Precipitation Titrations Chemistry Tutorial

Key Concepts

• A precipitate is an insoluble solid formed when solutions are mixed.
• Silver nitrate (AgNO_3(aq)) can be used to precipitate out halide ions in an aqueous solution⁽¹⁾:

  Ag⁺_(aq) + Cl^- → AgCl_(s)

  Ag⁺_(aq) + Br^- → AgBr_(s)

  Ag⁺_(aq) + I^- → AgI_(s)

• In a precipitation titration using silver nitrate, a suitable indicator is used to indicate the end point of the titration by changing colour when there is excess Ag⁺ in solution.⁽²⁾

  Indicator	Colour Change (Ag+ limiting to excess Ag+)	Use
  potassium chromate (K2CrO4(aq))	lemon-yellow to reddish-brown	Ag+(aq) + Cl-(aq) → AgCl(s) Ag+(aq) + Br-(aq) → AgBr(s)
  fluorescein	greenish to reddish	Ag+(aq) + Cl-(aq) → AgCl(s)
  eosin	greenish to magenta	Ag+(aq) + Br-(aq) → AgBr(s) Ag+(aq) + I-(aq) → AgI(s)

• The results of a precipitation titration experiment can be used to determine the concentration of halide ions in water samples as shown in the steps below:

1. For the precipitation reaction: Ag⁺_(aq) + X^-_(aq) → AgX_(s)
2. Volume of AgNO_3(aq) added is recorded (titre) and the average titre calculated (in L).

   V(AgNO_3(aq)) = average titre in L

3. Moles of AgNO_3(aq) added is calculated using known concentration and average titre:

   n(AgNO_3(aq)) = c(AgNO_3(aq)) × V(AgNO_3(aq))

4. Moles of halide ion (X^-_(aq)) present in solution is calculated using stoichiometric ratio (mole ratio):

   n(X^-_(aq)) = n(Ag⁺_(aq)) = n(AgNO_3(aq))

5. Concentration of halide ion, [X^-_(aq)], is calculated using known volume of solution containing halide ion (in L):

   [X^-_(aq)] = n(X^-_(aq)) ÷ V(X^-_(aq))

• A precipitation titration curve can also be used to determine volume of titrant required for complete reaction with the halide ion solution.
• A precipitation titration can be used to determine the concentration of chloride ions in water samples, in seawater for example.⁽³⁾

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Precipitation Titration Curves

From the solubility rules we know that if we add an aqueous solution of silver nitrate, AgNO_3(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:

The solubility product, K_sp, for the dissociation of silver chloride into its ions is very, very, small:

AgCl_(s) ⇋ Ag⁺_(aq) + Cl^-_(aq)     K_sp = 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 AgNO_3(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 AgNO_3(aq) from the burette to the NaCl_(aq) in the conical flask, then we can calculate:

1. moles of AgNO_3(aq) added:

   n(AgNO_3(aq)) = c(AgNO_3(aq)) × V(AgNO_3(aq))

   n(AgNO_3(aq)) = moles of AgNO_3(aq)

   c(AgNO_3(aq)) = concentration of AgNO_3(aq) in mol L^-1 = 0.100 mol L^-1

   V(AgNO_3(aq)) = volume of AgNO_3(aq) in L = 1.0 mL = 1.0 mL ÷ 1000 mL/L = 0.0010 L

   n(AgNO_3(aq)) = 0.100 mol L^-1 × 0.0010 L = 0.00010 mol

2. moles of NaCl_(aq) reacted:

   AgNO_3(aq) + NaCl_(aq) → AgCl_(s) + NaNO_3(aq)

   stoichiometric ratio (mole ratio) AgNO_3(aq) : NaCl_(aq) is 1:1

   n(NaCl_(aq) reacted) = n(AgNO_3(aq) added) = 0.00010 mol

3. 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

4. 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.

