SOLUBILITY CURVES The solubility of a solute is the maximum amount of solute that can be dissolved in a given amount of solvent at a given temperature.
A saturated solution is a solution in which no more solute can be dissolved at a given temperature.
An unsaturated solution is a solution into which more solute can be dissolved.
A supersaturated solution is an unstable solution as it contains more dissolved solute than a saturated solution.
One way of measuring the solubility is to determine the maximum mass of solute that can be dissolved in 100 g of solvent at a given temperature. This is given in grams of solute per 100 g of solvent or g per 100 g. Solubility curves show the relationship between solubility and temperature. An example of a solubility curve is given below.
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Each point on the curve represents a saturated solution, any point below the curve an unsaturated solution and any point above the curve a supersaturated solution.
We can use solubility curves to:
Determine how much solid will dissolve at a given temperature.
Obtain a general idea about the ability of a solute to dissolve in water, at different temperatures.
Compare the different solubilities of different solutes.
Determine the mass of crystals deposited, when a saturated solution so cooled from a higher to a lower temperature.
Note:
When a solution is cooled, the solute tends to become less soluble and no longer remains dissolved. The solute comes out of the solution as crystals and this process is known as crystallisation.
Unlike most solids, gases become less soluble as the temperature increases. They are, however, more soluble as pressure increases.
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QUESTION 8 (a)
What is the solubility of KNO3 at 60°C?
(b)
A solution of KNO3 contains 27 g of KNO3 in 50 g of water at 40C. Give a name to this type of solution.
(c)
110 g of KNO3 was dissolved in 100 ml of water at 60°C. If this solution was cooled to 20°C, what mass of KNO3 would crystallise out of this solution upon cooling?
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QUESTION 9 75 g of NaNO3 was dissolved in 50 ml of water at 80°C. If this solution was cooled to 40°C, what mass of crystals would appear? Solution
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CONCENTRATIONS OF SOLUTIONS The concentration of a solution describes the relative amounts of solute and solvent present. 1.
A high Solute:Solvent ratio gives a concentrated solution.
2.
A low Solute:Solvent ratio gives a dilute solution.
The term concentration also describes the extent to which particles are spread apart from each other. One can express concentration in a variety of ways.
PERCENTAGE BY MASS (W/W) Percentage by mass (w/w) indicates that the % is based on the masses of both the solute and solution. It describes the mass of solute (g) present in 100 g of solution. For example:
0.8% w/w indicates that there are 0.8 g of solute dissolved in 100 g of solution.
Concentration (w / w)
mass of solute ( g ) 100 mass of solution ( g )
When using this formula, it is important that the masses of both species are in the same units. QUESTION 10 Twenty grams of a salt solution contains 4.0 grams of salt. What is the concentration (w/w) of salt in this solution? Solution
Concentrat ion
4 100 20% w / w 20
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QUESTION 11 How much salt is required to prepare 500 g of a 10% w/w salt solution? Solution
x 100% 10% 500
x 0.1 500
x 50 g QUESTION 12 The label on a can of tomato soup states that the salt content is 300 mg per 100 g of soup. What is the percentage by mass of salt in the soup? Solution
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PERCENTAGE BY VOLUME (V/V) Percentage by volume (v/v) indicates that the % is based on the volumes of both the solute and solution. It describes the volume of solute (ml) present in 100 ml of solution. For example:
11% alcohol v/v indicates that there are 11 ml of alcohol dissolved in 100 ml of solution.
Concentration (v / v)
volume of solute (ml) 100 volume of solution (ml)
When using this formula, it is important that the masses of both species are in the same units. This formula is best used when the solute is a liquid. QUESTION 13 A 170 ml glass of fruit drink contains 15.0% (v/v) of pure orange juice. What volume of pure orange juice is present in this solution? Solution
x 100 15% 170
x 25.5 ml QUESTION 14 A bottle of wine contains alcohol at a concentration of 14% (v/v). What volume of alcohol is present in 250 ml of wine? Solution
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PERCENTAGE BY MASS/VOLUME (W/V) Percentage mass per volume (w/v) describes the mass of solute (g) dissolved in 100 ml of solution. For example:
2% (w/v) indicates that there are 2 g of solute dissolved in 100 ml of solution.
Concentration ( w / v)
mass of solute ( g ) 100 volume of solution (ml)
When using this formula it is important that mass and volume are of the same order, i.e. g and ml, kg and L. QUESTION 15 The concentration of Mg 2 in water is 10% (w/v). Calculate the mass of Mg 2 in 2.00 L. Solution
10% w / v 10g / 100 ml x 2.0 102 g QUESTION 16 A solution contains sugar at a concentration of 5% (m/v). What mass of sugar is present in 500 ml? Solution
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PPM AND PPB For species in an aqueous environment, ppm (parts per million) and ppb (parts per billion) describe the relative masses of materials. For example: A 5 ppm solution of Fe 3 describes a solution containing: 5 g of Fe 3 in 1 10 6 g of solution. 5 kg of Fe 3 in 1 10 6 kg of solution. An 8 ppb solution of Fe 3 describes a solution containing: 8 g of Fe 3 in 1 10 9 g of solution. 8 kg of Fe 3 in 1 10 9 kg of solution.
