Question

Suppose you have 1.00 Liter of an aqueous buffer containing 60.0 mmol benzoic acid (pKa=4.20) and 40.0 mmol benzoate.

Calculate the pH of this buffer.

What volume of 2.00M NaOH would be required to increase pH to 4.93?

Answer 1

Given,

pKa = 4.20

Benzoic Acid = 60 mmol

Benzoate (salt) = 40 mmol

We know that,

pH of a buffer = pKa + log (Salt / Acid)

=> pH = 4.2 + log (40 / 60) = 4.2 - 0.176 = 4.024

pH after adding NaOH = pKa + log (salt + X / Acid - X)

where,

X = mmol of NaOH added

=> 4.93 = 4.2 + log (40 + X / 60 - X)

=> 0.73 = log (40 + X / 60 - X)

=> (40 + X / 60 - X) = 5.37

=> X = 44.3

=> We will need 44.3 mmol of NaOH

Molarity of NaOH = 2

Volume = Moles / Molarity

=> Volume = 44.3 / 2 = 22.15 mL of NaOH is required