When we add 5.0 mL of AgNO_3(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, K_sp = 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:

K_sp = [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:

After the equivalence point the Ag⁺_(aq) from further additions of AgNO_3(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 (K_sp = 1.7 × 10^-10 at 25°C):

For example, if we add 1.0 mL more of the AgNO_3(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 AgNO_3(aq), then we can calculate:

1. moles of AgNO_3(aq) added:

   n(AgNO_3(aq)) = c(AgNO_3(aq)) × V(AgNO_3(aq))

   n(AgNO_3(aq)) = moles of AgNO_3(aq)

   c(AgNO_3(aq)) = concentration of AgNO_3(aq) in mol L^-1 = 0.100 mol L^-1

   V(AgNO_3(aq)) = volume of AgNO_3(aq) in L = 6.0 mL = 6.0 mL ÷ 1000 mL/L = 0.0060 L

   n(AgNO_3(aq)) = 0.100 mol L^-1 × 0.0060 L = 0.00060 mol

2. moles of Ag⁺_(aq) in excess:

   n(AgNO_3(aq) excess) = n(AgNO_3(aq) available) - n(AgNO_3(aq) reacted)

   n(AgNO_3(aq) available) = 0.00060 mol (see above)

   n(AgNO_3(aq) reacted) = n(Cl^-_(aq) initial) = 0.00050 mol (see first section)

   n(AgNO_3(aq) excess) = 0.00060 - 0.00050 = 0.00010 mol

3. concentration of excess Ag⁺_(aq) in solution:

   c(AgNO_3(aq) excess) = n(AgNO_3(aq) excess) ÷ V(total)

   n(AgNO_3(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(AgNO_3(aq) excess) = 0.00010 mol ÷ 0.0160 L = 0.00625 mol L^-1

4. concentration of Cl^-_(aq) in solution as a result of the dissociation of the AgCl_(s) precipitate:

   c(Cl^-_(aq)) = K_sp ÷ [Ag⁺_(aq)]

   K_sp = 1.7 × 10^-10 (25°C)

   [Ag⁺_(aq)] = c(AgNO_3(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 AgNO_3(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.

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 = -log₁₀K_c

Then pCl = -log₁₀[Cl^-_(aq)]

The table below shows the results of these calculations:

We can draw a graph of these results as shown below:

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:

Reading off the graph, the equivalence point for this precipitation titration occurs when 5.0 mL AgNO_3(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, K₂Cr₂O_4(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, Ag₂CrO_4(aq), as shown in the equation below:⁽⁵⁾

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 AgNO_3(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 AgNO_3(aq) is recorded (titre).
The experiment is repeated until 3 concordant titres are obtained.

The results are tabulated, and the average titre calculated:

The value of the average titre in L, V_(av), and the known concentration of the AgNO_3(aq) in mol L^-1 is used to calculate the moles of AgNO_3(aq) used to precipitate all the Cl^-_(aq) as AgCl_(s):

n(AgNO_3(aq)) = c(AgNO_3(aq)) × V_(av)

Using the balanced chemical equation:

AgNO_3(aq) + NaCl_(aq) → AgCl_(s) + NaNO_3(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(AgNO_3(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))

Worked Example: Determining the Concentration of Chloride Ions in Seawater Using a Precipitation Titration

The problem:

Chris the Chemist has been asked to determine the concentration of chloride ions in a sample of seawater.

First Chris filters the seawater sample to remove any solid impurities.

Then Chris dilutes the filtered seawater by pipetting 20.00 mL of the sample into a 100.0 mL volumetric flask, then filling it up to the mark with de-ionised water.⁽⁶⁾

Next Chris pipettes 10.00 mL of this diluted solution into a 250 mL conical flask and adds about 50 mL of de-ionised water and 1 mL of K₂CrO_4(aq) indicator.

Chris fills a 50.00 mL burette with 0.100 mol L^-1 standardised AgNO_3(aq) ⁽⁷⁾

Chris slowly adds AgNO_3(aq) to the diluted seawater in the flask until the first permanent red-brown colour emerges.
This titration is repeated several times.
The average titre was calculated and found to be 9.62 mL

What is the concentration of chloride ions in seawater in mol L^-1 ?

The solution: (based on the StoPGoPS approach to problem solving)

1. What is the question asking you to do?

   Calculate the concentration of chloride ions in seawater in mol L^-1

   c(Cl^-_(aq)) = ? mol L^-1

2. What information (data) has been given in the question?

   Dilution of original seawater sample:

   V_(i) = 20.00 mL = 20.00 mL ÷ 1000 mL/L = 0.02000 L

   V_(f) = 100.00 mL = 100.00 mL ÷ 1000 mL/L = 0.1000 L

   Titration results:

   V(Cl^-_(aq)) = 10.00 mL = 10.00 mL ÷ 1000 mL/L = 0.01000 L

   c(AgNO_3(aq)) = 0.100 mol L^-1

   V(AgNO_3(aq)) = 9.62 mL = 9.62 mL ÷ 1000 mL/L = 0.0096200 L

   NOTE: the addition of more water to the flask AFTER the 10.00 mL of seawater was added to it does NOT change the moles of chloride ion in solution so we are ignoring it.

