Instructions - Lab 3 - Titration Curves (PDF Instructions + Graph Paper) PDF

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CHEM 115: EXPERIMENT THREE TITRATION CURVES Introduction This experiment is the second of which has you doing a quantitative evaluation of acid-base equilibrium systems. Acid-base equilibria are ubiquitous in nature, playing important roles in physiological buffering systems, ocean and lake chemistry, soil chemistry, and acid rain. In this experiment you will be examining some acid-base equilibria involving weak acids, weak bases, and indicators. In the first part of the lab, you will be studying is the first ionization of ascorbic acid (better known as vitamin C). The chemical formula for ascorbic acid is H2C6H6O6 (and its molar mass is 176.12 g/mol). Ascorbic acid is a powerful antioxidant in its anionic form (HC6H6O6-). Ascorbic acid is actually a diacid (is capable of losing two protons), but we will be examining only the first ionization step in the lab today. The Ka for this first ionization has a literature value of 8.0 ×10-5.

Acid-Base Titration Curves This material is discussed in much greater detail in the text. A brief review only will be presented here. The purpose of performing an acid-base titration is to determine when an exact stoichiometric amount of acid and base have been mixed. This is called the equivalence point of the titration. To this point you have used indicators to determine these equivalence points. It is also possible to measure the pH of a solution as a function of the volume of titrant added. This produces a titration curve. Typically the pH will change only very gradually until the equivalence point is approached at which point it will change dramatically.

Strong Acid-Strong Base Titrations Consider, for example, the titration of a strong acid with a strong base (see Figure 1 below). At the start of the titration, the acid will be in excess and the pH will be less than 7. At the equivalence point, equal amounts of acid and base will have reacted, leaving a neutral solution with pH = 7. (H3O+ + OH- → 2H2O) Beyond the equivalence point the base will be in excess and the pH will be greater than 7. Strong acidstrong base titrations are characterized by very sharp, steep equivalence points.

Figure 1 It is relatively easy to calculate pH at any point along a strong acid-strong base titration curve. This is because the acid and base react completely, and pH is therefore determined simply by how much excess acid or base is in the solution at a given point in the titration. Simple stoichiometry can be used to do these calculations. (Note: The most common place that students make errors in this type of calculation is simply to forget to include the effects of dilution as the solution volume increases during the titration.)

Weak Acid/Strong Base or Weak Base/Strong Acid Titrations In these cases the situation is somewhat complicated by the fact that the weak acid or weak base is only partially ionized in aqueous solution. Consider the case of the titration of a weak acid, HA, with a strong base (See Figure 2 below). When HA is dissolved in water, it is partially ionized as follows: H 3O + + A -

HA + H2O

H O  A  

Ka 

(1)

3



eq

eq

(2)

HAeq

Now let’s consider what happens at various points during the titration.

Figure 2 a) At the Start of the Titration Since the acid is only partially ionized, the [H3O+] will be smaller and so the pH is higher than for a strong acid of the same molarity. If we assume that all of the H3O+ in the solution comes from ionization of HA (Equation 1), then [H3O+] = [A-] and [HA] = [HA]0 – [A-] = [HA]0 – [H3O+]

In the above (and subsequent) equations, the subscript “0” as in [HA]0 denotes the initial concentration before equilibration. Substituting into Equation (2) we obtain:

H O 

 2

Ka 

3

HAo  [ H 3O  ]

(3)

If the acid is very weak, [H3O+]...