Weak Acid - Strong Base Titration Curve Chemistry Tutorial
Key Concepts
- It is possible to calculate the pH of a solution when a weak acid is titrated with a strong base:
⚛ Before any strong base is added to weak acid :
[H+(aq)] ≈ √Ka[weak acid]
pH = −log10[H+(aq)]
⚛ Addition of strong base while weak acid is in excess :
R.I.C.E. Table to determine equilibrium concentrations of anion and weak acid
[OH-(aq)] = (Kb[anion])/[weak acid]
pOH = −log10[OH-(aq)]
pH = 14.00 − pOH (at 25°C)
⚛ When all the weak acid has been neutralised by all the strong base we calculate the pH of the aqueous salt solution:
[OH-(aq)] ≈ √Kb[anion])
pOH = −log10[OH-(aq)]
pH = 14.00 − pOH (at 25°C)
⚛ When the strong base is in excess:
[OH-(aq)] = n(OH- in excess) ÷ V(total)
pOH = −log10[OH-(aq)]
pH = 14.00 − pOH (at 25°C)
- The titration curve for a weak acid - strong base titration has a characteristic shape in which the following features can be identified:
⚛ Initial pH (pH < 7) : before any base is added the pH of the weak acid is dependent on:
(a) concentration of weak acid
(b) value of Ka (which is also temperature dependent)
⚛ Buffer zone (buffer region) of titration curve :
(a) addition of strong base while weak acid is in excess
(b) weak acid is in equilibrium with its conjugate base
⚛ Point at which pH = pKa :
(a) occurs when half the weak acid has been neutralised by strong base
(b) pH at the point V(base) = ½V(base added to neutralise acid)
(c) also corresponds to the pH of the most effective buffer solution
⚛ Equivalence point (pH > 7):
(a) just enough strong base has been added to exactly neutralise all the weak acid
(b) pH determined by hydrolysis of anion and is dependent on:
(i) concentration of anion
(ii) Kb = Kw/Ka (which is temperature dependent)
(ii) Kw (which is temperature dependent)
Note: an appropriate indicator for the titration is one for which pKIn ± 1 = pH
KIn is the dissocation constant for the indicator
pKIn = −log10KIn
pH is the pH at the equivalence point of the neutralisation reaction.
⚛ After equivalence point (pH > 7): strong base is in excess and pH is dependent on:
(a) concentration of excess base
(b) Kw (which is temperature dependent)
- We can draw a reasonable sketch of a weak acid - strong base titration using these 4 calculations:
(1) initial pH of weak acid before base is added
(2) pH of the salt solution at the equivalence point
(3) pH of the most effective buffer solution (pH = pKa)
(4) pH of the solution when strong base is in excess (approximately approaches pH of the strong base solution).
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Calculating a Titration Curve for a Weak Acid - Strong Base Titration
An aqueous solution of acetic acid (ethanoic acid), CH3COOH(aq), is an example of a weak acid.
An aqueous solution of sodium hydroxide, NaOH(aq), is an example of a strong base.
Imagine an experiment in which a 10.00 mL aliquot of 0.200 mol L-1 CH3COOH(aq) is transferred to a clean conical flask (erlenmeyer flask).
A burette (buret) is then filled with 0.200 mol L-1 NaOH(aq)
The equipment would be set up as in the diagram below:
The NaOH(aq) is added 1.00 mL at a time to the CH3COOH.
We can calculate the pH of the solution in the conical flask after each addition of NaOH(aq).
You might like to visit the tutorial on calculating the pH of a solution after mixing weak acid and strong base before continuing with the following calculations.
Initially, before we add any sodium hydroxide to the acetic acid (ethanoic acid), we have 10.00 mL of 0.200 mol L-1 CH3COOH(aq).
