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# Solubility curve

Solubility curve:

The solubility of a given solute in water, as done above, can be determined at many different temperatures, say 0oC, 10oC, 20oC,…..100oC. The result so obtained can be plotted on a graph by taking temperatures along the X-axis, and solubility along the Y-axis. By joining the points so obtained, you get a curve, called the solubility curve. The solubility curves of six solutes in water are given in the graph.

Observe the graph carefully and notice the following, regarding the solubility of these six solutes in water.

1. a) Sodium nitrate has the highest solubility at 0o
2. b) Potassium chlorate has the lowest solubility at 0o
3. c) Potassium nitrate has the highest solubility at 100o
4. d) The solubility of potassium nitrate is low at 0oC but increases drastically with increase in temperature.
5. e) The solubility of sodium chloride is very slightly affected by the increase in temperature; the increase in solubility is only 4.1 grams from 0oC to 100o

DEDUCTIONS of solubility curve:
a) It gives a general idea about the ability of a solute to dissolve in water, at different temperatures.

1. b) The solubility of the solute at any temperature, between the given lowest and highest temperatures, can be more or less accurately read.
2. c) The mass of crystals deposited, when a saturated solution so cooled from a higher to a lower temperature, can be calculated.

For e.g.,

As per the given graph, the solubility of potassium nitrate at 60oC= 110 g and at 50oC = 91 g.

Hence, by cooling a saturated solution of potassium nitrate containing 100 g of water from 60oC to 50oC, 19 g of the solute (110 g – 91 g) is deposited in the form of crystals.

Applications of Solubility Curves

1. a) They are used in pharmaceutical companies to determine the amounts of drugs that must be dissolved in a particular quantity of solvent to give a prescribed drug mixture.
2. b) They are used by chemists and other research workers to determine the most suitable solvents that can be used to extract essential oils from their natural sources at various temperatures. Example of an essential oil is the eucalyptus oil.
3. c) It is applied in fractional crystallization.

Calculations on Solubility

Mathematically, solubility can be expressed as:

Solubility (mol dm^3) = No. of moles/Volume (dm3)…………(i)

= (No. of moles/Volume (cm3)) x 1000………(ii)

Equations (i) and (ii) are used when the volume is given in dm^3 and cm^3 respectively.

Also,

Solubility (g dm^-3) = Reacting mass/Volume (dm3)…………(iii)

= (Reacting mass/Volume (cm3)) x 1000………(iv)

but,

No. of moles = Reacting mass/Molar mass…………(v)

Substituting equation (v) into (i) and (ii), then;

Solubility (mol dm-3)
= (Reacting mass/Molar mass) x 1/Volume (dm3)……(vi)

= (Reacting mass/Molar mass) x 1000/Volume (cm3)……(vii)

Equations (vi) and (vii) are used when the molar mass with the reacting mass is given.

.

Examples

Question 1
In an experiment to determine the solubility of a given salt Z, the following data were obtained:
Mass of empty dish = 14.32g
Mass of dish + saturated solution of salt Z = 35.70g
Mass of dish + salt Z = 18.60g
Temperature of solution = 30°C
Molar Mass of salt Z = 100
Density of solution Z = 1.00g cm-3
Determine the solubility of salt Z in (i) g dm-3 (ii) mol dm-3

Mass of saturated solution
= Mass of dish + saturated solution of salt Z – Mass of empty dish
= 35.70g – 14.32g
= 21.38g

Mass of salt Z = Mass of dish + salt Z – Mass of empty dish
= 18.60g – 14.32g
= 4.28g

Molar Mass of salt Z = 100

Amount in moles of salt Z = Mass of salt Z/Molar mass of salt Z
= 4.28/100
= 0.0428 mole

Density of solution Z = 1.00g cm-3

Volume of solution = Mass of solution/Density of solution
= 21.38g /1g cm-3
= 21.38 cm3

Therefore, solubility of salt Z in:
i) g dm-3 = (Mass/Volume (cm3)) x 1000
= (4.28/21.38) x 1000
= 200.19g dm-3

1. ii) mol dm-3 = (Moles/Volume (cm3)) x 1000
= (0.0428/21.38) x 1000
= 2.00mol dm-3

Question 2
16.55g of lead (II) trioxonitrate (V) was dissolved in 100g of distilled water at 20°C, calculate the solubility of the solute in mol per dm3. [Pb = 207, N = 14, O = 16]

Mass of Pb(NO3)2 = 16.55g
Molar mass of Pb(NO3)2 = (1xPb) + 2[(1xN) + (3xO)]
= (1 x 207) + 2[(1 x 14) + (3 x 16)]
= 207 + 2(14 + 48)
= 207 + 124
= 331g mol-1
Volume of water = 100cm3 (density of water = 1g cm-3)

Solubility (mol dm-3) = (Reacting mass/Molar mass) x 1000/Volume (cm3)
= (16.55/331) x (1000/100)
= 0.05mol dm-3

Question 3
Potassium trioxochlorate (V) has a solubility of 1.8 mol per dm3 at 35°C. On cooling this solution to a temperature of 20°C, the solubility was found to be 0.3 mol per dm3. What mass of KClO3 was crystallized out? [K = 39, Cl = 35.5, O = 16]

Molar mass of KClO3 = (1xK) + (1xCl) + (3xO)
= (1 x 39) + (1 x 35.5) + (3 x 16)
= 39 + 35.5 + 48
= 122.5g mol-1

Solubility of salt at 35°C = 1.8mol dm-3
Solubility of salt at 20°C = 0.3mol dm-3
Amount in moles of KClO3 crystallized out on cooling from 35°C – 25°C = 1.8 – 0.3
= 1.5 moles

Therefore, mass of salt crystallized out on cooling = Amount in moles x Molar mass
= 1.5 X 122.5
= 183.75g

Question 4
The solubility of a salt of molar mass 101g at 35°C is 0.35 mol per dm3. If 3.50g of the salt is dissolved completely in 250cm-3 of water in beaker, what will the resulting solution be – unsaturated, saturated or supersaturated?

