Lesson 6
Solutions
A solution is a homogenous mixture of two or more substances. The concentration of a solution is
a measure of the amount (moles) of solute dissolved in a solution. The solute is the substance
being dissolved and the solvent is the substance the solute is dissolved in. In aqueous solutions the
solute is water. Solids and gases can also be solutes. A solution is different from a suspension (fine
particles of a solid in a liquid) because it is transparent, does not settle out and cannot be separated
by filtration.
The solubility of a substance is the amount (moles) that will dissolve to form a certain volume of
solution. Water is assumed to be the solvent unless another solvent is stated.
A solution with a small amount of solute is called dilute and if it contains a large amount of solute it
is said to be concentrated. A saturated solution is one in which no more solute will dissolve at that
temperature and excess solute is present. Sometimes, temporarily the concentration of a solute
can exceed its solubility. In this case, the solution is referred to as supersaturated. This can occur if
the temperature of the solution is changed, or more commonly, if the substance is produced in a
chemical reaction. In this case the excess solid will eventually separate from the solution as a
precipitate.
Concentration (c) is the amount of substance (n) contained within a given volume (V). The symbol
for concentration is often written in shorthand notation as either: c, [ ] or [HCl]. The relationship
between concentration, moles and volume is expressed using the following formula.
c
=
n
v
where: c is the concentration in mol dm-3
n is the amount of the solute in mol
v is the volume of the solution in dm3 (or L)
In many problems involving concentration you also need to use the equation:
n = m
M
Units
mol dm-3 is the SI unit and the preferred unit for concentration. Other common units for
concentration are mol L-1 and M (molarity).
c = 1 mol dm-3 = 1 mol L-1 = 1 M
dm3 (or L) is the SI unit for volume. Other common units for volume are cm3, L and ml.
1 dm3 = 1 L
1 cm3 = 1 ml
1 cm3
= 1 ÷ 1000 dm3
and
1 dm3
1
= 1 x 1000 cm3
Lesson 6
1 mol dm-3 = 1 mol/dm3
because x-3 = 1
x3
In the diagram below Solution B is the most concentrated. It has twice as many moles of sugar per
cubic decimeter as A and C. Solutions A and C have the same concentration even though they are
made using different amounts of sugar and different volumes of water.
A
Add
1 dm3
water
1 dm3
sugar solution
0.1 mol sugar
water
sugar
solution
B
0.2 mol sugar
c = n = 0.1 = 0.1 moldm-3
v 1
0.2 dm3
sugar solution
Add
1 dm3
water
water
sugar
solution
C
0.2 mol sugar
c = n = 0.2 = 0.2 moldm-3
v 1
v 1
0.1 dm3
sugar solution
Add
2 dm3
water
water
sugar
solution
c = n = 0.2 = 0.1 moldm-3
v 2
v 2
2
Lesson 6
Dilution
Adding more of the solvent to a solution will dilute it causing the concentration to decrease.
Add
900 cm3
water
0.3 mol of sugar in
100 cm3 water
c = n = 0.3 = 3
v 0.1 moldm-3
0.3 mol of sugar in
1000 cm3 (1 dm3)
water
water
sugar
solution
c = n = 0.3 = 0.3
v 1 moldm-3
Notice in the example above when a solution is diluted the number of moles remains the same, but
the concentration decreases and the volume increases.
Often in a lab a chemist does not want to make the same solution over and over again. To reduce
the workload they create a stock solution. The stock solution is usually much more concentrated
than the required useable solution so all the chemist needs to do is dilute the stock solution using
the following equation:
Ci x Vi = Cf x Vf
Where: Ci is the intial concentration of the stock solution in mol dm-3
Vi is the initial volume of the stock solution required
Cf is the final concentration of the diluted stock solution needed
Vf is the Final volume of the required solution needed
Volume of water that needs to be added for the dilution = V f - Vi
Worked Example
Mr Salim, the HS science assistant requires 50. cm3 of a 0.0500 mol dm-3 solution of sodium
hydrogen phosphate, NaHPO4 for an IB Chemistry lab. He has a stock solution in the cabinet of
2.000 mol dm-3. What volume of stock solution does he need to make a useable solution?
