Titration is the controlled neutralization of a acid and a base. If we consider the titration of a weak acid by a strong base, we can analyze the situation completely and determine all the concentrations of the aqueous species at any volume addition of the titrant.

Notice, from the figure below, that the titration of a weak acid with a strong base is different from that of a strong acid/base titration, primarily in the begiinig of the titration, i.e., before the equivalence point. Moreover, the equivalence point itself is shifted upward in pH.



We can calculate our own weak acid/strong base titration curve. Assume, for simplicity, some numbers. Let's say we have 25.00 mL of 0.100 M of a hypothetical acid HA (Ka = 1.0 x 10-5) and we titrate this with 0.100 M NaOH. The titration table below can then be grouped into regions of similar calculations.
The entries in the tables are linked to the calculation.

25.0 mL of 0.100M HA (Ka=1 x 10-5), titrated with with 0.100M NaOH:
Vol. NaOH Total Volume % Diss n[H+]e n[OH-]e pH
0 25 (a) (a) (b) (a)
10 35 (c) (c) (b) (c)
20 45 (c) (c) (b) (c)
22 47 (c) (c) (b) (c)
24 49 (c) (c) (b) (c)
25 50 (d) (e) (e) (e)
26 51 (d) (g) (f) (g)
28 53 (d) (g) (f) (g)
30 55 (d) (g) (f) (g)
40 65 (d) (g) (f) (g)
50 75 (d) (g) (f) (g)

Hints for the On-line quiz...

What, in general, is the pH at the equivalence point for this type of titration?
What about at 1/2 equivalents base added?

Calculation (a):
Before any base is added to the acid, the extent of the dissociation of the acid determines the H+(aq) concentation through the reaction:

HA = H+(aq) + A-(aq)

Recognizing that the number o moles of H+(aq) is equal to the number of moles of A-(aq), and each is equal to the number of moles of HA that dissociates gives the following expression:

The number of moles of HA present before titration is just the initial volume times the initial concentration or 0.025L x 0.100 M= 2.50 mmol of HA. Solving for the nH+ by successsive approximation or the quadratic formula will yield a result of 2.488 x 10-2 millimoles of H+(aq).
Since the volume is 25.0 mL, the concentration of H+(aq) of 9.950 x 10-4 molar (which is approximately 1.0 x 10 -3)
The pH is easily calculated from the H+(aq) concentration through its' definition:

pH = -log10([H+(aq)])
The pH of the acid before the base is added is 3.002. The percent dissociation is just (xe/nHA0) x 100 or 1.0%.

In this case the concentrated weak acid approximation is adequate to predict the pH, namely because the % dissociation is so low. The result of this approximation is:

pH = (1/2){-log(Ka{nHA0/V})} = 3.000




Calculation (b):
Once the pH is known, the pOH is easily calculated at 25oC by the relation


pOH = 14 - pH
The OH-(aq) concentration is calculated from the pOH as:

[OH-(aq)] = 10- pOH
And finally the moles of OH-(aq) is:

nOH-(aq) = 10- pOH * Vtotal

Calculation (c)
During the beginning of the titration, the number of moles of HA in solution will be reduced stoichiometrically by reaction with a strong base through the reaction:

Therefore, the addition of strong base to the weak acid creates a buffer just like mixing the salt of the weak acid to the acid. Is it clear how the equilibrium constant for this neutralization is related to the pKa of the acid HA (1014-pKa). After this step takes place the equilibrium that must be considered is the same one as is step (a)

HA = H+(aq) + A-(aq)

but there are significant quantities of Both HA and A- in solution. Now the equilibrium expression is:


As a numerical example, consider the point at which 10.0 mL of base is added.
V = 35mL, nHA0 = 1.5 mmol; n(A-)0 = 1.0 mmol