CH3COOH(aq) is a weak acid so it exists in equilibrium with its ions; acetate ions (ethanoate ions), CH3COO-(aq), and hydrogen ions (protons), H+(aq), as shown in the balanced chemical equation below:
CH3COOH(aq) ⇋ CH3COO-(aq) + H+(aq) Ka = 1.80×10-5 (at 25°C)
Use a R.I.C.E. Table to determine the equilibrium concentration of H+(aq)
| Reaction |
CH3COOH(aq) |
⇋ |
CH3COO-(aq) |
+ |
H+(aq) |
Initial concentration
(mol L-1) |
0.200 |
|
0 |
|
0 |
Change in concentration
(mol L-1) |
−x |
|
+x |
|
+x |
Equilibrium concentration
(mol L-1) |
0.200 − x ≈ 0.200
(Assume x << 0.200)(1)
|
|
0 + x = x |
|
0 + x = x |
Use the expression for the equilibrium constant for the dissociation of acetic acid to calculate x, which is the concentration of hydrogen ions, [H+(aq)]:
| Ka |
= |
[CH3COO-(aq)][H+(aq)]
[CH3COOH(aq)] |
| 1.80×10-5 |
= |
[x][x]
[0.200] |
| √(0.200 × 1.80×10-5) |
= |
√[x]2 |
| 1.90×10-3 mol L-1 |
= |
[x] |
| 1.90×10-3 mol L-1 |
= |
[H+(aq)] |
Calculate the pH of the solution before any base has been added:
pH = −log10[H+(aq)] = −log10[1.90×10-3] = 2.72
So, when 0.00 mL of NaOH(aq) has been added to 10.00 mL of 0.200 mol L-1 CH3COOH(aq) the pH of the solution is 2.72.
Now, let's add 1.00 mL of 0.200 mol L-1 from the burette to the 10.00 mL of 0.200 mol L-1 CH3COOH(aq) in the conical flask.
A neutralisation reaction occurs between the acid and base according to the following balanced chemical equation:
| general word equation: |
acid |
+ |
base |
→ |
salt |
+ |
water |
| word equation: |
acetic acid
(ethanoic acid) |
+ |
sodium hydroxide |
→ |
sodium acetate
(sodium ethanoate) |
+ |
water |
| balanced chemical equation: |
CH3COOH(aq) |
+ |
NaOH(aq) |
→ |
CH3COONa(aq) |
+ |
H2O(l) |
Use this equation to calculate the moles of each substance consumed and produced:
| balanced chemical equation: |
CH3COOH(aq) |
+ |
NaOH(aq) |
→ |
CH3COONa(aq) |
+ |
H2O(l) |
initial concentration
(mol L-1 |
0.200 |
|
0.200 |
|
0 |
|
|
initial volume
(L) |
10.00×10-3 |
|
1.00×10-3 |
|
0 |
|
|
moles
n = c × V |
2.00×10-3 |
|
2.00×10-4 |
|
0 |
|
|
mole ratio
(stoichiometric ratio) |
1 |
: |
1 |
: |
1 |
: |
1 |
| |
CH3COOH is in excess, NaOH is the limiting reagent. |
moles AFTER reaction:
= n(initial) − n(reacted) |
2.00×10-3 − 2.00×10-4 = 1.80×10-3 |
|
2.00×10-4 − 2.00×10-4 = 0 |
|
2.00×10-4 |
|
|
volume AFTER reaction:
= V(acid) + V(base added) |
10.00 + 1.00 = 11.00 mL
(11.00×10-3 L) |
|
10.00 + 1.00 = 11.00 mL
(11.00×10-3 L) |
|
10.00 + 1.00 = 11.00 mL
(11.00×10-3 L) |
|
|
concentration AFTER reaction:
c = n ÷ V
(mol L-1) |
0.1634 |
|
0 |
|
0.01818 |
|
|
The acetate ion (ethanoate ion), CH3COO-(aq), undergoes hydrolysis (reacts with water) according to the following equation:
CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)
Use a R.I.C.E. Table to determine the equiliubrium concentrations of each species:
| Reaction |
CH3COO-(aq) |
+ |
H2O(l) |
⇋ |
CH3COOH(aq) |
+ |
OH-(aq) |
Initial concentration
(mol L-1) |
0.01818 |
|
|
|
0.1632 |
|
0 |
Change in concentration