For species in a gaseous phase, ppm and ppb describes the relative volumes of gas. For example: A 2 ppm H 2 sample contains 2 ml of hydrogen gas in a total volume of 1 10 6 ml. When concentration units are stated as mg/kg, mg/L, µg/g, µg/ml, ng/mg or g/tonne these units are also ppm. i.e. 40 mg/kg = 40 ppm or 1.5 µg/ml = 1.5 ppm. For example: 2.6 mg of salt in a 200 ml solution is how many ppm? 2.6 mg in 200 ml
X = 13 mg in 1L 13 mg/L = 13 ppm
X mg in 1000 ml
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QUESTION 17 A solution contains iron at a concentration of 12 ppm. Calculate the mass of iron in 250 g of this solution. Solution
QUESTION 18 Water contains copper ions at a concentration of 6 ppm. What mass of copper would be 8 present in 4 10 g ?
Solution
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OTHER MEANS OF EXPRESSING CONCENTRATION MASS OF SOLUTE PER LITRE OF SOLUTION Units: g / L or mg / L This measure involves expressing concentration in terms of mass of solute ( g or mg ) per litre ( L ) of solution.
MASS OF SOLUTE PER GRAM OF SOLUTION Units: mg / g or g / g This measure involves expressing concentration in terms of mass of solute ( mg or g ) per gram ( g ) of solution.
QUESTION 19 A solution contains sugar at a concentration of 30 g/L. What mass of sugar is present in 200 ml? Solution
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QUESTION 20 Brine is a concentrated salt solution. If a sample of brine contains 90 g of salt in 115 ml of water, what is its concentration in: (a)
%( w / v )
(b)
g/L
(c)
mg / L
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(d)
mg / g
(e)
g / g
Note: The density ( d ) of all aqueous solutions is 1 g / ml . Given that d that m V for such solutions.
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m , we can say V
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CONVERTING CONCENTRATION UNITS The following relationships will be required to convert concentration units:
1000 1000 1000 1000 ng g mg g kg 1 109 g 1106 g 1103 g 1103 g Mass Conversions
1109 g 1 106 g 1 103 g ng g mg g 1000 1000 1000
1103 g kg 1000
1000 1000 l ml L
1106 L
1103 L
Volume Conversions
1106 L 1103 L l ml L 1000 1000 IMPORTANT UNITS Mass Units
Volume Units
1 kg 1 10 3 g
1 L 1000 ml 1000 cm 3 1 10 6 l
1 g 1 109 ng 1 106 g 1 103 mg
1 ml
1 L 1103 L 100
1 l
1 L 1 10 6 L 1,000,000
1 mg
1 g 1 10 3 g 1000
1 g
1 g 1 10 6 g 1,000,000
1 ng
1 g 1 10 9 g 1,000,000,000
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METHOD: Step 1: Draw a flow diagram to illustrate the unit changes required. Consider one unit change at a time. For example: Converting mol / L to ppm .
mol / L mol / 1000 ml
mol / ml mol / g mol / 1 10 6 g g / 1 10 6 g Step 2: For each unit change, assess whether the quantity has been altered in terms of size, or whether it has been simply expressed in a different form. If the change has altered the size of the quantity, correct the other side of the expression by the same factor.
Quantity on left must also decrease 10 fold.
10 mol / L x mol / 100 ml
Quantity size has changed i.e. decreased 10 fold.
If the change has simply expressed the quantity in a different form (of identical size), do not alter the other side of the expression.
Do not alter the left hand side.
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10 mol / L x mol / 1000 ml
Quantity size has not changed. It has simply been expressed in a different form.
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QUESTION 21A 2 The concentration of glucose ( C6 H12O6 ) in water is 0.50 10 M . Calculate the mass of glucose in a 80.00 ml solution. Solution Steps:
mol / L g / 80 ml
mol / L
0.50 102 mol / L
mol / 1000 ml
0.50 102 mol /1000 ml
mol / 80 ml
/80 ml
Volume has decreased
1000 12.5 80 the other side by
g / 80 ml
fold, therefore, divide
12.5 as well.
2
0.50 10 mol / 80 ml 12.5 Convert
4.00 104 mol / 80 ml
mol
x g / 80 ml
to mass
4.00 104 180 g / 80 ml 7.2 102 g / 80 ml Answer is 7.2 102 g
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