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Precipitation Titration Curves
From the solubility rules we know that if we add an aqueous solution of silver nitrate, AgNO3(aq), to an aqueous solution of sodium chloride, NaCl(aq), a white precipitate of silver chloride, AgCl(s), is produced.
This precipitation reaction can be represented by the following balanced chemical equations:
word equation: |
silver nitrate |
+ |
sodium chloride |
→ |
silver chloride |
+ |
sodium nitrate |
molecular equation: |
AgNO3(aq) |
+ |
NaCl(aq) |
→ |
AgCl(s) |
+ |
NaNO3(aq) |
ionic equation: |
Ag+(aq) + NO3-(aq) |
+ |
Na+(aq) + Cl-(aq) |
→ |
AgCl(s) |
+ |
Na+(aq) + NO3-(aq) |
net ionic equation: |
Ag+(aq) |
+ |
Cl-(aq) |
→ |
AgCl(s) |
|
|
The solubility product, Ksp, for the dissociation of silver chloride into its ions is very, very, small:
AgCl(s) ⇋ Ag+(aq) + Cl-(aq) Ksp = 1.7 × 10-10
So the equilibrium position lies very far to the left of the dissociation equation, that is, the formation of silver chloride from its ions is favoured, in other words, the reverse reaction goes practically to completion as shown in the balanced chemical equation below:
Ag+(aq) + Cl-(aq) → AgCl(s) K = 1/Ksp = 1 ÷ (1.7 × 10-10) = 5.9 × 109
So we could set up a titration experiment using the equipment below to slowly add AgNO3(aq) to NaCl(aq):
|
- Burette holds 0.100 mol L-1 AgNO3(aq)
- Conical flask holds 10.0 mL of 0.0500 mol L-1 NaCl(aq)
- A white precipitate of AgCl(s) will form in the conical flask.
|
Before the experiment begins, the conical flask contains only 10.0 mL of 0.0500 mol L-1 NaCl(aq).
We can calculate the moles of NaCl present in the conical flask:
n(NaCl(aq)) = c(NaCl(aq)) × V(NaCl(aq))
where:
n(NaCl(aq)) = moles of NaCl
c(NaCl(aq)) = concentration of NaCl in mol L-1 = 0.0500 mol L-1
V(NaCl(aq)) = volume of NaCl in L = 10.0 mL = 10.0 mL ÷ 1000 mL/L = 0.0100 L
n(NaCl(aq)) = 0.0500 mol L-1 × 0.0100 L = 0.00050 mol
If we add 1.0 mL of 0.100 mol L-1 AgNO3(aq) from the burette to the NaCl(aq) in the conical flask, then we can calculate:
- moles of AgNO3(aq) added:
n(AgNO3(aq)) = c(AgNO3(aq)) × V(AgNO3(aq))
n(AgNO3(aq)) = moles of AgNO3(aq)
c(AgNO3(aq)) = concentration of AgNO3(aq) in mol L-1 = 0.100 mol L-1
V(AgNO3(aq)) = volume of AgNO3(aq) in L = 1.0 mL = 1.0 mL ÷ 1000 mL/L = 0.0010 L
n(AgNO3(aq)) = 0.100 mol L-1 × 0.0010 L = 0.00010 mol
- moles of NaCl(aq) reacted:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
stoichiometric ratio (mole ratio) AgNO3(aq) : NaCl(aq) is 1:1
n(NaCl(aq) reacted) = n(AgNO3(aq) added) = 0.00010 mol
- moles of Cl-(aq) in excess in the conical flask:
n(Cl-(aq) initial) = n(NaCl(aq)) = 0.00050 mol
n(Cl-(aq) reacted) = 0.00010 mol
n(Cl-(aq) excess) = n(Cl-(aq) initial) - n(Cl-(aq) reacted) = 0.00050 - 0.00010 = 0.00040 mol
- concentration of Cl-(aq) now in the conical flask:
c(Cl-(aq)) = n(Cl-(aq)) ÷ V(Cl-(aq))
n(Cl-(aq)) = 0.00040 mol
V(Cl-(aq)) = 10.0 mL + 1.0 mL = 11 mL = 11 mL ÷ 1000 mL/L = 0.0110 L
c(Cl-(aq)) = 0.00040 mol ÷ 0.0110 L = 0.0364 mol L-1
We can continue these calculations right up until the equivalence point, the point at which all the available Cl-(aq) has reacted with Ag+(aq).
