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Instructions for making a buffer solution state to mix 18.5 mL of a 0.130 M solution of acetic acid, CH₃COOH, with 36.6 mL of a 0.165 M solution of sodium benzoate, CH₃COONa.

What is the pH of this solution?

Answer :

Final answer:

The pH of the buffer solution is approximately 5.16.

Explanation:

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it. It is made up of a weak acid and its conjugate base, or a weak base and its conjugate acid. In this case, we have a mixture of acetic acid (CH3COOH) and sodium benzoate (CH3COONa).

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

pH = pKa + log([A-]/[HA])

First, we need to determine the pKa value for acetic acid. The pKa value for acetic acid is approximately 4.76.

Next, we need to calculate the concentrations of the conjugate base ([A-]) and the weak acid ([HA]).

Given:

  • Volume of acetic acid solution (CH3COOH): 18.5 mL
  • Concentration of acetic acid solution (CH3COOH): 0.130 M
  • Volume of sodium benzoate solution (CH3COONa): 36.6 mL
  • Concentration of sodium benzoate solution (CH3COONa): 0.165 M

To calculate the concentrations, we can use the formula:

Concentration (M) = moles/volume (L)

First, let's calculate the moles of acetic acid:

Moles of acetic acid = concentration of acetic acid * volume of acetic acid

Moles of acetic acid = 0.130 M * 0.0185 L

Moles of acetic acid = 0.002405 mol

Next, let's calculate the moles of sodium benzoate:

Moles of sodium benzoate = concentration of sodium benzoate * volume of sodium benzoate

Moles of sodium benzoate = 0.165 M * 0.0366 L

Moles of sodium benzoate = 0.006039 mol

Now, we can calculate the concentrations:

Concentration of acetic acid ([HA]) = moles of acetic acid / total volume of solution

Concentration of acetic acid ([HA]) = 0.002405 mol / (0.0185 L + 0.0366 L)

Concentration of acetic acid ([HA]) = 0.043 M

Concentration of sodium benzoate ([A-]) = moles of sodium benzoate / total volume of solution

Concentration of sodium benzoate ([A-]) = 0.006039 mol / (0.0185 L + 0.0366 L)

Concentration of sodium benzoate ([A-]) = 0.108 M

Now, we can substitute the values into the Henderson-Hasselbalch equation:

pH = 4.76 + log(0.108/0.043)

pH = 4.76 + log(2.51)

pH = 4.76 + 0.400

pH = 5.16

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