Solution
The first step is to identify the known and unknown variables.
Ci = 2.000 mol dm-3
Vi = ? unknown
Cf = 0.0500 mol dm-3
Vf = 50. cm3 = 50. ÷ 1000 = 0.050 dm3 (2SF)
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Lesson 6
The second step involves rearranging the formula to make the unknown the subject.
Ci x Vi = Cf x Vf
Vi = Cf x Vf
Ci
The third step involves substituting the known variables into the equation and calculating the
answer.
Vi = 0.0500 mol dm-3 x 0.050 dm3 = 0.025 = 1.2 x 10-3 dm3 (2SF)
2.000 mol dm-3
2.000
Problems
1.
State all the factors that affect the concentration of a solution?
2.
Determine the concentrations of the following solution produced by dissolving
a)
5.0 g of magnesium sulfate heptahydrate (MgSO4.7H2O) in 250. cm3 of water
b)
0.75 moles of NaC2H3O2 in a 115 cm3 solution
3.
Determine the number of moles in 7.0 dm3 of sulfuric acid of concentration 0.30 mol dm-3
4.
What volume of solution could be produced from:
a)
1.0 mol dm-3 of copper (II) chloride (CuCl2) from 0.40 moles of the solid
b)
0.020 mol dm-3 of sodium nitrate (NaNO3) starting from 5.0g of the solid
5.
Sodium thiosulfate, Na2S2O3 is used in developing photographic film. It is known to
photographers as ‘hypo’ or ‘fixer’. How many grams of Na2S2O3 are needed to make
100.cm3 of a 0.250 mol dm-3 solution of hypo?
6.
Mr Williams AP Biology students were asked to make a 500.0 cm3 solution of a 0.20 mol dm3 solution of sucrose, C H O for an osmosis lab. What mass of sucrose would they need
12 22 11
to dissolve?
7.
What total volume of water would be need to make a 0.50 mol dm-3 solution of sulfuric acid,
H2SO4 if you had to dilute a 20.0 cm3 of a concentrated 18 mol dm-3 of the solution?
8.
Determine the number of moles in 0.015 dm3 of a solution made by dissolving 5.8g of zinc
chloride (ZnCl2) in 2.5 dm3 of solution. HINT: The concentration of the 0.015dm3 and 2.5dm3
solutions is the same.
9.
Sulfuric acid and sodium hydroxide react together according to the reaction:
H2SO4 (aq)
+
2 NaOH (aq)  Na2SO4 (aq) + 2 H2O (l)
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Lesson 6
Calculate the volume of 0.250 mol dm-3 sodium hydroxide in dm3 and cm3 that is required to
exactly neutralize 25.0 cm3 of 0.125 mol dm-3 sulfuric acid?
10.
What mass of solid will remain when 2.00 dm3 of a 0.400 moldm-3 solution of sucrose
(C12H22O11) is evaporated to dryness?
11.
Professor Ying needs to have 0.500 liters of a 0.01125 M solution. She has as much stock
solution of 1.000 M that is required. What volume of the stock solution does she need to
dilute? How much water does she need to add to the initial stock solutions to have the
proper final volume?
12.
Doctor Myung Soo at Seoul University dilutes 225 cm3 of a solution of unknown
concentration. Noticing his error, he determines the final solution (2.000 dm 3) has a final
concentration of 0.0113 mol dm-3.
a)
What was the initial concentration of the unknown solution?
b)
If the original compound was iron (III) oxide, how many moles of the
compound were in the original solution and how many moles were in the final
solution?
c)
If all the water of the solution were evaporated, how many grams of iron (III) oxide
would remain?
13.
Pharmacist Kate requires a patient to drink 0.0050 mol of a drug to treat a stomach ulcer.
The stock solution is much too thick (viscous) for the patient to drink properly so she dilutes
a 2.50 M stock solution to prepare the drinkable solution.
a)
Determine what her original volume of stock solution needs to be if she wants to
make a 250. ml of drinkable solution, calculate the concentration of the drinkable
solution.
b)
Determine how much water needs to be added to the stock solution to get the
correct final volume.
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