James Richard Fromm
The titration curve for an acid-base titration is a plot of the solution pH, normally on the vertical axis, against the volume of titrant added. We will investigate these curves by following a particular example in detail.
Example. Let us calculate the pH values along the course of a particular titration, the titration of exactly 100 mL of 0.100 molar acetic acid with 0.100 molar sodium hydroxide. In practice, we would probably titrate 25.00 mL instead for convenience, but these volumes are selected to simplify some of the calculations. In the course of the calculations and discussion, we will answer such questions as the titrant to use, the indicator to select, and what manner of solution is present at various points along the titration curve.
First, consider the titrant solution. Acetic acid is a weak acid, so a base is required to obtain any acid-base reaction. To get a pH break which is as large as possible at the equivalence point we should choose a strong base. The common strong bases are NaOH and KOH, so we choose one of these. It is normal to choose a concentration of base that will give us a reasonable volume to add, so that the volume can be accurately measured and yet not be too large. Here we choose 0.100 molar NaOH and will therefore require about 100 mL of it, since the stoichiometric reaction is
CH3COOH + Na+ + OH- CH3COO- + Na+ + H2O.
At the start of the titration the solution contains only the weak acid CH3COOH, for which
Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH]; Ka = [H3O+]2/[CH3COOH]
We know that [H3O+] is approximately equal to [CH3COO-], from the stoichiometric dissociation of the weak acetic acid and the fact that it is the major source of hydrated protons in the solution. Solving:
[H3O+]2 = 1.75 x 10-5 x 0.1, [H3O+] = 1.32 x 10-3, pH = 2.88
The pH slowly rises as the NaOH added reacts with the acetic acid. The acetic acid has been one-quarter titrated when 25 mL of the NaOH solution have been added. At this point 1/4 of the original moles of CH3COOH have been titrated to CH3COO-, while 3/4 of the original moles of CH3COOH remain, so 3[CH3COO-] = [CH3COOH] and
Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] = [H3O+]/3
[H3O+] = 3 x 1.75 x 10-5 = 5.25 x 10-5, pH = 4.28
Now consider the situation when half of the acetic acid has been titrated. On the volume axis, that would be at 1/2 x 100 mL = 50 mL of added NaOH. In the solution at that point, half of the original acetic acid has been titrated and half has not, so mol CH3COOH = mol CH3COO- and, since both are present in the same solution, [CH3COOH] = [CH3COO-]. As a consequence,
Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] and
[H3O+] = 1.75 x 10-5, pH = 4.76.
The acetic acid is three-quarters titrated when 75 mL of the NaOH solution have been added. At this point 3/4 of the original moles of CH3COOH have been titrated to CH3COO-, while 1/4 of the original moles of CH3COOH remain, so [CH3COO-] = 3[CH3COOH] and
Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] = 3[H3O+]/1
[H3O+] = (1/3)(1.75 x 10-5) = 5.83 x 10-6, pH = 5.23
The equivalence point is reached after 100 mL of the NaOH solution have been added, just enough to react with all of the acetic acid present. Since the reaction is that of a weak acid plus a strong base to yield a weak base and water, the solution at the equivalence point is simply a solution of the weak base CH3COONa, sodium acetate. By definition of the equivalence point no excess of either weak acid or strong base can be present. The situation is therefore just as it would be for a solution of the weak base sodium acetate in water. Since there were originally 100 mL of 0.1 molar CH3COOH, or 10 mmol CH3COOH, there are now 10 mmol of CH3COONa. These are contained in 200 mL of solution because we started with 100 mL and added another 100 mL, so the formal concentration of acetate is 10 mmol/200 mL = 0.05 molar.
The equilibrium constants are Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] and Kb = 5.77 x 10-10 = [OH-][CH3COOH]/[CH3COO-]. It can no longer be assumed that [H3O+] is approximately equal to [CH3COO-], because the solution is now basic, containing the weak base acetate ion. The major source of the [H3O+] is not the dissociation of CH3COOH since there is virtually no CH3COOH left to dissociate. However, it can now be assumed that [OH-] is approximately equal to [CH3COOH], because the major source of hydroxide ion is the weak base CH3COO- which hydrolyzes, giving CH3COOH and OH- in a stoichiometric 1:1 ratio, following the reaction equilibrium
CH3COO- + H2O CH3COOH + OH- .
To a good approximation, Kb = [OH-]2/[CH3COO-]. Since [CH3COO-] is about 0.05 molar,
[OH-]2 = Kb[CH3COO-], [OH-] = 5.37 x 10-6
pOH = -log(5.37 x 10-6) = 5.27, pH = 14.00 - 5.27 = 8.73
The equivalence point pH having been determined as 8.73, an indicator can now be chosen for this titration. Those indicators given in a Table of Acid-Base Indicators for which pKa is approximately equal to the pH at the equivalence point, 8.73, are thymol blue, whose pKa is 8.9, and cresol purple, whose pKa is 8.3. Either would be satisfactory for this titration. For thymol blue, the color change would be from the yellow color of the acid form to the blue color of the base form.
Notice that the pH slowly rises throughout that part of the titration curve prior to the equivalence point. If an indicator such as bromocresol green (pKa = 4.7) had been chosen it would have changed from its acid color, yellow, to green well before a chemist was 3/4 of the way to the equivalence point. The titration results would be meaningless, because the endpoint at which color change occurs would not be easily relatable to the equivalence point.
As NaOH addition continues beyond the equivalence point, the pH continues to rise. When twice as much NaOH has been added as is required, in other words when 200 mL of the NaOH solution have been added, the resulting solution is then a mixture of the strong base NaOH and the weak base CH3COONa. In any mixture of a strong base and a weak base of comparable concentration, the strong base controls the pH, so that [OH-] is approximately equal to the excess concentration of added NaOH. There were added 200 mL x 0.1 molar = 20 mmol NaOH, but of this 10 mmol was used to react with the 10 mmol of acetic acid originally present. As a consequence, there are now 20 - 10 = 10 mmol NaOH present in 100 + 200 = 300 mL solution. Then [OH-] = 10 mmol/300 mL = 0.033 molar,
[OH-] = 3.3 x 10-2, pOH = 1.48, pH = 14.00 - 1.48 = 12.52
Example. A total of 50 mL of aqueous ammonia, NH3, solution which is approximately 0.01 molar is to be titrated. Let us consider what it should be titrated with, what indicator should be used, and what color change would then be seen.
To titrate this weak base, we could choose any strong acid, such as HCl, with a convenient concentration of about 0.01 molar. For ammonium ion, Ka = 5.60 x 10-10 = [H3O+][NH3]/[NH4+].
At the equivalence point, the solution is essentially a solution of NH4Cl. On hydrolysis one ammonium ion yields one hydrogen ion and one ammonia;
NH4+ NH3 + H3O+.
To a good approximation, [H3O+] = [NH3], so Ka = [H3O+]2/[NH4+]; [H3O+] = Ka[NH4+]. There were 0.01 mmol/L x 50 mL = 0.5 mmol NH4+; this is in 50 + 50 = 100 mL solution so [NH4+] = 0.5 mmol/100 mL = 0.005 mmol/L NH4+.
[H3O+]2 = (5.60 x 10-10)(5 x 10-3) = 2.80 x 10-12
[H3O+] = 1.67 x 10-6, pH = 5.78
Since the pH should be approximately equal to pKa of the indicator, a reasonable choice would be either chlorophenol red or methyl red. The change is from base color to acid color, which is from red to yellow for chlorophenol red and yellow to red for methyl red.