CHEM1011 Lecture Notes - Lecture 10: Magnesium Chloride, Potassium Chromate, Calcium Chloride
2 Jul 2018
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molarity = moles of solute or moles of solute = molarity × volume (in L)
litres of solution
TOPIC 10.
CHEMICAL CALCULATIONS IV - solution stoichiometry.
Calculations involving solutions.
Frequently reactions occur between species which are present in solution. One type of
chemical analysis called VOLUMETRIC ANALYSIS makes use of the fact that
volumes are easier and faster to measure than mass. Volumetric analysis is done using
specialised glassware in a process called TITRATION.
In order to deduce the amount of a dissolved species (called the SOLUTE) which is
present in a given volume of a solution, it is necessary to know the
CONCENTRATION of the solution. Concentration is most commonly expressed as
how much solute is present per unit volume (e.g. per mL or L) of the solution. The
concentration of a solution is therefore independent of the volume taken and to
calculate the amount of solute in any given volume of solution, the concentration must
be multiplied by that volume. There are many ways of expressing concentrations, for
example % m/v means "the mass of solute in 100 mL of solution". In chemical
calculations, by far the most commonly used concentration unit is the number of moles
of solute present per litre of solution, and this is termed the MOLARITY of the
solution, abbreviated as M.
Thus a 1 molar solution (written as 1 M) of a compound would contain 1 mole of that
compound dissolved in solvent so that the total volume of solution is 1 litre, while a 2
molar solution (2 M) would have 2 moles of the compound per litre of solution and a
10 M solution contains 10 moles of compound per litre of solution.
As an example, a 1 M solution of sodium chloride, NaCl (1 M), would contain 1 mole
of NaCl (= 22.99 + 35.45 g) dissolved in enough water so that the final volume of the
solution was 1 litre.
Note: The molarity applies to the solute formula. Thus, while the amount of sodium
chloride, NaCl, dissolved is 1 mole, there are actually 1 mole of Na+ ions and 1 mole
of Cl– ions present in that solution. Similarly, a 1 M solution of barium chloride,
BaCl2 , contains 1 mole of the solute per litre of solution which would provide 1 mole
of Ba2+ ions and 2 mole of Cl– ions per litre of solution.
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The following examples show how molarity, volume and moles are related. If two of
these quantities are known, then the third can be deduced.
Example 1. Calculate the concentration in moles per litre of a solution containing
45.2 g of magnesium chloride, MgCl2, in a total volume of 800 mL.
The first step is to calculate the molar mass (gram formula weight) of MgCl2.
Molar mass = (24.31 + 2 × 35.45) = 95.21 g mol–1
i.e. 95.21 g of MgCl2 is exactly 1 mole.
Then calculate how many moles are in 45.2 g of MgCl2.
Moles of MgCl2 in 45.2 g = 45.2 = 0.475 mol
95.21
Finally, from its definition, calculate the molarity.
Concentration of MgCl2 = moles / litres = 0.475 = 0.594 M
0.800
[ Note that the concentration always applies to the solute specified, MgCl2 here, not its
component ions if the solute is ionic. Thus in this example, while the concentration of
MgCl2 dissolved is 0.594 M, the solution actually contains Mg2+ ions at a
concentration = 0.594 M and Cl– ions at a concentration = 2 × 0.594 M = 1.19 M,
because there are 1 Mg2+ and 2 Cl– ions in each formula unit of the compound.]
Example 2. What mass of sodium chloride is present in 500 mL of NaCl (2.00 M)
solution?
molarity = moles or moles = molarity × litres
litres
i.e. moles of NaCl = 2.00 × 0.500 mol = 1.00 mol
and mass of NaCl = moles × molar mass = 1.00 × (22.99 + 35.45) g = 58.4 g
Example 3. What volume of 0.450 M sodium carbonate solution contains 10.0 g of
the solute?
Firstly, it is necessary to convert the mass of sodium carbonate to moles.
Moles of Na2CO3 = 10.0 = 0.0943 mol
(2 × 22.99 + 12.01 + 3 × 16.00)
As molarity = moles of solute
volume of solution in litres
then the volume containing a specified amount of solute is given by the expression
volume in litres = moles of solute = 0.0943 = 0.210 L or 210 mL
molarity 0.450
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The next set of examples shows how stoichiometric calculations can be carried out
when solutions are involved.
Example 4. A solution of sodium hydroxide of unknown concentration is titrated
against a STANDARD sulfuric acid solution (i.e. one of known concentration). The
volume of 0.104 M sulfuric acid needed for complete reaction with 25.00 mL of the
sodium hydroxide solution was 20.05 mL. Calculate the concentration of the sodium
hydroxide solution. [This reaction was studied in Topic 6 - “Reactions of acids with
oxides and hydroxides of metals”.]
The procedure is essentially the same as in earlier examples.
Step 1: Write a balanced formula equation.
H2SO4 + 2NaOH Na2SO4 + 2H2O
Step 2: Write down the mole ratios of the known and unknown species.
1 mol 2 mol
Step 3: From step 2, deduce number of moles of the unknown (NaOH) that requires 1
mole of the standard (H2SO4) for complete reaction. This number is known as the
EQUATION FACTOR.
1 mole of standard requires 2 moles of unknown. ∴ equation factor = 2
Step 4: Calculate the moles of standard reacting.
Moles of H2SO4 in 20.05 mL = volume (in litres) × molarity
= 0.02005 × 0.104 mol = 2.085 × 10–3 mol
Step 5: Using the equation factor, calculate the moles of unknown reacting.
As 1 mole of H2SO4 uses 2 moles of NaOH,
then moles of NaOH needed = 2 × 2.085 × 10–3 mol = 4.170 × 10–3 mol
Step 6: From the volume of NaOH solution used, its concentration can now be
deduced.
As 4.170 × 10–3 mole of NaOH are in 25.00 mL, then
concentration of NaOH = moles = 4.170 × 10–3 = 0.167 M
litres 0.02500
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