The results are shown in the table below, but you should verify the calculations for yourself.
Volume of AgNO3(aq) added (mL) |
[Cl-(aq)] (mol L-1) |
0.0 |
0.0500 |
1.0 |
0.0364 |
2.0 |
0.0250 |
3.0 |
0.0154 |
4.0 |
0.00714 |
When we add 5.0 mL of AgNO3(aq) to the NaCl(aq) we will have reached the equivalence point of the reaction, the moles of Ag+(aq) we add is exactly the same as the moles of Cl-(aq) in the solution.
When these two solutions are mixed, the resulant solution is saturated, and the precipitate exists in equilibrium with its ions in solution:
AgCl(s) ⇋ Ag+(aq) + Cl-(aq)
We can use the solubility product, Ksp = 1.7 × 10-10, for this reaction to determine the concentration of chloride ions and silver ions in the resultant solution at this point of the reaction:
Ksp = [Ag+(aq)][Cl-(aq)] = 1.7 × 10-10
At the equivalence point:
moles(Ag+(aq)) = moles(Cl-(aq))
All the ions are present in the same volume of solution
Therefore: [Ag+(aq)] = [Cl-(aq)] = x
[Ag+(aq)][Cl-(aq)] = 1.7 × 10-10
x2 = 1.7 × 10-10
√x2 = √(1.7 × 10-10)
x = 1.3 × 10-5
We can add these values to our table:
Volume of AgNO3(aq) added (mL) |
[Cl-(aq)] (mol L-1) |
0.0 |
0.0500 |
1.0 |
0.0364 |
2.0 |
0.0250 |
3.0 |
0.0154 |
4.0 |
0.00714 |
5.0 |
0.000013 |
After the equivalence point the Ag+(aq) from further additions of AgNO3(aq) will be in excess.
This means that the concentration of Ag+(aq) in the resultant solution after mixing will increase, shifting the equilibrium position for the dissociation of AgCl(s) to the left.
We will see the concentration of Cl-(aq) (as a result of the dissociation of AgCl(s)) decrease.
We can determine the concentration of Cl-(aq) that will be in solution as a result of the dissociation of the precipitated AgCl(s) using its solubility product (Ksp = 1.7 × 10-10 at 25°C):
Ksp |
= |
[Ag+(aq)][Cl-(aq)] |
1.7 × 10-10 |
= |
[Ag+(aq)][Cl-(aq)] |
1.7 × 10-10 [Ag+(aq)] |
= |
[Ag+(aq)][Cl-(aq)] [Ag+(aq)] |
1.7 × 10-10 [Ag+(aq)] |
= |
[Cl-(aq)] |
For example, if we add 1.0 mL more of the AgNO3(aq) after the equivalence end point we will have added a total volume of 5.0 + 1.0 mL = 6.0 mL of 0.100 mol L-1 AgNO3(aq), then we can calculate:
- moles of AgNO3(aq) added:
n(AgNO3(aq)) = c(AgNO3(aq)) × V(AgNO3(aq))
n(AgNO3(aq)) = moles of AgNO3(aq)
c(AgNO3(aq)) = concentration of AgNO3(aq) in mol L-1 = 0.100 mol L-1
V(AgNO3(aq)) = volume of AgNO3(aq) in L = 6.0 mL = 6.0 mL ÷ 1000 mL/L = 0.0060 L
n(AgNO3(aq)) = 0.100 mol L-1 × 0.0060 L = 0.00060 mol
- moles of Ag+(aq) in excess:
n(AgNO3(aq) excess) = n(AgNO3(aq) available) - n(AgNO3(aq) reacted)
