Question

2) Two students want to prepare 2.0 L of a buffer solution which must be buffered at a pH of 4.75. Student A wants to use acetic acid (CH3COOH, Ka = 1.8 x 10-5) and sodium acetate to prepare the buffer solution, while student B wants to use formic acid (HCOOH, Ka = 1.8 x 10-4) and sodium formate.

Which student chose the better conjugate acid-base pair to prepare the buffer? Why?

Explain the step-by- step procedure that the students will use to prepare the buffer solution:

Answer 1

To solve this question we just need to calculate the pka

pka = -log Ka

so for acetic acid/sodium ac. system

Pka = -log (1.8x 10^-5) = 4.744

so for formic acid/sodium form. system

Pka = -log (1.8x 10^-4) = 3.744

for the buffer solution apply hendersson hasselbach equation:

PH = Pka + log (salt / acid)

4.75 = 4.744 + log (salt / acid )

4.75 - 4.744 =  log (salt / acid )

0.006 = log (salt / acid)

10^0.006 = salt / acid = 1.013911

moles of salt = 1.013911 moles of acid

so what you need to do is to add 1.013911 moles of acid for every mole of sodium acetate you add