A buffer is made by mixing 2.0 grams of sodium acetate and 10 mL of 3.0 M acetic acid. Then 50 mL of water is added to the mixture. Answer the questions below.

Question

A buffer is made by mixing 2.0 grams of sodium acetate and 10 mL of 3.0 M acetic acid. Then 50 mL of water is added to the mixture.

1. Write the equilibrium expression
2. Calculate the molarity of the acid and the conjugate base
3. Find the pH of the buffer
4. Use Henderson-Hasselbalch equation (pH=pKa+log([conj base]/[acid]

Answer 1

Equilibrium expressions:

CH3COOH + OH^- <==> H2O + CH3COO^-

CH3COO^- + H⁺ <==> CH3COOH

Molar mass sodium acetate (CH3COONa) = 82.0 g / mole

Acid = acetic acid (CH3COOH)

Conjugate base = sodium acetate

moles acetic acid = 10 ml x 1 L / 1000 ml x 3.0 mol / L = 0.03 moles

moles sodium acetate = 2.0 g x 1 mol / 82.0 g = 0.0244 moles

Final volume = 10 ml + 50 ml = 60 ml = 0.060 L

[acetic acid] = 0.03 moles / 0.060 L = 0.50 M acetic acid

[sodium acetate] = 0.0244 moles / 0.060 L = 0.407 M sodium acetate

pH = pKa + log [sodium acetate] / [acetic acid]

Neither the Ka nor the pKa was provided in the question, so looking it up we find the pKa to be 4.75.

pH = 4.75 + log [0.0.407/0.50]

pH = 4.75 + log 0.813

pH = 4.75 + (-0.0897)

pH = 4.66