3. What is the relationship between the information you have been given and what you need to calculate?

   Titration results (concentration of Cl^-_(aq) in 10 mL of diluted sample):

   Ag⁺_(aq) + Cl^-_(aq) → AgCl_(s)

   At the equivalence point: n(Cl^-_(aq)) = n(Ag⁺_(aq)) = n(AgNO_3(aq))

   n(Cl^-_(aq) diluted) = n(AgNO_3(aq)) = c(AgNO_3(aq)) × V(AgNO_3(aq))

   c(Cl^-_(aq) diluted) = n(Cl^-_(aq) diluted) ÷ V(Cl^-_(aq))

   Concentration of Cl^-_(aq) in original seawater sample:

   c_i × V_i = c_f × V_f

   c_f = c(Cl^-_(aq) diluted) mol L^-1

   V_f = V(sample after dilution) = 0.1000 L

   V_i = V(sample before dilution) = 0.02000 L

   c_i = c(undiluted sample) = ? mol L^-1

4. Substitute the known values into the equations and solve for c(Cl^-_{(undilted sample)}):

   Titration results (concentration of Cl^-_(aq) in 10 mL of diluted sample):

   Ag⁺_(aq) + Cl^-_(aq) → AgCl_(s)

   At the equivalence point: n(Cl^-_(aq)) = n(Ag⁺_(aq)) = n(AgNO_3(aq))

   n(Cl^-_(aq) diluted) = n(AgNO_3(aq)) = c(AgNO_3(aq)) × V(AgNO_3(aq)) = 0.100 × 0.0096200 = 9.620 × 10^-4 mol

   c(Cl^-_(aq) diluted) = n(Cl^-_(aq) diluted) ÷ V(Cl^-_(aq)) = 9.620 × 10^-4 mol ÷ 0.01000 L = 0.0962 mol L^-1

   Concentration of Cl^-_(aq) in original seawater sample:

   c_i × V_i = c_f × V_f

   c_f = c(Cl^-_(aq) diluted) mol L^-1 = 0.0962 mol L^-1

   V_f = V(sample after dilution) = 0.1000 L

   V_i = V(sample before dilution) = 0.02000 L

   c_i = c(undiluted sample) = ? mol L^-1

   c_i × 0.02000 L = 0.0962 mol L^-1 × 0.1000 L

   c_i × 0.02000 L = 0.00962 mol

   c_i = 0.00962 mol ÷ 0.02000 L = 0.481 mol L^-1

5. Is your answer plausible?

   Have we answered the question that was asked?
   Yes, we have calculated the concentration of chloride ions in the original, undiluted, sample of seawater.

   Work backwards: use our c(Cl^-_(aq)) to determine the volume of AgNO_3(aq) required to precipitate out all the chloride ion in seawater.
   c(Cl^-_(aq) in seawater) = 0.481 mol L^-1
   n(Cl^-_(aq) in 20.00 mL seawater) = c × V = 0.481 × 0.02 = 0.00962 mol
   c(Cl^-_(aq) in diluted sample) = n ÷ V = 0.00962 ÷ 0.100 L = 0.0962 M
   n(Cl^-_(aq) in diluted sample in 10 mL aliquote) = c × V = 0.0962 × 0.01 L = 0.000962 mol
   n(AgNO_3(aq)) for complete reaction = n(Cl^-_(aq)) = 0.000962 mol
   V(AgNO_3(aq)) = n(AgNO_3(aq)) ÷ c(AgNO_3(aq)) = 0.000962 mol ÷ 0.100 M = 0.00962 L = 9.62 mL
   Since this value agrees with average titre given in the question we are confident our value for the concentration of chloride ions in undilted seawater is correct.

6. State your solution to the problem: concentration of chloride ions in seawater

   [Cl^-_(seawater)] = 0.481 mol L^-1