Question

how many ml of 1.0 M NaOH should be added to 500 ml 0.1 M acetic acid in order to make a buffer of pH 5.5?

Answer 1

The pK_a of acetic acid is 4.76. Use the Henderson-Hasslebach equation to find the ratio of acetic acid and sodium acetate in the buffer at pH 5.50. Plug in values and obtain

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

===> 5.50 = 4.76 + log [sodium acetate]/[acetic acid]

===> log [sodium acetate]/[acetic acid] = 5.50 – 4.76 = 0.74

===> [sodium acetate]/acetic acid] = antilog (0.74) = 5.4954

===> [sodium acetate] = 5.4954*[acetic acid]

The total acetic acid concentration is given as

[sodium acetate] + [acetic acid] = 0.1 M

===> 5.4954*[acetic acid] + [acetic acid] = 0.1 M

===> 6.4954*[acetic acid] = 0.1 M

===> [acetic acid] = (0.1 M)/6.4954 = 0.015395 M

The concentration of sodium acetate is [sodium acetate] = 5.4954*0.015395 M = 0.08460 M.

Since we have 500 mL buffer solution, hence, the mole(s) of sodium acetate in the buffer = (500 mL)*(1 L/1000 mL)*(0.08460 mol/L) = 0.0423 mol.

Sodium acetate is formed by the action of NaOH on acetic acid. The balanced chemical equation is

CH₃COOH (aq) + NaOH (aq) ------> CH₃COO^-Na⁺ (aq) + H₂O (l)

As per the stoichiometric equation,

1 mole NaOH = 1 mole CH₃COO^-Na⁺

Therefore, mole(s) of NaOH required = 0.0423 mole.

Volume of NaOH required = (0.0423 mole)/(1.0 mol/L) = 0.0423 L = (0.0423 L)*(1000 mL/1 L) = 42.3 mL (ans).