Solubility of salt at 35°C = 0.35mol dm-3
Molar mass of salt = 101g
Mass of salt in solution = 3.50g
Volume of water = 250cm3

Solubility of salt in saturated solution = (Reacting mass/Molar mass) x 1000/Volume (cm^3)
= (3.5/101) x (1000/250)
= 0.138mol dm^-3

Since the calculated solubility of the salt solution in the beaker at 35°C, is less than the standard solubility of the salt at the same temperature, it means that the salt has not reached its solubility limit at that given temperature. Hence, the resulting solution will be unsaturated.

Question 5
The solubility of Na3AsO4.12H2O is 38.9g per 100g H2O. What is the percentage of Na3AsO4 in the saturated solution? [As = 75, Na = 23, O = 16, H = 1]

Molar mass of Na3AsO4.12H2O = (3xNa) + (1xAs) + (4xO) + 12[(2xH) + (1xO)]
= (3 x 23) + (1 x 75) + (4 x 16) + 12[(2 x 1) + (1 x 16)]
= 69 + 75 + 64 + 12(2 + 16)
= 208 + 216
= 424g mol-1

From the above, molar mass of Na3AsO4 = 208g mol-1

Mass of Na3AsO4.12H2O in 100g of water = 38.9g
Volume of water = 100cm3 (density of water – 1g cm-3)

Therefore, solubility of Na3AsO4.12H2O (g dm-3)
=  (Mass/Volume (cm3 )) x 1000
= (38.9/100) x 1000
= 389g dm-3

but,

424g of Na3AsO4.12H2O contains 208g of Na3AsO4
Therefore,
389g of Na3AsO4.12H2O will contain x of Na3AsO4
x = (389 x 208)/424
= 190.83g of Na3AsO4

Percentage of Na3AsO4 in the saturated solution
= (Mass of Na3AsO4 in solution/Mass of Na3AsO4.12H2O in solution) x 100
= (190.83/389) x 100
= 49.1%

Question 6
80g of a salt, XCl2, was placed in
40cm3 of water to give a saturated solution at 25°C. If the solubility of the salt is 8 mol dm-3 at that temperature, what is the mass of the salt that will be left undissolved at the given temperature? [X = 24, Cl = 35.5]

Molar mass of XCl2 = (1xX) + (2xCl)
= (1 x 24) + (2 x 35.5)
= 24 + 71
= 95g mol-1
At 25°C,
Solubility of XCl2      = 8.0mol dm-3
i.e, 1 dm3 of solution contains 8 x 95 = 760g

If, 1000cm3 of water = 760g of XCl2
Then, 40cm3 of water = x g of XCl2
x = 760 x 40/1000
= 30.4g of XCl2

From the above, based on the solubility of the salt, 40cm3 will only dissolve 30.4g

Therefore, the mass of the salt that will be left undissolved is 80 – 30.4 = 49.6g

Question 7
The solubility of potassium trioxochlorate (V) is 1500g per 1000g water at 80°C and 600g per 1000g water at 45°C. Calculate the mass of KClO3 that will crystallize out of solution if 200g of the saturated solution at 80°C is cooled to 45°C.

Masses at 80°C,
1000g of water + 1500g of KClO3 = 2500g of saturated solution

Masses at 45°C,
1000g of water + 600g of KClO3  = 1600g of saturated solution

Mass of salt deposited on cooling from 80°C to 45°C
= 2500 – 1600 = 900g

If 2500g of saturated solution deposit 900g of solute on cooling
Then, 200g of the saturated solution will deposit x g of solute
x = 900 x 200/2500
= 72g of KClO3.

Do These

Question 1
If 24.4g of lead (II) trioxonitrate (V) were dissolved in 42g of distilled water at 30°C; calculate the solubility of the solute in g per dm3

Question 2
Calculate the solubility in mol per dm3 of 50g of CuSO4 dissolved in 100g of water at 120°C. [Cu = 64, S = 32, O = 16]

Question 3
If 10.5g of lead (II) trioxonitrate (V) in 20cm3 of distilled water at 18°C, what is the solubility of the solute in mol per dm^3?

Question 4
117g of NaCl was dissolved in 1dm3 of distilled water at 25°C. Determine the solubility in mol per dmof sodium chloride at that temperature. [Na = 23, Cl = 35.5]

Question 5
a) What do you understand by terms saturated solution and solubility?
b) When 50cm3 of a saturated solution of sugar of molar mass 345g at 50°C was evaporated to dryness, 34.5g of dry solid was obtained. Calculate the solubility of sugar ay 50°C in a) g per dm3 b) mol per dm3

Question 6
3.06g of a sample of potassium trioxochlorate (V), KClO3, was required to make a saturated solution with 10cm3 of water at 25°C. What is the solubility of the salt at that temperature?

REVISION EXERCISES (POST ANSWERS USING QUESTION BOX BELOW FOR EVALUATION AND DISCUSSION. ADD QUESTION TITLE)

1. Differentiate between true solution and false solution.
2. Define solubility.
3. What is meant by the term saturated solution ?
4. What is meant by supersaturated solution?
5. What is the solubility of a solute in a given solvent?

6. 25.0g of a solute are required to saturate 80.0g of water at 25oC.Calculate its solubility

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