The equilibrium expression above has but one unknown, so we may solve for the number of moles of H+(aq). For this numerical example, then:

n(H+) = 5.25 x 10-4, pH = 4.82


Can you think of the approximations that are made at this point? Well we can make a useful one by observing that since this is a buffer, we can use the Henderson-hasselbalch equation for the pH of a Buffer. (What approximations are implicit in the use of 'The Buffer equation'? (nH+ << nA-, nHA))

pH = pKa + log10([A-]/[HA])


The amount the acid has dissosciated is almost, but not exactly, the ratio of the moles of base added to the moles of acid initially present. Thus, for our example:
% Diss = n(A-)0 / nHA0 x 100 = (1.0mmol/2.5mmol) = 40%

Calculation (d)
At equivalence, the number of moles of base added is equal to the number of moles of HA acid initially present, so all of the acid is converted to its conjugate base anion. Therefore the dissociation is complete


Calculation (e)
At equivalence, the solution is an aqueous salt. In fact, it is identical to a 0.050M solution of the salt NaA(aq). So, the pH is determined from the equilibrium of a weak base, the conjugate base of HA

A-(aq) + H2O = OH-(aq) + HA(aq)

The equilibrium constant of this reaction is related to the pKa of the acid HA and the solution of the equilibrium expression for this reaction will yield the hydroxide concentration, [OH-(aq)].


We may use the concentrated weak base approximation, in exact analogy to the concentrated weak acid approximation, to determine the [OH-(aq)], whcih is tantemount to ignoring y in the denominator of the equilibrium expression. Therefore:

y = [OH-(aq)] = (Kb [A-])1/2 = 7.07 x 10-6 M

And, of course, the pOH is then:

pOH = -log10([OH-(aq)]) = 5.15

and the pH is:

pH = 14 - pOH = 8.85

The moles of OH-(aq) and H+(aq) are then:

nOH-(aq) = [OH-(aq)] * Vtotal

nH+(aq) = [H+(aq)] * Vtotal


Calculation (f)
Now we consider only the concentration of OH- coming from the excess strong acid


Calculation (g) Of course, as the titration has changed into a simple strong base dilution problem, you have considered and calculated the pOH first and use it to determine the pH and n(H+).

pH = 14 - pOH

n(H+) = 10-pH * Vtotal


A Summary of Numerical Results for our Example
Vol. NaOH

Added [mL]

Total Volume

[mL]

% Diss

Acid

nHA nA- n[H+]e

mmol

n[OH-]e

mmol

pH
0 25 1.0 2.5 0 2.488 x 10-2 2.513 x 10-10 3.00
10 35 40 1.5 1.0 5.245 x 10-4 2.335 x 10-8 4.82
12.5 37.5 50 1.25 1.25 3.748 x 10-4 3.777 x 10-7 pKa = 5.00
20.0 45.0 80 0.500 2.000 1.125 x 10-4 1.801 x 10-7 5.602
22.0 47.0 88 0.300 2.200 6.408 x 10-5 3.448 x 10-7 5.865
24.0 49.0 96 0.100 2.400 2.041 x 10-5 1.176 x 10-6 6.380
25 50 100 3.54 x 10-4 2.5 7.07 x 10-8 3.54 x 10-4 8.85
26 51 100 2.5 x 10-9 2.5 2.60 x 10-10 0.1 11.29
28 53 100 8.3 x 10-9 2.5 9.363 x 10-11 0.3 11.75
30 55 100 5.0 x 10-9 2.5 6.050 x 10-11 0.5 11.96
40 65 100 1.7 x 10-9 2.5 2.82 x 10-11 1.5 12.36
50 75 100 1.0 x 10-9 2.5 2.25 x 10-11 2.5 12.52

The shape of the titration curve for a weak acid with a strong base will depend on the value of its' Ka.

The Titration of a Weak Acid with a Strong Base is calculated in 4 steps:

See below for 25 mL of 0.100 M acid (HA) titrated with 0.100 M NaOH (with three different values of the pKa of HA):


What does the titration curve for a weak base titrated with a strong acid look like?


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