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Titration

Part 2: Preliminaries

2.2. Constructing the Mathematical Model

As described in the preceding section, our goal is to model the relationship in an acid-base titration between volume of titrant added and pH of the solution. We will base our model on the properties of the chemical reactions occurring in an acid-base titration. In this project, we assume that the acid is a weak acid. That is, the acid does not completely break up into H+ and A- in water. More will be said about this later.

Let:
a0  =  initial volume of acid solution (in liters),
A0  =  initial concentration of acid (in M = moles/liter),
c    =  concentration of base (in M = moles/liter),
v    =  total volume of base added to acid solution (in liters).

Of these four quantities, only v has a value that will vary throughout the titration. The other three are fixed parameters of the model -- a0 and c are accurately-known prior to the titration, while A0 is the only unknown parameter.

When the base is dripped into the acid solution, the following chemical reaction occurs between the weak acid and the base:

Acid-Base Reaction:    HA (acid) + OH- (base) harpoon A- + H2O,

where HA denotes a generic weak acid, OH- denotes the generic base and A- denotes an anion of the acid. This reaction states that 1 molecule of the acid reacts with 1 molecule of the base to produce 1 molecule of the anion of the acid and 1 molecule of water.

When a total volume of v liters of base has been added to the acid solution, one can determine the amount of each substance present at any step prior to equivalence by tracking values in the following table based on the acid-base reaction.

TABLE 1

  HA (acid) OH-(base) A-
# moles prior to reaction a0 A0 vc 0
# moles after reaction a0A0 - vc 0 vc

Referring to the variable and parameter definitions, we see that prior to the reaction there are a0A0 moles of acid (a0 L acid x A0 moles/L acid) and vc moles of base (v L base x c moles/L base). From the acid-base reaction, we know that 1 molecule of acid HA reacts with 1 molecule of base OH- to produce 1 molecule of the anion A-. (We don't need to keep track of the molecules of water because the concentration of water remains relatively constant during the course of the reaction.) Since for every molecule of base that goes into the reaction a molecule of acid is lost and a molecule of A- is gained, after the reaction there will be a0A0 - vc moles of acid, 0 moles of base, and vc moles of A- present in the solution, as noted in the second row of Table 1. At steps prior to equivalence, all of the base is "used up" while there is still acid remaining (i.e., vc < a0A0) because the acid-base reaction can be considered to go to completion. It follows from our definition in Part 1 that the equivalence point is that point at which enough base has been added so that vc = a0A0.

Recall that the measurement taken from the solution after each drop of base has been added is the pH of the solution, where

pH = -log[H+].

We are interested now in making the connection between the number of moles of each substance present after the acid-base reaction and the resulting pH of the solution.

We begin by investigating the relationship between the hydrogen ion concentration of the solution, H+, and the amount of acid present. This relationship was discussed briefly in Part 1. Once again, at any step in the titration, this relationship is characterized by the following chemical reaction which represents the dissociation of the weak acid, HA:

Dissociation Reaction:    HA  H+ + A-.

Some substances, when dissolved, break up into ions, a process known as dissociation. Since HA is a weak acid, it does not dissociate completely before reaching equilibrium. In other words, not all of the acid present in the solution will dissociate to form hydrogen ions and anions of the acid according to the dissociation reaction. Therefore, we cannot assume (as we did in forming Table 1) that all of any substance gets "used up" in this reaction.

To illustrate this point, suppose that the acid HA did dissociate completely. Then each molecule of the HA would form 1 hydrogen ion H+ for each molecule of HA present, and consequently, the number of moles of HA present in the solution would equal the number of moles of H+. This, in fact, is what happens for a limited number of so-called "strong acids".

However, the weak acid HA does not dissociate completely. So, we need to form a table based on the dissociation reaction similar to Table 1, except that for this one we must introduce an unknown quantity representing the amount of dissociation that actually does take place. This unknown quantity is given by the variable,

x = [H+] at equilibrium.