(mol L-1) |
−x |
|
|
|
+x |
|
+x |
Equiliubrium concentration
(mol L-1) |
0.01818 −x ≈ 0.01818 |
|
|
|
0.1632 +x ≈ 0.1632 |
|
x |
| |
(Assume x << 0.01818 and therefore x << 0.1632) |
and we can write an expression for the hydrolysis equilibrium constant, Kh, (which is equivalent to the base dissociation constant, Kb) and calculate the value of Kb (since Kb = Kw ÷ Ka):
| Kb |
= |
[CH3COOH(aq)][OH-(aq)]
[CH3COO-(aq)] |
= |
1.00×10-14
Ka |
= |
1.00×10-14
1.8×10-5 |
= |
5.56×10-10 |
and use the equilibrium concentrations of each species to determine [OH-(aq)]:
[CH3COOH(aq)][OH-(aq)]
[CH3COO-(aq)] |
= |
5.56×10-10 |
| [OH-(aq)] |
= |
5.56×10-10[CH3COO-(aq)]
[CH3COOH(aq)] |
| [OH-(aq)] |
= |
5.56×10-10[0.01818]
[0.1632] |
| [OH-(aq)] |
= |
6.19×10-11 |
which we can use to calculate the pOH :
pOH = −log10[OH-(aq)] = −log10[6.19×10-11] = 10.21
and therefore the pH (at 25°C):
pH = 14.00 − pOH = 14.00 − 10.21 = 3.79
We could continue these calculations for the addition of 2.00 mL, 3.00 mL, 4.00 mL, ... up to 9.00 mL of NaOH(aq).
But once we add 10.00 mL of 0.200 mol L-1 NaOH(aq) to 10.00 mL of 0.200 mol L-1 CH3COOH(aq) the acid will no longer be in excess, and the base is no longer the limiting reagent, we have reached the equivalence point for this neutralisation reaction.
The only substance in solution at the equivalence point is an aqueous solution of sodium acetate, so we need to calculate the pH of this salt solution.
The salt, sodium acetate (sodium ethanoate), CH3COONa(aq) is soluble in water (all acetates are soluble); it fully dissociates into acetate ions (ethanoate ions), CH3COO-(aq), and sodium ions, Na+(aq), as shown by the chemical equation below:
CH3COONa(aq) → CH3COO-(aq) + Na+(aq)
The sodium ion, Na+(aq), does not undergo hydrolysis (does not react with water), but the acetate ions, CH3COO-(aq), will undergo hydrolysis (will react with water) according to the chemical equation below:
CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)
In Arrhenius terms this is a hydrolysis reaction, in Brønsted-Lowry terms this is a proton transfer reaction in which the acetate ions, CH3COO-, are acting as a base with the conjugate acid being acetic acid, CH3COOH.
The expression for the equilibrium constant for this hydrolysis reaction is given the symbol Kh, and equally it can also be given the symbol Kb :
| Kh = Kb |
= |
[CH3COOH(aq)][OH-(aq)]
[CH3COO-(aq)] |
and since Kb(acetate ions) = Kw ÷ Ka(acetic acid)
at 25°C, Kb(acetate ions) = (1.00×10-14) ÷ (1.80×10-5) = 5.56×10-10
Calculate the intial concentration of acetate ions using the neutralisation reaction between acetic acid and sodium hydroxide:
| balanced chemical equation: |
CH3COOH(aq) |
+ |
NaOH(aq) |
→ |
CH3COONa(aq) |
+ |
H2O(l) |
concentration before reaction
(mol L-1) |
0.200 |
|
0.200 |
|
0 |
|
|
volume before reaction
(L) |
10.00×10-3 |
|
10.00×10-3 |
|
0 |
|
|
moles before reaction
(mol) |
2.00×10-3 |
|
2.00×10-3 |
|
0 |
|
|
mole ratio
(stoichiometric ratio) |
1 |
