Sample Calculation:

Problem:
Prepare 500 mL of a 0.1 M buffer solution with a pH of 5.0 using acetic acid and sodium acetate.

Solution:

The Mass Method will be used here.  The conjugate acid-base pair to be used is acetic acid, C2H3CO2H and acetate ion, C2H3CO2-, to be represented as HA and A-, respectively.  From the Henderson-Hasselbach Equation we have:

where pKa is the pKa of the conjugate acid, acetic acid,  in this case.  Therefore, pKa = -log(1.86 x 10-5) = 4.74.  Substitution into the equation above gives:

and

We must prepare a solution with this ratio of concentrations.  Note,  that this is also a mole ratio:

The total concentration of  conjugate acid and conjugate base is 0.1 M.  Therefore:

and from the previous equation:

nA- = 1.82nHA, and therefore,

Solving for nHA gives:

 nHA = 0.0177 mol.

and

nA- = 0.0177 x 1.82 = 0.0323 mol

In this case,  nHA and nA- are the moles of acetic acid and  acetate ion, respectively, that must be added to 500 mL of solution.  Note that although the conjugate base is acetate ion, it must be weighed out as sodium acetate.

The molar masses of acetic acid and sodium acetate are 60.05 g mol-1 and 82.03 g mol-1, respectively.  The masses of acetic acid and sodium acetate that must be dissolved in 500 mL of solution are:

mass acetic acid = 0.0177 mol x 60.05 g mol-1 = 1.06 g

mass sodium acetate = 0.0323 mol x 82.03 g mol-1= 2.65 g

The procedure for preparing this buffer solution is to add the above masses of acetic acid and sodium acetate to a 500 mL volumetric flask, and diluting to the mark with deionized water.