Problem 225 - A Complete Resource Book in Chemistry for JEE Main

Question

Acetic acid solution was \(66.6 \%\) neutralized by adding a base. If \(\mathrm{pK}_{\mathrm{a}}\) of acetic acid is \(4.7\), the \(\mathrm{pH}\) of the above solution is approximately

Step-by-Step Solution

Answer

The pH of the acetic acid solution is approximately 5.0.

Step 1: Understanding Neutralization

The neutralization of acetic acid by a base means that acetic acid ( ext{CH}_3 ext{COOH}) is converted into its conjugate base, acetate ions ( ext{CH}_3 ext{COO}^-). If the solution is 66.6% neutralized, this means that 66.6% of the acetic acid has been converted to acetate ions.

Step 2: Finding Concentrations

Let the initial concentration of acetic acid be ext{[HA]}. After 66.6% neutralization, 33.4% of the original acetic acid concentration remains. So, we have ext{[HA]} = 0.334 imes ext{[Initial HA]} and ext{[A}^-] = 0.666 imes ext{[Initial HA]}.

Step 3: Using the Henderson-Hasselbalch Equation

For a solution composed of a weak acid and its conjugate base, the pH is given by the Henderson-Hasselbalch equation: \[ ext{pH} = ext{pK}_{ ext{a}} + ext{log}rac{[ ext{A}^-]}{[ ext{HA}]} \]For this solution, ext{pK}_{ ext{a}} = 4.7. Substituting the concentrations found in the previous step, we have: \[ ext{pH} = 4.7 + ext{log}rac{0.666}{0.334} \]

Step 4: Calculating the Logarithm Value

Compute the value of the logarithm: \[ rac{0.666}{0.334} \ \approx 2 \]. Thus, \[ ext{log} 2 \approx 0.3010 \].

Step 5: Finding the pH

Substitute the value of the logarithm back into the Henderson-Hasselbalch equation to find the pH:\[ ext{pH} = 4.7 + 0.3010 = 5.001 \].The pH of the solution is approximately 5.0.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a vital tool in chemistry that simplifies the process of calculating the pH of a solution containing a weak acid and its conjugate base. This equation is expressed as: \[ \text{pH} = \text{pK}_{\text{a}} + \log \frac{[\text{A}^-]}{[\text{HA}]} \] Here, \(\text{pH}\) represents the acidity of the solution, \(\text{pK}_{\text{a}}\) is the negative logarithm of the acid dissociation constant, \([\text{A}^-]\) is the concentration of the conjugate base, and \([\text{HA}]\) is the concentration of the acid. This equation allows us to see how the pH of the solution is affected by the ratio of the concentrations of the base and the acid. When the concentrations of the conjugate base and the acid are equal, the \(\text{pH}\) equals the \(\text{pK}_{\text{a}}\), because the log of 1 is zero. Understanding this correlation can help students grasp why some solutions buffer or resist changes in pH upon the addition of small amounts of acid or base.

Neutralization Reaction

Neutralization reactions are a fundamental concept in acid-base chemistry, involving the reaction between an acid and a base to produce water and a salt. When applying this concept to acetic acid (\(\text{CH}_3\text{COOH}\)), when it is partially neutralized by a base, it forms acetate ions (\(\text{CH}_3\text{COO}^-\)). In the given exercise, the acetic acid solution was 66.6% neutralized. This means 66.6% of the acetic acid has converted into acetate ions, leaving 33.4% of the original acetic acid concentration unchanged.

• Percentage Neutralization: This is key in calculating the new concentrations of the acid and its conjugate base.
• Reaction Formula: It shows the transformation of \(\text{CH}_3\text{COOH}\) into \(\text{CH}_3\text{COO}^-\) and how this impacts the resultant solution.

Neutralization helps us calculate the concentrations necessary to use in the Henderson-Hasselbalch equation. Understanding this reaction bridges the gap between calculating the amount of base or acid added and the resultant pH.

pH Calculation

Calculating the pH is often one of the first steps students take when they begin learning about acids and bases. In this exercise, using the previously mentioned concentrations after 66.6% neutralization, we apply these values within the Henderson-Hasselbalch equation. Given the relation: \[ \text{pH} = \text{pK}_{\text{a}} + \log \frac{[0.666]}{[0.334]} \] Finding \(\text{pH}\) begins with determining the concentration ratio of the conjugate base to the remaining acid. Calculating the value of \(\frac{0.666}{0.334}\) gives approximately 2, and taking the logarithm of 2 returns approximately 0.3010. Adding this to the \(\text{pK}_{\text{a}}\) value of 4.7 yields a final pH of about 5.0.

• The natural logarithm values are used here to find the precise pH change stemming from acid neutralization.
• It’s important to remember that pH is a logarithmic scale, meaning small changes in ratio lead to significant pH changes.

Mastering pH calculations helps in predicting how acidic or basic a solution will become in practical scenarios like titrations or buffer solutions.