Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2013

Lecture Notes 38: 6 May

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Buffer calculations, cont.

Example: What ratio of acetate ion to acetic acid would you need to make up an acetate buffer with a pH of 5.25? pKa = 4.74.

Easiest way to approach this is to use the Henderson-Hasselbalch equation to find the ratio of HOAc to OAc-

pH = pKa + log[A-] / [HA]

Rearranging,

log[A-] / [HA] = pH - pKa

Substituting,

log[A-] / [HA] = 5.25 - 4.74 = 0.51

Now get rid of logs by raising both sides to power of ten:

10log[A-] / [HA] = 100.51

[A-] / [HA] = 3.236

[A-] = 3.236 [HA]

Extra example: How many moles of sodium acetate must be added to 0.120 moles of acetic acid and sufficient water to make a liter if we want a buffer with a pH of 4.50? pKa = 4.74.

Again use the Henderson-Hasselbalch equation to find the ratio of HOAc to OAc-

pH = pKa + log[A-] / [HA]

Rearranging and substituting,

log[OAc-] / [HOAc] = pH - pKa

Substituting,

log[OAc-] / [HOAc] = 4.50 - 4.74 = - 0.24

Raise both sides to power of 10:

[OAc-] / [HOAc] = 0.575

[OAc-] =0.575 [HOAc]

[OAc-] = 0.069 mol

Note that pH is essentially insensitive to concentration within a wide range of concentrations, that is, dilution of a buffer with water gives a new, dilute buffer with the same pH! Look at the Henderson-Hasselbalch equation and prove this to yourself.

Acid/Base Titration Curves

Titration curves examine the reaction of an acid with a strong base (most common and the examples we will look at) or a base with a strong acid (these titrations will look like mirror images of the titration with base reflected around the y {pH}-axis).

Strong acid/strong base titration: (plot drawn on board, Zumdahl Fig 15.1, p 717) In this case we can observe the titration by simply calculating the pH at each point by assuming 100% reaction. For example, assume an initial volume of 100 mL of 0.10 M HCl titrated with 0.10 M NaOH, and calculate the pH with 20 mL additions (drawn on board from data below):

Volume NaOH added Moles H+ remaining [H+] pH
0 0.010 0.10  1
20 0.0080 0.0080/0.12 = 0.0667 1.18
40 0.0060 0.0060/0.14 = 0.043 1.37
60 0.0040 0.0040/0.16 = 0.025  1.6
80 0.0020 0.0020/0.18 = 0.011 1.95
100 ª0  10-7 7
120  ª0  10-11.96  11.96
140  ª0  10-12.22 12.22
 200  ª0  10-12.70 12.70

Notice that once the acid has reacted the pH changes very rapidly. Also notice that the pH levels off at high pH. This is simply due to the fact that we are approaching pure 0.10 M NaOH, which would have a pH of 13.00.

Note that the titrtion of a strong base with a strong acid is just the reflection of the titration above, as seen in text Figure 15.2, p 717 (drawn on board).

Titration of a weak acid with a strong base: (lecture slide, Zumdahl Fig 15.3, p 722)

The curve below shows the titration of 1M acetic acid with 1M NaOH. In this case the pH is no longer a simple function of the concentration of H+, since the acid only partially dissociates in water. It does however react completely with any added strong base until the acid has been consumed, so adding base has little effect on pH initially.

plot of the titration of acetic acid with strong base

Note the following:

Looking more carefully at the titration curve of 1M acetic acid with 1M NaOH, The reaction is:

HC2H3O2 + OH- Equilibrium double arrow C2H3O2- + H2O

titration curve with labels for inflection points and acid/base compositions

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© R A Paselk

Last modified 6 May 2013