Wednesday 21 August 2013

Metathesis reactions in chemistry


Metathesis Reactions
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 ® [Na(H2O)6]+ + [Cl(H2O)6]-
AgNO3(s) + 12 H2O ® [Ag(H2O)6]+ + [NO3(H2O)6]-
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 Olifin Metathesis. Metathesis reaction is a type of chemical reactions, which include combination, decomposition, and displacement.

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, aprecipitate 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).

Gravimetric Analysis

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.
Example 1:
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.

Hint -
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
he sample is 0.2579 g / 1.0000 g * 100 % = 25.79 % MgSO4.
T
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 study purposes.

Skill learned
Perform quantitative analysis is an important skill, and this link gives the procedures.
Example 2:
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, 163.4.

Hint -
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.
163.4 g
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 into moles
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 %

Skill learned:
Determine the weight and mole percentages of a mixture.

Confidence Building Problems

  1. What is the product when solids of AgNO3 and NaCl are mixed? 

    Skill:
    Solids do not react until moisture is present.
  2. What is the product when solutions of AgNO3 and NaCl are mixed? 

    Skill:
    Metathesis reaction takes place in solution!
  3. An impure AgNO3 sample weighing 1.00 g dissolving in water is treated with NaCl to give 0.600 g AgCl. Calculate the percentage of AgNO3 in the sample. 
    1 mol AgCl 1 mol AgNO3 169.9 g AgNO3
    0.600 g AgCl ------------ ----------- ------------- = ? g AgNO3
    143.4 g AgCl 1 mol AgCl 1 mol AgNO3

    Skill:
    Determine the percentage of an impure substance.
  4. Is there any reaction between AgNO3 and NaNO3 solution? 
    The resulting solution consists of Ag+, Na+, and NO3- ions.

    Skill:
    Explain the species of an electrolyte.
  5. When solutions of H2SO4 and NaCl are mixed, what is evolved in the vapour? Give the formula 
    HCl has a much higher vapour pressure than H2SO4.

    Skill:
    Use this reaction to make HCl.
  6. A 1.140 g mixture of NaCl and CaCl2 dissolved in water is mixed with sufficient solution of AgNO3 to give 2.868 g of dry AgCl. Calculate the WEIGHT percentage of NaCl? Use two significant digits. (Na, 23.0; Cl, 35.5; Ca, 40; Ag, 107.9) 
    2.868 g AgCl = 0.0200 mol (Cl- or AgCl)
    Assume x g NaCl, then you have (1.140 - x) g of CaCl2.
    The equation:

    # mol NaCl + 2 # mol CaCl2 = # mol Cl-
    ead to
    l
    x g NaCl 2 (1.140 - x) CaCl2
    ------------- + -------------------- = 0.0200 mol
    - 58.5 g /mol 111.1 g/mol
    Solve for x = ? g


    Skill:
    Determine the percentage of a mixture by one measurement.
  7. A 1.140 g mixture of NaCl and CaCl2 dissolved in water is mixed with sufficient solution of AgNO3 to give 2.868 g of dry AgCl. Calculate the MOLE percentage of NaCl? (Na, 23.0; Cl, 35.5; Ca, 40; Ag, 107.9) 

    Skill:
    Determine the mole percentage of a mixture.
  8. When 0.10 mol each of NaCl and CaCl2 dissolved in water is treated with AgNO3, how many mole of AgCl should be collected? 

    Skill:
    Apply the limiting reagent concept in chemical analysis?
  9. What is the weight of 0.30 mol AgCl? (formula wt. AgCl, 163.4) 

    Skill:
    Use AgNO3 as a reagent for gravimetric analysis for chloride?
    © CChieh@UWaterloo.ca

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