Since x represents a concentration, we will switch to keeping track of concentrations of substances (i.e., divide amounts by total volume = a0 + v) rather than amounts as we did in the first table. The resulting table is partially filled-in below. Complete the table, keeping in mind the relationship described in the dissociation reaction between the hydrogen ion concentration, [H+], and the amount of acid present in the solution: When 1 molecule of HA dissociates, 1 hydrogen ion and 1 anion are formed. (Note: We are still looking specifically at the step in the titration where a total volume of v liters of base has been added into the original solution.)

TABLE 2

  HA (acid) H+ (hydrogen ion) A-
concentration prior to dissociation 0
concentration at equilibrium   x  

We've already mentioned that since HA is a weak acid, it does not dissociate completely. But what percentage of the molecules of HA are dissociated? This depends upon the identity of HA. The extent of dissociation is an important characterization property of a weak acid. In a particular concentration of solution, one weak acid may dissociate 3% of its molecules while a different weak acid may dissociate .002%.

The magnitude of the dissociation constant for a particular weak acid indicates the extent of dissociation for that acid. The dissociation constant, Ka, for a weak acid is defined as the quotient of concentration of products over concentration of reactants from the dissociation reaction:

 
(2)

Clearly then, the larger the magnitude of Ka, the higher the percentage of dissociated molecules. Many chemistry textbooks list the Ka values for a large variety of weak acids. In fact, the Ka value can be used as a basis for the identification of a weak acid.

Using the values from Table 2 (at equilibrium), write out an expression for Ka in terms of x,v,c,a0, and A0 (do not simplify yet!):

(3)

Notice that we are closing in on our desired mathematical model. Equation (3) gives us the relationship between v = volume of base added and x = [H+], and it is a simple step from there to relate v to pH = -log[H+]. However, if you were to take equation (3) and solve for x in terms of v (and all the parameters of the model), you would obtain a very complicated modeling function. It is not necessary to create such a complicated model because the following simplifying assumptions can be made without any noticeable loss of accuracy:

Simplifying Assumptions:

(a)
The magnitude of  x = [H+]  is insignificant compared to the magnitude of

In applications involving the use of our model, we will be considering acid solutions of large enough concentration so that the percentage of dissociation that occurs is very small, thus making  x  very small in comparison with

So, this assumption is valid for any value of  v  prior to the equivalence point.

(b)
The magnitude of  x = [H+]  is insignificant compared to the magnitude of

This assumption is valid for any value of  v  prior to the equivalence point, except for  v  very close to  0  (i.e., when very little base has been added).

These two assumptions allow us to formulate the following statements without any noticeable loss of accuracy:

(4a)

and

(4b)

Given the statements in (4), rewrite your expression for the dissociation constant Ka from equation (3):

(5)

Now simplify your expression for  Ka  from equation (5), and solve for  x  in terms of  v  (and all the parameters of the model --  a0,  A0,  c,  Ka):

(6)

Using your expression for  x  from equation (6), verify that the pH of the solution can be expressed as the following function of volume of base added. (We call the modeling function   H(v).  It gives the pH of the solution corresponding to  v  liters of base added.)

Titration Model:   H(v) = log(cv) - log(a0A0 - CV) - log Ka, for 0 < equivalence point

Question: Why do we restrict the domain of our model to 0 < equivalence point?

Note: This model has four fixed parameters:

a0 = initial volume of acid solution (measured in liters),
A0 = initial concentration of acid (measured in M = moles/liter),
c = concentration of base (measured in M = moles/liter),
Ka = dissociation constant of acid.

A is usually the only unknown parameter. (Sometimes the identity of the acid, and therefore  Ka,  is unknown, but it can easily be calculated -- as you will see in the Project.) Thus, to determine the concentration of acid in a solution, one can perform an acid-base titration and simply find the value of  A for which the model  H(v)  best fits the titration data collected.

An Example

We formulate the Titration Model for the following specific acid-base titration:

Suppose that a titration is performed on  0.20  of a solution containing formic acid (a weak acid) with  0.1 M NaOH  (the base). Suppose that the concentration of formic acid in this solution is  0.074 M.  Formic acid has a dissociation constant of  Ka = 1.77 x 10-4.

Using appropriate values for the four parameters listed in the Note, write down the Titration Model for this particular acid-base titration. Then graph your model, and check that it has the general shape of a typical titration graph as in Figure 2.

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