Lecture 14 and 15
Chem 2210
Chapter 11: Precipitation Titrations
- Precipitation titrations are based upon reactions that yield ionic compounds of limited solubility.
- The most important precipitating reagent is silver nitrate (AgNO3).
1. Solubility Equilibrium
-
Precipitation: the formation of a solid from solution. The reverse of dissolution.
Dissolution: the process by which a substance dissolves. The reverse of precipitation.
Saturated Solution: one in which a dissolution - precipitation equilibrium exist between a solid
and its dissolved form.
Unsaturated Solution: one in which the concentration of dissolved solid is not sufficient to cause
precipitation.
1.1. Solubility (S)
The quantity of solute that dissolves in a given quantity of solvent at a particular temperature.
Solubility is often expressed as the mass of solute per volume (g/L) or mass of solute per mass of
solvent (g/g), or as the moles of solute per volume (mol/L: molar solubility). Even for very soluble
substances, however, there is usually a limit to how much solute can dissolve in a given quantity of
solvent. In general, the solubility of a substance depends on the temperature, the nature of solute or
solvent, and the pH.
1.2. Solubility Product Constant (Ksp)
The Solubility Product Constant, Ksp is the equilibrium constant for a solid substance dissolving
in an aqueous solution. It represents the level at which a solute dissolves in solution. A more a substance
dissolves, the higher the Ksp value it has.
For example:
To solve for the Ksp it is necessary to take the concentrations of the products (cC and dD) and multiply
them. If there are coefficients in front of any of the products, it is necessary to raise the product to that
coefficient power (and also multiply the concentration by that coefficient). This is shown here:
[ ]
[ ]
Please note that the reactant, aA, is not included in the Ksp equation because it is a solid. Solids are not
included when calculating any type of equilibrium constant expressions but must be somewhat present
in order for equilibrium to be established. The reason for not including solids is because since they are
solid, their concentration does not change the expression; any change in their concentration is
insignificant therefore left out.
Ksp represents the maximum amount of solid that can be dissolved in the aqueous solution.
Mathematically we can write Ksp expressions using molar solubility, S, in place of individual ion
concentrations. If we define S as the molar solubility of AgCl, then S moles of this salt dissolve in one
]
[
]
liter of solution. In a saturated solution of pure AgCl, [
and Ksp can be rewritten as:
[
and
]
[
]
√
Example 11.1
Calculate the solubility of AgCl by g/L if the Ksp of AgCl is 10-10. Where, Mwt(AgCl) = 143.5 g/mol
Example 11.2
Calculate the Ksp of Ag2CrO4 if the solubility of Ag2CrO4 = 2.5 x 10-2 g/L. Where, Mwt(Ag2CrO4) = 322
g/mol
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Lecture 14 and 15
Chem 2210
2. Precipitation Titrations
Precipitation titrations are volumetric methods based on the formation ionic compounds of
limited solubility. So the titrant, T, and the analyte, X, react to form an insoluble salt:
The most widely used and important precipitating reagent, silver nitrate (AgNO3), which is used for the
determination of the halogens, the halogen-like anions. Titrations with silver nitrate are sometimes
called argentometric titrations. The equivalence point is reached when an equivalent amount of the
titrant has been added. From the volume of the latter, the amount of the substance is calculated. The
precipitate must be sufficiently insoluble to ensure completion of the reaction and to ensure a marked
change in the concentration of the ions of precipitate at the equivalence point of the titration.
2.1. Titration Curves
Titration curves for precipitation reactions are derived in a completely analogous way to the
methods described for titrations involving strong acids and strong bases. p-functions are derived for
the pre-equivalence point region, the poste-quivalence point region, and the equivalence point for a
typical precipitation titraton.
Most indicators for argentometric titrations respond to changes in the concentration of silver
ions. As a consequence, titraton curves for precipitation reactions usually consist of a plot of pX or pT
versus volume of AgNO3.
It is convenient to define a general p-function as
is analogous to pH, which you of course already know as
[
[ ], or
].
[ ]. This
Let's consider the titration of an aqueous chloride solution with standard silver nitrate solution.
The titration reaction is:
You will recognize that this reaction is just the dissolution of solid AgCl written backwards. Hence, the
equilibrium constant, K, for the titration reaction is just the inverse of the solubility product, Ksp, for
AgCl, which has the value Ksp = 1.8 x 10-10. Hence, the equilibrium constant for the titration reaction is:
Very large equilibrium constants such as this are essential for the success of any titration.
As an example, let's consider the titration of 50 mL of 0.0125 M Cl- solution with 0.025 M Ag+.
Our object to is calculate pCl, as functions of the volume of added titrant (
).
Initial region
Note that prior to the addition of any titrant, there can be no Ag+ in solution, so that
[
]
[
]
Once we begin to add titrant, there are three different regions of the titration that must be
considered. These are:
i) Prior to the equivalence point
ii) At the equivalence point
iii) Past the equivalence point
The calculations will depend on what region of the titration we are in. As such, it is convenient to begin
by determining what the equivalence point volume, Veq, is. In this case, we have one-to-one
stoichiometry, so the equivalence point volume is defined as that volume of 0.025 M Ag+ which contains
exactly the same number of moles of Ag+ as the moles of Cl- in the 50 mL of 0.0125 M Cl- we are
titrating. Hence,
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Lecture 14 and 15
Chem 2210
Pre-Equivalence region:
Now let's consider the addition of 10 mL of titrant. This is in the region prior to the
equivalence. At this point we have added 10 x 10-3 x 0.025 = 0.25 x 10-3 moles of Ag+. We'll start by
considering that every Ag+ ion we have added precipitates one Cl- ion. We started with 50 x 10-3 x
0.0125 = 0.625 x 10-3 moles of Cl-. After the addition of 0.25 x 10-3 moles of Ag+, we will have 0.625 0.25 = 0.375 x 10-3 moles of Cl- remaining. The total volume is now the 10 mL we added plus the 50
mL we started with, or 60 mL, so
[
]
[
]
[
where,
]
[
]
Other pre-equivalence points can be determined in a similar way!
