Buffer Problems

The Common Buffer Problems

1. Compute pH given the actual concentrations of the conjugate acid and conjugate base.

The pH is easily calculated with:

2. The pH shift problem:

If you add 2.00 mL of 0.100 M HCl to 100 mL of a buffer consisting of 0.100 M HA and 0.200 M NaA, what will be the change in pH?

Ka of HA is 1.5x10-5. pKa = 4.82.

The initial pH is:

If we add a strong acid to a weak base, such as NaA, they REACT and form new products.

HCl + NaA --> HA + NaCl.

(The total ionic equation is H+ + Cl- + Na+ + A- --> HA + Na+ + Cl-)

mmoles HCl

mmoles HA
mmoles NaA
Volume, mL
Initial
0.200
10.0
20.0
100+2 mL
Final
0
10.2
19.8
102 mL
Molarities

10.2/102
19.8/102

3. The "make a buffer of given pH" problem:

We wish to prepare a buffer of pH 10 using common laboratory reagents. How many mL of 6M HCl would we have to add to a solution prepared by adding 100 mL of 15M NH3 to 200 mL of water?

Ka of NH4+ = 5.70x10-10, pKa = 9.24.

HCl + NH3 --> NH4Cl

mmoles HCl

mmoles NH4Cl
mmoles NH3
Volume
Initial
x
0
1500
x/6M + 300mL
Final
0
x
1500-x
x/6M + 300mL
Molarity




We didn't bother computing the molarities, since the volumes cancel out in the buffer equation anyway--a good thing, since the volumes have an "x" in them.

Finally, 222 mmoles HCl/6M = 37.0 mL of 6M HCl needed.

Note that the buffer produced will not actually have a pH of 10. The concentrations are so high that we would need to compute activities from the ionic strengths and generate a corrected value of the Ka for ammonium ion. In an actual lab, we would make up the buffer as calculated above, then use a pH meter and HCl or NH3 to adjust the buffer to 10.0.

David L. Zellmer, Department of Chemistry, California State University, Fresno. March 7, 1997