Question

$\text{The pH of an aqueous solution containing 1 M benzoic acid (} pK_a = 4.20 \text{) and 1 M sodium benzoate is 4.5.}$
\(\text{The volume of benzoic acid solution in 300 mL of this buffer solution is \_\_\_\_\_\_ mL.}\)

Answer 1

Correct Answer: 100

Solution: To find the volume of benzoic acid in the buffer solution, we can use the Henderson-Hasselbalch equation:

\[ \text{pH} = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \]

Where: \([A^-]\) is the concentration of the conjugate base (sodium benzoate) and \([HA]\) is the concentration of the weak acid (benzoic acid).

Given Information: pH = 4.5

\[ pK_a = 4.20, \quad [A^-] = 1 \, M \text{ (sodium benzoate)} \]

Total volume of buffer solution = 300 mL.

Calculate the ratio of base to acid:

\[ 4.5 = 4.20 + \log \left( \frac{[A^-]}{[HA]} \right) \]

\[ 0.30 = \log \left( \frac{1}{[HA]} \right) \]

\[ 10^{0.30} = \frac{1}{[HA]} \implies [HA] = \frac{1}{10^{0.30}} \approx 0.50 \, M \]

Calculate the volume of benzoic acid: Let \(V_a\) be the volume of 1M benzoic acid. The concentration in the total 300 mL buffer solution:

\[ [HA] = \frac{V_a}{200} \]

Setting the concentrations equal gives:

\[ 0.50 = \frac{V_a}{200} \implies V_a = 0.50 \times 200 = 100 \, mL \]

Total volume of benzoic acid solution: To find the total volume of benzoic acid needed to maintain the desired pH:

\[ V_a = 100 \, mL \quad \text{(rounded from calculations)} \]

Thus, the volume of benzoic acid solution in 300 mL of the buffer solution is: 100 mL

Answer 2

An aqueous buffer solution contains 1 M benzoic acid and 1 M sodium benzoate, with a pH of 4.5 and benzoic acid pK_a = 4.20. The total volume of the buffer is 300 mL. We need to find the volume of the 1 M benzoic acid solution used in preparing this buffer.

Concept Used:

For an acidic buffer composed of a weak acid (HA) and its salt (A^-), the pH is given by the Henderson-Hasselbalch equation: \[ \text{pH} = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right) \] The concentrations of the acid and salt in the final buffer solution depend on the volumes of the stock solutions mixed, assuming they have the same molarity.

Step-by-Step Solution:

Step 1: Apply the Henderson-Hasselbalch equation to the final buffer.

Let the final concentration of benzoic acid be [HA] and the final concentration of sodium benzoate be [A^-]. Substituting the given values:

\[ 4.5 = 4.20 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

Step 2: Solve for the ratio of salt to acid concentration.

\[ 4.5 - 4.20 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] \[ 0.30 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] \[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.30} \approx 2.0 \]

So, in the final buffer, [A^-] = 2[HA].

Step 3: Relate the final concentrations to the volumes of the stock solutions.

Let V_acid be the volume (in mL) of 1 M benzoic acid solution used, and V_salt be the volume of 1 M sodium benzoate solution used. The total volume is 300 mL, so:

\[ V_{\text{acid}} + V_{\text{salt}} = 300 \, \text{mL} \]

The final concentration of benzoic acid, [HA], is:

\[ [\text{HA}] = \frac{1 \times V_{\text{acid}}}{300} \]

The final concentration of sodium benzoate, [A^-], is:

\[ [\text{A}^-] = \frac{1 \times V_{\text{salt}}}{300} \]

Step 4: Use the concentration ratio from Step 2 to find the volume ratio.

\[ \frac{[\text{A}^-]}{[\text{HA}]} = \frac{V_{\text{salt}}}{V_{\text{acid}}} = 2 \] \[ V_{\text{salt}} = 2 V_{\text{acid}} \]

Step 5: Substitute into the total volume equation to find V_acid.

\[ V_{\text{acid}} + 2 V_{\text{acid}} = 300 \] \[ 3 V_{\text{acid}} = 300 \] \[ V_{\text{acid}} = 100 \, \text{mL} \]

Therefore, the volume of benzoic acid solution in the 300 mL buffer is 100 mL.