the acetate buffer solution, a neutralization reaction occurs where NaOH, a strong base, reacts with acetic acid, a weak acid: \( \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \). The amount of NaOH added will affect the concentrations of \( \text{CH}_3\text{COOH} \) and \( \text{CH}_3\text{COO}^- \) in the buffer solution.

Step 2: Using Henderson-Hasselbalch Equation

For buffer solutions, the pH can be calculated using the Henderson-Hasselbalch equation: \[ \mathrm{pH} = \mathrm{pK}_a + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]where \([\text{A}^-]\) is the concentration of acetate ion, \([\text{HA}]\) is the concentration of acetic acid, and \( \mathrm{pK}_a \approx 4.76 \) for acetic acid.

Step 3: Initial Concentrations for Case (a)

For the initial buffer: 0.10 M acetic acid and 0.10 M sodium acetate. \([\text{CH}_3\text{COOH}] = 0.10 \, \text{M} \), \( [\text{CH}_3\text{COO}^-] = 0.10 \, \text{M} \). Calculate pH before NaOH addition using Henderson-Hasselbalch: \[ \mathrm{pH}_{initial} = 4.76 + \log\left( \frac{0.10}{0.10} \right) = 4.76 \].

Step 4: Effect of NaOH Addition on Case (a)

Adding 1.0 mL of 1.0 M NaOH introduces 0.0010 mol of NaOH, which reacts with the acetic acid, decreasing its concentration and increasing acetate ion concentration. Updated concentrations after reaction are: \([\text{CH}_3\text{COOH}] = 0.099 \, \text{M} \) and \([\text{CH}_3\text{COO}^-] = 0.101 \, \text{M}\). Updated pH: \[ \mathrm{pH}_{final} = 4.76 + \log\left(\frac{0.101}{0.099}\right) = 4.78 \].

Step 5: pH Change for Case (a)

The change in pH for case (a) is \( \Delta \mathrm{pH} = \mathrm{pH}_{final} - \mathrm{pH}_{initial} = 4.78 - 4.76 = 0.02 \).

Step 6: Initial Concentrations for Case (b)

Initially, 0.010 M acetic acid and 0.010 M sodium acetate. Calculate pH before NaOH addition:\[ \mathrm{pH}_{initial} = 4.76 + \log\left(\frac{0.010}{0.010}\right) = 4.76 \].

Step 7: Effect of NaOH Addition on Case (b)

Addition of 1.0 mL of 1.0 M NaOH introduces 0.0010 mol of NaOH, fully neutralizing the acetic acid. Updated concentrations are now:\([\text{CH}_3\text{COOH}] = 0.009 \, \text{M} \) and \([\text{CH}_3\text{COO}^-] = 0.011 \, \text{M}\). Updated pH:\[ \mathrm{pH}_{final} = 4.76 + \log\left(\frac{0.011}{0.009}\right) = 4.88 \].

Step 8: pH Change for Case (b)

The change in pH for case (b) is \( \Delta \mathrm{pH} = \mathrm{pH}_{final} - \mathrm{pH}_{initial} = 4.88 - 4.76 = 0.12 \).

Step 9: Initial Concentrations for Case (c)

Initially, 0.0010 M acetic acid and 0.0010 M sodium acetate. Calculate pH before NaOH addition:\[ \mathrm{pH}_{initial} = 4.76 + \log\left(\frac{0.0010}{0.0010}\right) = 4.76 \].

Step 10: Effect of NaOH Addition on Case (c)

Adding 1.0 mL of 1.0 M NaOH adds 0.0010 mol NaOH, completely consuming the acetic acid and leaving no buffer capacity. Solution becomes 0.0010 M NaOH, thus: \[ \mathrm{pH}_{final} = 14 - \log(0.0010) = 11.00 \].

Step 11: pH Change for Case (c)

The change in pH for case (c) is \( \Delta \mathrm{pH} = \mathrm{pH}_{final} - \mathrm{pH}_{initial} = 11.00 - 4.76 = 6.24 \).

Key Concepts

Henderson-Hasselbalch equation

When dealing with buffer solutions, especially those involving weak acids like acetic acid, the Henderson-Hasselbalch equation proves to be invaluable. This equation provides a simple way to estimate the pH of a buffer solution. It relates the pH of the solution to the pKa of the acid and the ratio of the concentrations of the conjugate base, \[ \mathrm{pH} = \mathrm{pK}_a + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]Here,

• \([\text{A}^-]\) represents the concentration of the acetate ion, the conjugate base
• \([\text{HA}]\) represents the concentration of the undissociated acetic acid
• \( \mathrm{pK}_a \) is a constant value for the acid, which for acetic acid is approximately 4.76

By substituting the concentrations of the acid and its conjugate base into the equation, you can easily calculate the pH of the buffer prior to and after the addition of a strong base like NaOH. This equation highlights the importance of the ratio between the concentrations in determining the pH, rather than their absolute values.

acid-base reactions

An acid-base reaction is a chemical process that involves the transfer of hydrogen ions (\(\text{H}^+\)). In the context of the exercise, the addition of sodium hydroxide (\(\text{NaOH}\)) to the acetate buffer initiates such a reaction. Here’s what happens:- Sodium hydroxide, a strong base, dissociates completely in water to provide hydroxide ions (\(\text{OH}^-\)).- These hydroxide ions immediately react with the acetic acid (\(\text{CH}_3\text{COOH}\)), a weak acid, as follows: \[\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}\]During this neutralization reaction, the acetic acid is converted to acetate ions (\(\text{CH}_3\text{COO}^-\)), and water is formed. The reaction shifts the balance between the acid and its conjugate base, leading to changes in their concentrations. Understanding these reactions is crucial to predicting how the pH will change as a result of adding the base, which can be calculated through the Henderson-Hasselbalch equation.

buffer capacity

Buffer capacity refers to a buffer solution's ability to resist changes in pH upon the addition of small amounts of acids or bases. It is directly related to the concentrations of the acid and its conjugate base in the solution. - A higher concentration of both components ensures a greater buffer capacity, thereby better maintaining a stable pH. - Conversely, when the concentrations are too low, even a minute addition of a strong acid or base can significantly alter the pH. In the exercise:

• Case (a), with 0.10 M concentrations, displays a higher buffer capacity, showing minimal pH change (0.02) when NaOH is added.
• Case (b), with 0.010 M concentrations, already shows a noteworthy increase in pH change to 0.12.
• By case (c), with merely 0.0010 M concentrations, the buffer is overwhelmed by the NaOH, resulting in a drastic pH change of 6.24.

Understanding buffer capacity is crucial in predicting and rationalizing why some buffer solutions can effectively maintain their pH, whereas others are easily disrupted.