n(AgNO3(aq) available) = 0.00060 mol (see above)
n(AgNO3(aq) reacted) = n(Cl-(aq) initial) = 0.00050 mol (see first section)
n(AgNO3(aq) excess) = 0.00060 - 0.00050 = 0.00010 mol
- concentration of excess Ag+(aq) in solution:
c(AgNO3(aq) excess) = n(AgNO3(aq) excess) ÷ V(total)
n(AgNO3(aq) excess) = 0.00010 mol
V(total) = 10.0 mL + 6.0 mL = 16.0 mL = 16.0 L ÷ 1000 mL/L = 0.0160 L
c(AgNO3(aq) excess) = 0.00010 mol ÷ 0.0160 L = 0.00625 mol L-1
- concentration of Cl-(aq) in solution as a result of the dissociation of the AgCl(s) precipitate:
c(Cl-(aq)) = Ksp ÷ [Ag+(aq)]
Ksp = 1.7 × 10-10 (25°C)
[Ag+(aq)] = c(AgNO3(aq) excess) = 0.00625 mol L-1
c(Cl-(aq)) = (1.7 × 10-10) ÷ 0.00625 = 2.72 × 10-8 mol L-1
If we continue to add more AgNO3(aq) to the flask, then the moles of excess Ag+(aq) in solution increases, which shifts the equilibrium position to the AgCl(s) side of the chemical equation and the concentration of Cl-(aq) decreases.
The table below shows the results of these calculations. You should verify these calculations for yourself.
Volume of AgNO3(aq) added (mL) |
[Cl-(aq)] (mol L-1) |
0.0 |
0.0500 |
1.0 |
0.0364 |
2.0 |
0.0250 |
3.0 |
0.0154 |
4.0 |
0.00714 |
5.0 |
0.000013 |
6.0 |
2.72 × 10-8 |
7.0 |
1.45 × 10-8 |
8.0 |
1.02 × 10-8 |
In order to draw a titration curve, we are going to change the concentration of Cl-(aq) in mol L-1 to a new term, pCl.
Recall that pH = -log10[H+(aq)]
and pOH = -log10[OH-(aq)]
and pK = -log10Kc
Then pCl = -log10[Cl-(aq)]
The table below shows the results of these calculations:
Volume of AgNO3(aq) added (mL) |
[Cl-(aq)] (mol L-1) |
pCl = -log10[Cl-(aq)] |
0.0 |
0.0500 |
1.30 |
1.0 |
0.0364 |
1.44 |
2.0 |
0.0250 |
1.60 |
3.0 |
0.0154 |
1.81 |
4.0 |
0.00714 |
2.15 |
5.0 |
0.000013 |
4.89 |
6.0 |
2.72 × 10-8 |
7.57 |
7.0 |
1.45 × 10-8 |
7.84 |
8.0 |
1.02 × 10-8 |
7.99 |
We can draw a graph of these results as shown below:
pCl |
Precipitation Titration Curve volume of AgNO3(aq) (mL) |
If you have been given a precipitation titration curve, you can use it to find the equivalence point for the precipitation reaction in the same way that we have previously used an acid-base titration curve to determine the equivalence point of an acid-base titration.
This is shown on the graph below:
pCl |
Precipitation Titration Curve volume of AgNO3(aq) (mL) |
Reading off the graph, the equivalence point for this precipitation titration occurs when 5.0 mL AgNO3(aq) has been added.(4)
But we have a practical problem.
Imagine an experiment in which we need to determine the concentration of chloride ions in a sample of water by adding aqueous silver nitrate solution.