: |
1 |
: |
1 |
: |
1 |
| |
n(acid) = n(base) : equivalence point for neutralisation reaction |
moles aftere reaction
(mol) |
2.00×10-3 − 2.00×10-3 = 0 |
|
2.00×10-3 − 2.00×10-3 = 0 |
|
0 + 2.00×10-3 = 2.00×10-3 |
|
|
volume after reaction
= V(acid) + V(base)
(L) |
10.00×10-3 |
+ |
10.00×10-3 |
= |
20.00×10-3 |
|
|
concentration after reaction
(c = n ÷ V)
(mol L-1) |
0 |
|
0 |
|
0.100 |
|
|
Use a R.I.C.E. Table to determine the equilibrium concentrations of all species as a result of the hydrolysis of acetate ions:
| Reaction |
CH3COO-(aq) |
+ |
H2O(l) |
⇋ |
CH3COOH(aq) |
+ |
OH-(aq) |
Initial concentration
(mol L-1) |
0.100 |
|
|
|
0 |
|
0 |
Change in concentration
(mol L-1) |
−x |
|
|
|
+x |
|
+x |
Equilibrium concentration
(mol L-1) |
0.100 −x ≈ 0.100
(Assume x << 0.100) |
|
|
|
x |
|
x |
Use Kb and equilibrium concentrations to calculate concentration of hydroxide ions, [OH-(aq)], which is equal to x:
| Kb |
= |
[CH3COOH(aq)][OH-(aq)]
[CH3COO-(aq)] |
| 5.56×10-10 |
= |
[x][x]
[0.100] |
| √(0.100 × 5.56×10-10) |
= |
√[x]2 |
| 7.46×10-6 mol L-1 |
= |
[x] = [OH-(aq)] |
Calculate the pOH of this solution:
pOH = −log10[OH-(aq)] = −log10[7.46×10-6] = 5.13
Calculate the pH of this solution (at 25°C):
pH = 14.00 − pOH = 14.00 − 5.13 = 8.87
The pH of this titration experiment at the equivalence point is 8.87
After the equivalence point, further additions of NaOH(aq) will produce larger excess moles of NaOH(aq) but in increasing volumes of solution.
For example, if 11.00 mL of 0.200 mol L-1 NaOH(aq) is added to 10.00 mL of 0.200 mol L-1 CH3COOH(aq) :
| balanced chemical equation: |
CH3COOH(aq) |
+ |
NaOH(aq) |
→ |
CH3COONa(aq) |
+ |
H2O(l) |
concentration before reaction
(mol L-1 |
0.200 |
|
0.200 |
|
0 |
|
|
volume before reaction
(L) |
10.00×10-3 |
|
11.00×10-3 |
|
0 |
|
|
moles before reaction
(mol) |
2.00×10-3 |
|
2.20×10-3 |
|
0 |
|
|
mole ratio
(stoichiometric ratio) |
1 |
: |
1 |
: |
1 |
: |
1 |
| |
NaOH is in excess, CH3COOH is the limiting reagent |
moles after reaction
(mol) |
2.00×10-3 − 2.00×10-3 = 0 |
|
2.20×10-3 − 2.00×10-3 = 2.00×10-4 |
|
2.00×10-3 |
|
|
volume after reaction
=V(acid) + V(base)
(L) |
21.00×10-3 |
|
21.00×10-3 |
|
21.00×10-3 |
|
|
concentration after reaction
=moles + volume
(mol L-1) |
0 |
|
9.52×10-3 |
|
9.52×10-2 |
|
|
Because the value of the equilibrium constant for the hydrolysis of acetate ions is very small, Kh = Kb = 5.56×10-10, we will ignore its contribution to the hydroxide ion concentration, and only consider the complete dissociation of sodium hyroxide to produce hydroxide ions and sodium ions as shown in the chemical equation below:
NaOH(aq) → Na+(aq) + OH-(aq)
So, [OH-(aq)] = [NaOH(aq)(after reaction)] = 9.52×10-3 mol L-1
Calculate the pOH of this solution:
pOH = −log10[OH-(aq)] = −log10[9.52×10-3] = 2.02
Calculate the pH of this solution (at 25°C):
pH = 14.00 − pOH = 14.00 − 2.02 = 11.98
We could continue these calculations for the addition of 12.00 mL of 0.200 mol L-1 NaOH(aq), 13.00 mL, etc.