Equivalence:
Now consider the equivalence point itself. In this case, consider that all of the chloride and all of
the silver have precipitated, and then let the solid AgCl redissolve to reach equilibrium. In that case,
[
]
[
]
√
√
[
and
]
[
]
Post-Equivalence Region:
After the equivalence point, the titrant (Ag+) is in excess. We first calculate the concentration
of excess Ag+ and then use the Ksp expression to calculate the concentration of Cl–. For example, after
adding 35 mL of titrant:
[
]
[
]
[
and
]
[
]
[
]
[
]
Other post-equivalence points can be determined in a similar way!
Plotting values of pCl versus volume of titrant throughout the course of the titration yields the
following:
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Lecture 14 and 15
Chem 2210
pCl
4.872
Equivalence point
1.903
Veq = 25 mL
Volume AgNO3
Added (mL)
Figure 1: Titration curve for 50 mL of 0.0125 M KCl versus 0.025 M AgNO3
Precipitation titrations also can be extended to the analysis of mixtures provided that there is a
significant difference in the solubilities of the precipitates. Figure 2 shows an example of a titration
curve for a mixture of I– and Cl– using Ag+ as a titrant. Note that the equivalence point for I– is earlier
than the equivalence point for Cl– because AgI (Ksp = 10-16) is less soluble than AgCl (Ksp = 1.8 x 10-10).
pX
Equivalence point
for I-
Equivalence point
for Cl-
Volume AgNO3
Added (mL)
Figure 2: Titration curve for the titration of KCl and KI with AgNO3
A unique feature of this curve is that the first equivalence point comes at a “corner” (a cusp in
mathematical terms) formed at the point where the solubility product of AgCl is exceeded. When that
happens, precipitation of AgCl stops the steep rise in [Ag+]. This occurs abruptly as I– is
stoichiometrically exhausted. As shown in Figure 2 the second equivalence point occurs on a
conventional S-shaped portion of the titration curve. This equivalence point, corresponding to
stoichiometric precipitation of Cl– as solid AgCl, occurs at the midpoint of the second break.
A product that is very insoluble (small Ksp) will lead to a titration reaction that is very complete.
Figure 3 shows the effect of solubility on the shape of the titration curve. The largest change in pX
occurs in the titration of iodide ion (I-), which, among the ions considered forms the most insoluble
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Lecture 14 and 15
Chem 2210
silver salt. The smallest change in pX occurs with chloride ion (Cl-), which forms the least insoluble
silver salt. Reactions that produce silver salts with solubilities intermediate between these extremes
yield titration curves with equivalence point changes that are also intermediate in magnitude. This
effect is common to all reaction types.
pX
Equivalence point
for IEquivalence point
for BrEquivalence point
for Cl-
Volume AgNO3
Added (mL)
Figure 3: Effect of solubility on the shape of the titration curve.
Three classical methods, known as the Mohr, Volhard, and Fajans method, utilize color
indicators for the end-point. In general, these three methods involve a silver ion halide. Thus, the
selection of chemical reagents as indicators are for these reactions and cannot be arbitrarily applied to
other precipitation titrations. However, the basis of end-point detection in these three methods,
specifically, formation of a colored precipitate and formation of a color homogeneously or on a surface,
can be applied to other methods.
The Mohr method
Mohr’s method is a direct titration using standard solution of silver nitrate in a neutral medium.
Chloride is titrated with AgNO3 solution. A soluble chromate salt (K2CrO4) is added as the indicator.
This produces a yellow color solution. When the precipitation of the chloride is complete.
At the end point: The first excess of Ag+ reacts with the indicator to precipitate red silver chromate as a
second precipitate after precipitation of all chlorides as silver chloride.
Titrant
Indicator: Yellow
Precipitate: Red
The Mohr’s method must be performed at a pH about 8. This method is useful for determining
Cl- in neutral or unbuffered solutions such as drinking water.
The Volhard method
This is an indirect titration procedure for the determination of anions that precipitate with silver
like Cl-, Br-, I-, SCN-, and it is preferred in acid (HNO3) solution. A measured excess of AgNO3 is added
to precipitate the anion, and the excess of Ag+ is determined by back titration with standard NH4SCN
solution:
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Lecture 14 and 15
Chem 2210
The end point is detected by adding iron III (Fe3+) as ferric ammonium sulfate which forms a soluble
red complex with the first excess of titrant.
[
]
Red
These indicators must not form a compound with the titrant that is more stable than the precipitate or
the color reaction would occur on addition of the first drop of titrant.
The Fajans method
The indicator reaction takes place on the surface of the precipitate. The indicator, which is a dye,
exists in solution as the ionized form, usually an anion.
Iodide can be determined by direct titration with standard silver nitrate using an adsorption
indicator such as eosin, fluorescein or diiodofluorescein.
For example, consider this titration:
Before the equivalent point, I- is in excess and the primary layer is I-. This repulses the indicator anions;
and the more loosely held the secondary (counter) layer of adsorbed ions is cations.
Beyond the equivalent point (end point as well), Ag+ is in excess and the surface of the precipitate
becomes positively charged, with the primary layer being Ag+. This will now attract the indicator
anion and adsorb it in the secondary (counter) layer:
AgI:Ag+ : : indicatorThe color of the adsorbed indicator is different from that of the un-adsorbed indicator, and this
difference signals the completion of the titration.
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