Calculate the pH of a buffer solution prepared by dissolving 23.8 g of benzoic acid (HC₇H₅O₂) and 35.9 g of sodium benzoate in 201.0 mL of solution.

Question

High School

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Answer :

Tom056 Tom056 · Answer 1 · 2025-03-02T00:45:23-05:00

The pH of the buffer solution is approximately 4.30.

To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation:

[tex]pH = pKa + \log \left( \frac{[A^-]}{[HA]} \right)[/tex]

where:

First, we need to find the molar mass of benzoic acid (HC₇H₅O₂) and sodium benzoate (NaC₇H₅O₂):

Next, we calculate the number of moles of each component:

Moles of benzoic acid (HC₇H₅O₂) = [tex]\frac{23.8 \text{ g}}{122.12 \text{ g/mol}} \approx 0.195 \text{ mol}[/tex]
Moles of sodium benzoate (NaC₇H₅O₂) = [tex]\frac{35.9 \text{ g}}{144.11 \text{ g/mol}} \approx 0.249 \text{ mol}[/tex]

Now, we find the concentrations of each species in the 201.0 mL solution (0.201 L):

Concentration of benzoic acid (HC₇H₅O₂) = [tex]\frac{0.195 \text{ mol}}{0.201 \text{ L}} \approx 0.970 \text{ M}[/tex]
Concentration of sodium benzoate (NaC₇H₅O₂) = [tex]\frac{0.249 \text{ mol}}{0.201 \text{ L}} \approx 1.239 \text{ M}[/tex]

We also need the [tex]pKa[/tex] value for benzoic acid, which is approximately 4.20.

Using the Henderson-Hasselbalch equation, we can now calculate the pH:

[tex]pH = 4.20 + \log \left( \frac{1.239 \text{ M}}{0.970 \text{ M}} \right)[/tex]

[tex]pH = 4.20 + \log(1.278)[/tex]

[tex]pH = 4.20 + 0.106[/tex]

[tex]pH \approx 4.30[/tex]

Therefore, the pH of the buffer solution is approximately 4.30.