metathesis, precipitation, neutralization, gas formation
gravimetric and volumetric analyses
Skills to develop
- Explain and identify metathesis reactions
- Apply metathesis reactions in gravimetric analysis and volumetric analysis
- Apply metathesis reactions for making chemicals
A reaction such as
NaCl(aq) + AgNO3(aq)
= AgCl(s) + NaNO3(aq)
in which the cations and anions exchange partners is called
metathesis. In actual fact, the chemistry takes place in several steps.
When the chemicals (sodium chloride and silver nitrate) are dissolved,
they become hydrated ions:
NaCl(s) + 12 H2O ®
AgNO3(s) + 12 H2O ®
When the silver ions and chloride ions meet in solution, they combine and
form a solid, which appears as a white precipitate:
Ag+(H2O)6 + + [Cl(H2O)6]-
® AgCl(s) + 12 H2O
The above equation shows the net ionic reaction, whereas the
bystander ions Na+ and NO3- are not
shown. Bystander ions are also called spectator ions.
A sound movie is available from the Journal of Chemical Education page on
sodium chloride and silver nitrate reactions. This movie plays fine
on Polaris computers, but ear phones are required for sound movies.
Metathesis reactions not only take place among ionic compounds, they
occur among other compounds such as
Sigma Bond Metathesis and
Metathesis reaction is a type of
chemical reactions, which include combination, decomposition, and
Types of metathesis reactions
What happens when you pour two solutions of different electrolytes together?
The mixture will have all ions from the two electrolytes. Ions of the same
charge usually repel each other, but ions of opposite charge may form a
stable molecule or solid. When a solid is formed such as AgCl, a
precipitate is formed.
From the observation point of view, metathesis reactions
can be further divided into three classes:
Precipitation reaction: products formed are not soluble, forming
solids which we call precipitates. The solid silver chloride AgCl(s)
mentioned above is a precipitate. Since the solid can be collected
and dried, precipitation reactions are often used in gravimetric
analysis, chemical analysis by mass or weight.
Neutralization reaction: Products formed are neutral water
molecules, and the net ionic reaction is actually
H+ + OH- = H2O.
With proper indicators or pH monitoring, equivalence points are easily
detected. Thus, neutralization reactions are used for volumetric analysis,
quantitative determination by volume measurement.
Gas formation reaction: methesis reaction may lead to the formation
of a neutral molecule that has low boiling point as well as low solubility
in water. Thus, a gas is formed. For example:
2 H+ + CO32- = H2O + CO2(g)
Why do ions exchange partners?
Cations are always attracted to anions, but the hydration and hydrogen
bonding keep the ions of electrolytes in solution. When two solutions
are mixed, cations of one electrolyte meat anions of the other. If they
form a more stable substance such as a solid or neutral molecules,
exchange or metathesis reaction takes place.
The new couples form a precipitation, gas, or neutral molecules.
These reactions can be employed for gravimetric or volumetric analysis
(determine the quantities present in a sample).
What substances are soluble?
You have to work with these materials to know them well.
Here are two basic rules regarding solubility:
Most nitrates are soluble. So are alkali and ammonium halides.
Most carbonates, phosphates, sulfites, sulfides, Ca(OH)2,
and AgCl are some of the substances that are only sparingly soluble
(less than 0.1 g per 100-mL water).
The quantitative determination of a component by measuring the mass
of a compound formed with the component using a chemical reaction is
called gravimetric analysis. Some examples are given here to show
how gravimetric analysis are carried out.
To determine % of MgSO4 in epsom salts, you treat it with
BaCl2, because of the following reaction:
MgSO4 + BaCl2 --> MgCl2 + BaSO4(s)
You can dry the substance BaSO4 formed and weigh the resulting
solid to determine the quantity of MgSO4 (mol. wt. 120.37) formed.
Suppose you started with 1.0000 g of epsom salts, and got 0.5000 g of
BaSO4 (mol. wt. 233.39). Calculate the percentage of
MgSO4 in the sample.
Use the following one-line method to do the conversion quickly.
1 mol BaSO4 1 mol MgSO4 120.37 g MgSO4
0.500 g BaSO4 -------------- ----------- --------------
233.39 g BaSO4 1 mol BaSO4 1 mol MgSO4
= 0.2579 g MgSO4
The sample is 0.2579 g / 1.0000 g * 100 % = 25.79 % MgSO4.
The numerators and denominators of the factors are equivalent under
the condition of the problem. Thus, these are conversion factors, and the
factors convert the weight of BaSO4 to that of MgSO4.
Note the strategy of the analysis, and the methods of calculation for
quantitative analysis is an important skill, and this link gives
A sample weighing 3.77 g containing CaCl2 and AlCl3
dissolved in water was treated with AgNO3, and the dry AgCl
collected weighs 13.07 g. Calculate the weight and mole percentages of
CaCl2 in the sample.
Formula wt: CaCl2, 111.1; AlCl3, 133.5; AgCl,
Since both compounds contain Cl-, this problem required
some thinking. Consider all quantities in moles.
1 mol AgCl
13.07 g AgCl ---------- = 0.080 mol AgCl or Cl in the sample.
Here is a place for the application of the skills learned in algebra.
You can assume x be the weight (g) of CaCl2,
then (3.77 - x) g must be AlCl3. Converting these into moles, and
the sum of the moles of Cl- ions from both salt must equal to
the (0.080) moles observed. Thus, we have
x g CaCl2 2 mol Cl (3.77 - x) g AlCl3 3 mol Cl
----------------- ----------- + ------------------ -----------
111.1 g/mol CaCl2 1 mol CaCl2 133.5 g/mol AlCl3 1 mol AlCl3
= 0.080 mol Cl-
Simplifying the above equation to give
0.0180 x + 0.0847 - 0.0225 x = 0.080
The solution gives
x = 1.04 g CaCl2, and
3.77 - 1.04 = 2.73 g AlCl3.
By definition, the weight percent of CaCl2 = 1.04/3.77 = 27.59 %
In order to calculate mole percent, the quantities are converted
1 mol CaCl2
1.04 g CaCl2 -------------- = 0.0094 mol CaCl2
111.1 g CaCl2
2.73 g / (133.5 g/mol) = 0.0204 mol AlCl3
Thus, the mole percentage of CaCl2 = 0.0094 / (0.0294) = 31.5 %
Determine the weight and mole percentages of a mixture.
Confidence Building Problems