If you do this, you would end up with the pH values listed in the table below:
| |
CH3COOH(aq) in excess |
equivalence
point |
NaOH(aq) in excess |
Volume NaOH(aq)
(mL) |
0.00 |
1.00 |
2.00 |
3.00 |
4.00 |
5.00 |
6.00 |
7.00 |
8.00 |
9.00 |
10.00 |
11.00 |
12.00 |
13.00 |
14.00 |
|---|
| pH |
2.72 |
3.78 |
4.13 |
4.37 |
4.56 |
4.74 |
4.92 |
5.11 |
5.34 |
5.70 |
8.87 |
11.98 |
12.26 |
12.42 |
12.52 |
| |
acidic solution |
basic solution |
Plotting the points on a graph using the table above will result in a curve as shown below:
| pH of solution |
0.200 M CH3COOH - 0.200 NaOH Titration Curve
volume of NaOH(aq) added (mL)
|
Features of a Weak Acid - Strong Base Titration Curve
Let's consider the various important features of the acetic acid - sodium hydroxide titration curve we plotted above.
We have used different colours on the curve to differentiate the regions of the titration curve we will be discussing:
| pH of solution |
0.200 M CH3COOH - 0.200 NaOH Titration Curve
volume of NaOH(aq) added (mL)
|
First, consider the gold line on the titration curve.
When 0.00 mL of NaOH(aq) has been added to 0.200 mol L-1 CH3COOH(aq) the pH of the solution is due to the dissociation of this weak acid.
The pH is less than 7.00 (at 25°C) because this solution is acidic.
At 25°C the actual pH is determined by the concentration of the actic acid (and the value of its acid dissociation constant, Ka).
| |
lower
concentration |
→ |
→ |
higher
concentration |
[CH3COOH(aq)]
(mol L-1) |
0.001 |
0.010 |
0.100 |
1.000 |
| pH |
3.87 |
3.37 |
2.87 |
2.37 |
| |
higher pH |
← |
← |
lower pH |
As the concentration of dilute acetic acid, CH3COOH(aq), increases the pH of the solution decreases because there will be a greater concentration of hydrogen ions, H+(aq), in solution.
Next, consider the blue line of the titration curve.
In Arrhenius terms, this is the region in which the acetic acid is in excess, not enough sodium hydroxide has been added to "neutralise" all the acid.
In Brønsted-Lowry terms, we have a weak acid (CH3COOH(aq)) in equilibrium with its conjugate base (CH3COO-(aq)):
| acid |
+ |
base |
⇋ |
conjugate base |
+ |
conjugate acid |
| CH3COOH(aq) |
+ |
H2O(l) |
⇋ |
CH3COO-(aq) |
+ |
H3O+(aq) |
which means that, for as long as the acetic acid (CH3COOH(aq)) is in excess (that is, NaOH(aq) is the limiting reagent), we have established a buffer solution.
Adding more sodium hydroxide solution to this excess weak acid initially decreases the concentration of acetic acid, so we expect the pH to increase as the concentration of hydrogen ions due to the dissociation of the acid will also decrease.
But the addition of sodium hydroxide also initially increases the concentration of acetate ions, some of which will react to reform some undissociated acetic acid molecules, thereby increasing the concentration of acetic acid molecules by a bit more.
So a new equilibrium position is established in which :
- [CH3COOH(aq)] is less than before the addition of NaOH(aq) but not as small as predicted by the mole ratio (stoichiometric ratio) of the neutralisation reaction alone.
- [CH3COO-(aq)] is greater than before the addition of NaOH(aq) but not as large as predicted by the mole ratio (stroichiometric ratio) of the neutralisation reaction alone.
Since the acid dissociation constant for acetic acid is given by the following expression:
| Ka = |
[CH3COO-(aq)][H+(aq)]
[CH3COOH(aq)] |
We can rearrange this expression as shown below to determine the concentration of hydrogen ions in this buffer solution (and hence the pH) :
| [H+(aq)] = Ka× |
[CH3COOH(aq)]
[CH3COO-(aq)] |
Note that [H+(aq)], and hence the pH of the solution, is determined by the ratio of [CH3COOH(aq)] to [CH3COO-(aq)].
