Part B: Weak Acid and Strong Base
© Silvia Kolchens
Pima Community College
B. Titration of a Weak Acid with a Strong Base
When tritrating a weak acid with a strong base we obtain an acid-base titration curve with the following characteristic features (Figure 2):
How to calculate pH values for weak acid/strong base titration curves:
Example: Titrate 50 mL 0.1 M acetic acid (CH_{3}COOH) with 0.1 M NaOH solution:
CH_{3}COOH + NaOH ó
CH_{3}COO^{- }+ H_{2}O + Na^{+}
(1) What is the pH of the acetic acid before any base is added?
When dealing with weak acids in aqueous solutions, we can no longer assume total dissociation. In fact, the equilibrium lies predominantly on the left hand side of the equation:
CH_{3}COOH(aq) + H_{2}O(l) ç CH_{3}COO^{-} (aq) + H_{3}O^{+}(aq)
In this case, the hydronium ion concentration is not immediately known, but can be determined from an iCe table:
iCe table
CH_{3}COOH(aq) | H_{2}O(l) | CH_{3}COO^{-} (aq) | H_{3}O^{+}(aq) | |
i | 0.1 M | - | 0 | 0 |
C | -x | - | +x | +x |
e | 0.1-x | - | x | x |
We make the approximation that x is much smaller than the initial concentration (x<<0.1) and can be neglected. This will greatly simplify our expression and we can solve for "x", the hydronium ion concentration:
The value for K_{a} of acetic acid is 1.8x10^{-5 }and we obtain
x=[H_{3}O^{+}] = 0.003 M
The pH of the 0.1 M acetic acid solution is then
pH=-log 0.003 =2.52
(2) What is the pH value at the equivalence point:
At the equivalence point we have equivalent amounts of acids and bases are present. In our example this means that 50 mL of 0.1 M CH_{3}COOH and 50 mL of 0.1 M NaOH solution have been combined. The total volume is now 100 mL.
At this point, all acid has been neutralized, i.e. all CH_{3}COOH has been converted to CH_{3}COO^{-}. The acetate ion (CH_{3}COO^{-}) itself is a weak base and will react with water (hydrolysis):
CH_{3}COO^{-} (aq) + H_{2}O(l) ó CH_{3}COOH + OH^{-}(aq)
This hydrolysis reaction will produce hydroxide ions, hence the equivalence point will be higher than pH 7. We can calculate the pH at the equivalence point using an iCe table and the base dissociation constant (K_{b}) for the acetate ion:
First, we have to calculate the molar concentration of the acetate ion:
Now we can use this concentration in the iCe table:
CH_{3}COO^{-} (aq) | H_{2}O(l) | CH_{3}COOH | OH^{-}(aq) | |
i | 0.05 | - | 0 | 0 |
C | -x | - | +x | +x |
e | 0.05-x | - | x | x |
We assume that x is much smaller than 0.05, and 0.05-x~0.5. The expression for K_{b} can then be written
K_{b} for the acetate ion is 5.6x10^{-10} and we solve for x;
x= [OH^{-}] = 5.29 x 10^{-6}
pOH = 5.27
pH = 14-pOH = 8.72
The equivalence point is in the basic range.
(3) What is the pK_{a} value of the acid?
What is a pK_{a} value? The pK_{a} value is the negative logarithm of the acid dissociation constant K_{a}:
pKa = -log K_{a}
The midpoint can be determined from the equivalence point: it is located where just half the volume of titrant has been used to neutralize the acid. Consequently the solution contains equal amounts of acid (CH_{3}COOH) and its conjugated base (CH_{3}COO^{-}) or
[CH_{3}COOH] = [CH_{3}COO^{-}]
Using this equation in the equilibrium constant and solving for H_{3}O^{+} yields
Taking the negative logarithm on both sides yiels
Which can also be expressed as
or
This last equation is known as the Henderson Hasselbach equation. What does the equation mean?
If the concentrations of acid (CH_{3}COOH) and its conjugated base (CH_{3}COO^{-}) are the same, then the ratio CH_{3}COO-/CH_{3}COOH will be "1". Since the logarithm of "1" equals zero, this equation simplifies to
pH = pKa
At the mid point the pKa value of an acid equals the measured pH. Thus we have a very simple method to determine acid dissociation constants.
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© copyright Silvia Kolchens, Pima Community College 2000
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