We will be able to visually observe the water sample become milky-white as a result of the formation of AgCl(s), but we won't be able to "see" that we have added just enough silver(1+) ions without adding too many!
Could we use a substance that performs the same function as an acid-base indicator during an acid-base titration?
Indicators for Precipitation Titrations
We are thinking about the following reaction:
Ag+(aq) + Cl-(aq) → AgCl(s)
While the Cl-(aq) is in excess, all the available added Ag+(aq) will be consumed in the formation of the precipitate AgCl(s).
As more Ag+(aq) is added after the equivalence point, there will be excess Ag+(aq) in solution.
We can use an aqueous solution of lemon-yellow potassium chromate, K2Cr2O4(aq), to indicate when the Ag+(aq) is in excess and hence determine the end point of the titration, because of the formation of a reddish-brown precipitate of silver chromate, Ag2CrO4(aq), as shown in the equation below:(5)
Ag+(aq) |
+ |
Cr2O42-(aq) |
→ |
Ag2Cr2O4(s) |
colourless |
|
pale yellow |
|
reddish-brown |
We can set up the precipitation titration experiment as shown below:
|
- Burette holds 0.100 mol L-1 AgNO3(aq)
- Conical flask holds 10.0 mL of 0.0500 mol L-1 NaCl(aq) and 1 mL K2CrO4(aq) (indicator). The solution has a lemon-yellow colour.
- Initially, as Ag+(aq) is added to the solution of Cl-(aq), the resultant solution becomes cloudy, milky-white (perhaps with a pale lemon tinge due to the chromate ions), as the AgCl(s) forms.
- Past the equivalence point, the excess Ag+(aq) reacts with the CrO42-(aq) to produce the reddish-brown precipitate of Ag2CrO4(s), and the mixture in the flask takes on a faint, very slight, reddish-brown hue.
- Overshooting the end point and adding too much Ag+(aq) results in the reddish-brown colour of the mixture in the flask deepening.
|
There are other indicators you could use for this precipitation titration. For example:
- fluorescein: greenish cloudy solution turns reddish at the end point.
- dichlorofluorescein: greenish cloudy solution turns reddish at the end point.
The indicator used will depend on the precipitation reaction and the nature of the ion in excess.
For example, if you want to determine the concentration of bromide ions in an aqueous solution, you could use potassium chromate as an indicator as described above, or could use eosin as an indicator (end point is reached when until the reddish mixture turns magenta).
For example, if you want to determine the concentration of iodide ions in an aqueous solution, you could use eosin as an indicator, or you could use di-iododimethylfluorescein (end point is indicated by a change of colour from orange-red to blue-red).
Precipitation Titration Calculations
For the titration in which AgNO3(aq) is gradually added to NaCl(aq), the equivalence point of the reaction is indicated by the colour change of the indicator used at the end point of the titration.
The volume of AgNO3(aq) is recorded (titre).
The experiment is repeated until 3 concordant titres are obtained.
The results are tabulated, and the average titre calculated:
Trial |
Volume (mL) |
Trial 1 |
V(1) |
Trial 2 |
V(2) |
Trial 3 |
V(3) |
Average Titre = |
V(1) + V(2) + V(3) 3 |
The value of the average titre in L, V(av), and the known concentration of the AgNO3(aq) in mol L-1 is used to calculate the moles of AgNO3(aq) used to precipitate all the Cl-(aq) as AgCl(s):
n(AgNO3(aq)) = c(AgNO3(aq)) × V(av)
Using the balanced chemical equation:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
we can determine the moles of Cl-(aq) that had reacted at the equivalence point of the reaction:
n(Cl-(aq)) = n(NaCl(aq)) = n(AgNO3(aq))
Using the known volume of NaCl(aq) (in L) used in the titration, we can calculate the concentration of Cl-(aq) in the original solution:
c(Cl-(aq)) = n(Cl-(aq)) ÷ V(NaCl(aq))