As the [CH3COOH(aq)] decreases by a bit, and the [CH3COO-(aq)] increases by a bit, the [H+(aq)] decreases by a bit so the pH increases by a bit.
Because this region of the titration curve relates to the formation of a buffer (a weak acid in equilibrium with its conjugate base), it is referred to as the "buffer zone" or the "buffer region".
Let's consider the green cross on the dark red line, x, on the titration curve above.
This cross represents the equivalence point for this neutralisation reaction: just enough NaOH(aq) has been added to neutralise all the available CH3COOH(aq), neither reactant is in excess, neither reactant is the limiting reagent.
This means that the solution at the equivalence point is made up of the salt named sodium acetate (sodium ethanoate), CH3COONa, dissolved in water.
The pH of this solution at the equivalence point is due to the hydrolysis of the acetate ion (ethanoate ion), CH3COO-(aq) which releases hydroxide ions, OH-(aq), so the solution is basic:
CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)
The pH can be calculate as follows:
| [OH-] = |
Kb[CH3COO-(aq)]
[CH3COOH(aq)] |
pOH = −log10[OH-(aq)]
pH = 14.00 − pOH (at 25°C)
The greater the concentation of acetic acid used in the experiment, the greater the concentration of acetate ions produced at the equivalence point, so the greater the hydroxide ion concentration.
Higher [OH-(aq)] means a lower pOH and therefore a higher pH.
| |
lower
concentration |
→ |
→ |
higher
concentration |
| [CH3COONa(aq)] |
0.001 |
0.010 |
0.100 |
1.000 |
| pH |
7.87 |
8.37 |
8.87 |
9.37 |
| |
lower pH |
→ |
→ |
higher pH |
Now, we choose an appropriate indicator for this titration experiment based on the predicted pH at the equivalence point for the neutralisation reaction.
Our chosen indicator must change colour at the equivalence point for the neutralisation reaction, put another way, the end point as indicated by the indicator must occur in the same pH region as the pH at the equivalence point for the neutralisation reaction.
So, for 0.001 mol L-1 CH3COONa(aq), the pH at the equivalence point is 7.87, so we need an indicator that changes colour around pH = 7.87.
If we have 1.00 mol L-1 CH3COONa(aq), the pH at the equivalence point is 9.37, so we need an indicator that will change colour around pH = 9.37.
An indicator is just a weak acid (or weak base) for which the acidic form (HIn) is a different colour to the basic form (In):
KIn is the value of the dissociation constant of an indicator:
| KIn |
= |
[H+][In-]
[HIn] |
or |
KIn |
= |
[H3O+][In-]
[HIn] |
At the end point, [HIn] = [In-], so
KIn = [H+(aq)]
and so,
−log10KIn = −log10[H+(aq)]
pKIn = pH
Since we choose an appropriate indicator for a titration as one which changes colour at the equivalence point for the neutralisation reaction, we can say that the best indicator will have an end point at the same pH as the reaction's equivalence point, therefore
pKIn = pH = equivalence point for the neutralisation reaction
So, given a list of possible indicators such as the one below (in which the effective range is assumed to be ± 1):
| indicator name |
pKIn |
effective pH range |
|---|
| bromothymol blue |
7.1 |
6.1 to 8.1 |
| phenolphthalein |
9.4 |
8.4 to 10.4 |
| thymolphthalein |
10.0 |
9.0 to 11.0 |
| alizarin yellow R |
11.2 |
10.2 to 12.2 |
we could choose a suitable indicator for each of our CH3COONa(aq) solutions resulting from the titration of acetic acid with sodium hydroxide:
| [CH3COONa(aq)] |
0.001 |
0.010 |
0.100 |
1.000 |
| pH (equivalence point) |
7.9 |
8.4 |
8.9 |
9.4 |
| pKIn (indicator) |
7.9 |
8.4 |
8.9 |
9.4 |
| suitable indicator |
bromothymol blue
(pKIn = 7.1 ± 1) |
phenolphthalein
(pKIn = 9.4 ± 1) |
phenolphthalein
(pKIn = 9.4 ± 1) |
phenolphthalein
(pKIn = 9.4 ± 1)
or thymolphthalein
(pKIn = 10.0 ± 1) |
Finally, we consider the indigo line on the titration curve above.
