Lec XIV

  Chem 1200

Angel C. de Dios

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Strong acid (HCl) versus strong base (NaOH)

At the beginning of an acid-base titration, when no NaOH has been added yet, the pH of the solution inside the beaker corresponds to the pH of an aqueous solution that contains a strong acid. At this point [H3O+] equals the concentration of HCl(aq) inside the beaker. Here is the graph with data values.

With no base added yet, pH = 1.00 because the concentration of HCl is initially 0.100 M. As we add base, OH- from the NaOH solution will react with the H3O+ inside the beaker containing the solution of HCl. This reaction essentially reaches completion, because this is simply the opposite of the autoprotolysis of water;

H3O+(aq)  +  OH-(aq) -->  HOH(l)  +  HOH(l)   K = 1.00 X 10^14

For example, when we add 10.0 mL of 0.100 M NaOH(aq) into the beaker containing 50.0 mL of 0.100 M HCl(aq), this is equivalent to adding 1.00 millimole of OH- (10.0 mL of 0.100 M NaOH means 10 times 0.100 millimole or 1.00 millimole). Inside the beaker, we have 5 millimoles of H3O+ (50.0 mL times 0.100 M of HCl = 5 millimoles). 1.00 millimole of OH- will react with the same number of H3O+, reducing the amount of H3O+ inside the beaker from 5 to 4 millimoles. Thus, upon adding 10.0 mL of 0.100 M NaOH, we now have 4 millimoles of H3O+ left inside a 60.0 mL of solution (50.0 mL was initially inside the beaker and we added 10.0 mL). The new concentration of H3O+ inside the beaker is therefore 4 millimole divided by 60.0 mL, or 0.0667 M. The pH is the negative logarithm of the concentration of H3O+, -log (0.0667) = 1.18, which agrees with the above data. You can do a similar calculation for the other points, up till 49.0 mL of 0.100 M NaOH added. Beyond this, the solution becomes a very dilute solution of an acid, and as we have seen previously, that situation is quite complicated. We can, however, calculate the pH at the point when 50.0 mL of 0.0100 M NaOH have been added. At this point, we have added exactly the same number of millimoles (5.00) of NaOH as the number of millimoles of HCl inside the beaker. This is called the equivalence point. At this point, all we have is an aqueous solution containing Na+ and Cl- ions, which are both neutral. The pH therefore is 7.00 at the equivalence point of NaOH versus HCl. Past the equivalence point, we have already added an excess of NaOH. For example, with 51.0 mL of 0.100 M NaOH added, we have an excess of 0.100 mmole of OH-. The concentration of OH- at this point is 0.100 mmole in 101.0 mL of solution, that is [OH-] = 0.100/101.0 = 0.000990 M. The pOH is therefore 3.00 and the pH is 11.00 at this point. One thing you should notice is that near the equivalence point, the pH is changing dramatically.

The titration curve for NaOH versus a weak acid like acetic acid looks different. The equivalence point is not at pH = 7, since at the equivalence, what we have is an aqueous solution of sodium acetate, and acetate is basic. The pH at the equivalence point is therefore slightly above 7. Also, the half-equivalence point (where we have added half the amount of NaOH required to completely neutralize the acetic acid inside the beaker) has a pH that is equal to the pKa of the weak acid. We will see soon why this happens.

Weak acid - strong base

First, we need to learn how to calculate the pH when we add to a solution containing a weak acid, either a strong base or its conjugate base.

At the initial point (when no NaOH has been added), all we have is an aqueous solution of a weak acid. Thus, at this point, we only have to pay attention to the ionization of the weak acid and this, as we have seen before, gives an approximate solution to the concentration of H+:

Once we add NaOH, then this reaction happens and it is complete:

Therefore, for every NaOH we add, an CH3COOH molecule gets converted into an CH3COO- ion. We can therefore estimate the number of mmoles of both acetic acid and acetate. The number of mmoles of acetic acid should be equal to what remains in solution, that is, mmoles of acetic acid equals initial mmoles of acetic acid minus mmoles of base added. We call this "mmoles of acid left". The number of mmoles of acetate, on the other hand, should equal the number of mmoles of NaOH we added since  it is NaOH reacting with acetic acid that forms the acetate ion.

In the above, what one should expect when we added enough NaOH to neutralize half of the acetic acid, the number of mmoles of acetate should equal the number of mmoles of acetic acid. This is the half equivalence point. We reach this point when we have neutralized half of the acid. At this point, pH becomes equal to pKa since [acetate]=[acetic acid].

At the equivalence point, we have converted all of the acetic acid into acetate so all we have in solution are acetate ions. Acetate is a weak base, so we simply apply the equation we have learned when dealing with an aqueous solution of a weak base.

What sets a weak acid - strong base titration from a strong acid - strong bae titration is the fact that the weak acid has to be in equilibrium with its conjugate base. There is an equilibrium between acetic acid and the acetate ion:

And as soon as we add NaOH, [H+] is no longer equal to [A-]