In this region of the titration curve there is an excess of sodium hydroxide solution, and because sodium hydroxide is a strong base that completely dissociates in water, we can ignore the contribution to the hydroxide ion concentration contributed by the hydrolysis of acetate ions (Kb is very small!) and we can ignore the contribution to the hydroxide ion concentration as a result of the dissociation of water molecules (Kw is also very small).
As we add more NaOH(aq) past the equivalence point for the neutralisation reaction, the moles of OH-(aq) in this added volume of solution are all in excess, so the pH of the solution increases, but as we keep adding more and more NaOH(aq) the line levels off as it approaches the pH of the NaOH(aq) being added (for 0.200 mol L-1 NaOH(aq) pH = 13.3).
There is one more region of interest on this titration curve that we have not yet considered.
When half the acetic acid has been neutralised then the concentration of the excess acetic acid is the same as the concentration of acetate ions in solution, and this leads to an interesting position:
| Ka |
= |
[CH3COO-(aq)][H+(aq)]
[CH3COOH(aq)] |
when [CH3COO-(aq)] = [CH3COOH(aq)] then
and if we take the negative log of both sides, then
pKa = pH
So, a titration curve can be used to determine the pKa (an hence Ka) of a weak acid.
For the titration of our 10.00 mL of 0.200 mol L-1 CH3COOH(aq) with 0.200 mol L-1 NaOH(aq), the equivalence point occurred when 10.00 mL of NaOH(aq) had been added.
Therefore, half the acid had been neutralised when half this volume of base had been added, that is, when 5.00 mL of NaOH(aq) had been added.
Let's add a blue line to the titration curve to represent the addition of 5.00 mL of NaOH(aq), and then read off the pH (see the dark red line ):
| pH of solution |
0.200 M CH3COOH - 0.200 NaOH Titration Curve
volume of NaOH(aq) added (mL)
|
Half the acetic acid has been neutralised when 5.00 mL of sodium hydroxide has been added, and the pH = 4.74
Therefore the value for pKa for acetic acid is 4.74
We can therefore calculate the value for the acid dissociation constant, Ka, for acetic acid:
Ka = 10-pKa = 10-4.74 = 1.8×10-5
Note that this pH is also the pH of the most effective buffer solution, when [CH3COO-(aq)] = [CH3COOH(aq)]
How to Sketch a Weak Acid - Strong Base Titration Curve
If you need to sketch a rough titration curve for a weak acid - strong base titration, you will need to perform 4 calculations to locate 4 key features of the titration curve:
(1) initial pH of weak acid before base is added
(2) pH of the salt solution at the equivalence point
(3) pH of the most effective buffer solution (pH = pKa)
(4) pH of the solution when strong base is in excess (approximately approaches pH of the strong base solution).
Formic acid (methanoic acid), HCOOH(aq), is a weak acid (Ka = 1.80×10-4)
Sodium hydroxide, NaOH(aq), is a strong base.
Let's sketch a curve for the titration in which 0.100 mol L-1 NaOH(aq) is added to a conical flask containing 20.00 mL of 0.200 mol L-1 HCOOH(aq).
Step 1: initial pH of weak acid before base is added
dissociation of formic acid: HCOOH(aq) → H+(aq) + HCOO-(aq) Ka = 1.80×10-4
assume very little formic acid dissociates so [HCOOH(initial)] ≈ [HCOOH(equilibrium)], then
Ka = [H+(aq)]2/[HCOOH(aq)] = 1.80×10-4
[H+(aq)] = √(1.80×10-4 × 0.200) = 6.00×10-3 mol L-1
pH = −log10[H+(aq)] = −log10[6.00×10-3] = 2.22
Step 2: pH of the salt solution at the equivalence point
neutralisation reaction: HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
n(HCOOH available) = n(NaOH added) = n(HCOONa(aq) produced)
n(HCOOH available) = c × V = 0.200 × 20.00/1000 = 4.00 × 10-3 mol
n(HCOONa) = 4.00 × 10-3 mol
n(NaOH added) = 4.00 × 10-3 mol
V(NaOH added) = n(NaOH added)/[NaOH] = (4.00 × 10-3)/0.100 = 0.0400 L
V(total) = V(HCOOH) + V(NaOH) = 0.0200 + 0.0400 = 0.0600 L
[HCOONa] = n(HCOONa)/V(total) = (4.00 × 10-3)/0.0600 = 0.067 mol L-1
Hydrolysis of HCOO- (Na+ does not hydrolyse)
HCOO-(aq) + H2O(l) ⇋ HCOOH(aq) + OH-(aq) Kb = Kw/Ka = (1.00×10-14)/(1.80×10-4) = 5.56×10-11 (25°C)
Kb is small so very little HCOO-(aq) hydrolyses:
[OH-(aq)] = √Kb[HCOO-(aq)] = √(5.56×10-11 × 0.067) = 1.93×10-6 mol L-1
pOH = −log10[OH-(aq)] = −log10[1.93×10-6] = 5.71
pH = 14.00 − pOH = 14.00 − 5.71 = 8.29
Step 3: pH of the most effective buffer solution (pH = pKa)
pH = pKa = −log10Ka = −log10(1.80×10-4) = 3.74
which occurs when half the acid has been neutralised, so the volume of NaOH(aq) added = half the volume at neutralisation:
V(NaOH(aq)added) = ½ × 0.0400 L = 0.0200 L
Step 4: pH of the solution when strong base is in excess (approximately approaches pH of the strong base solution)
At the equivalence point, 0.0400 L (40.00 mL) of NaOH(aq) had been added, if we add 1 more mL (0.001 L) then:
n(NaOH in excess) = c × V(excess) = 0.100 × 0.001 = 1.00 × 10-4 mol
V(total) = V(acid) + V(all base) = 0.0200 + 0.041 = 0.061 L
[NaOH final] = n(NaOH in excess)/V(total) = (1.00 × 10-4)/0.061 = 1.64×10-3 mol L-1
[NaOH final] = [OH-(aq)] = 1.64×10-3 mol L-1
pOH = −log10[OH-(aq)] = −log10[1.64×10-3] = 2.79
pH = 14.00 − pOH = 14.00 − 2.79 = 11.2
Note that the line will have a gentle slope of increasing pH that cannot exceed 14 −log[0.100] = 13
Using these 4 points we could sketch a rough titration curve as shown below:
| pH of solution |
Sketch of Titration Curve
volume of NaOH(aq) added (mL)
|
Footnotes:
(1) If we don't assume that x is negligible then we must solve the quadratic equation to determine the value of x
| Ka |
= |
[CH3COO-][H+]
[CH3COOH] |
| 1.8 × 10-5 |
= |
[x][x]
[0.200 − x] |
| 1.8 × 10-5 |
= |
x2
[0.200 − x] |
| (1.8 × 10-5) × [0.200 − x] |
= |
x2 |
| (3.60 × 10-6) − (1.8 × 10-5)x |
= |
x2 |
| 0 |
= |
x2 + (1.8 × 10-5)x − (3.60 × 10-6) |
| x = |
-b ± √(b2 − 4ac)
2a |
| x = |
-(1.8 × 10-5) ± √([1.8 × 10-5]2 − (4 × 1 × −3.60 × 10-6)
2 × 1 |
| x = |
-(1.8 × 10-5) ± √(1.44 × 10-5)
2 × 1 |
| x must be positive so |
| x = |
-(1.8 × 10-5) + √(1.44 × 10-5)
2 × 1 |
| x = |
-(1.8 × 10-5) + (3.79 × 10-3)
2 × 1 |
| x = |
1.89 × 10-3 |
[CH
3COOH] = 0.200 − (1.89 × 10
-3) = 0.198 mol L
-1 (compared to 0.200 mol L
-1 using the assumption that
x is negligible).
pH = −log
10[H
+] = −log
10[1.89 × 10
-3] = 2.72 (compared to pH = 2.72 using the assumption that
x is negligible).
