Definition of Terms
Solute
This is a substance that dissolves in a solvent to form a solution. It can be solid, liquid or gas. Examples include common salt, sugar, copper (II) tetraoxosulphate (VI) etc.
Solvent
This is a substance that dissolves a solute to form a solution. It can be liquid or gas. Examples are water, ethanol, benzene etc.
Solution
This is a mixture of solute and solvent, i.e,
Solute + Solvent = Solution
It can be homogeneous with a uniform composition, e.g, an unsaturated solution of a sodium chloride, or heterogeneous with non-uniform composition like an unsaturated solution of the same salt.
Saturated Solution
A saturated solution is one that holds as much solute as it can dissolve in the presence of undissolved solute particles at a given temperature. In a saturated solution, the dissolved and undissolved solutes are in equilibrium at a given temperature.
Unsaturated Solution
An unsaturated solution is one that can still dissolve more solutes at a given temperature until it becomes saturated.
For instance, the solubility limit of sodium chloride is 36g dm^-3, which means that as long as the mass of sodium chloride in 1dm^3 of its solution is below 36g, the solution will remain unsaturated. However, immediately the mass is 36g and above, the solution will be considered to be saturated because any mass in excess of 36g will be precipitated out of solution as undissolved particles. This implies that if the mass of sodium chloride in the solution is 37g; 1g which is the excess will be precipitated out as undissolved particles.
Supersaturated Solution
A solution which contains more of the solute than it can normally hold at a given temperature is said to be supersaturated. The supersaturated solution is unstable at a given temperature and the solute particles can crystallize out of the solution at the slightest touch or addition of a dry crystal of the solute. At times, dust particles can cause the crystallization of the solute particles. Also, merely scratching the inside of the glass vessel holding the solution with a glass rod or spatula can cause the solute particles to crystallize out. This is known as seeding, and is employed in crystallization as a separation technique.
Solubility
The solubility of a solute in a given solvent is the amount of a solute in moles or grams required to saturate 1 dm^3 of the solvent at a definite temperature. Its units are moles per dm^3 (mol dm^-3), g per dm^3 (g dm^-3) or g per 100g of solvent.
It is a way of comparing the level to which different solutes dissolve in a given solvent at a particular temperature.
Generally, the solubility of some substances increases with increase in temperature, especially most solids. This is why your hot water dissolves more sugar or beverages than cold water. However, the solubility of gases decreases with increase in temperature and increases with increase in pressure. Gases like hydrogen chloride and ammonia are highly soluble in water and do not follow this rule.
Calculations on Solubility
Mathematically, solubility can be expressed as:
Solubility (mol dm^3) = No. of moles/Volume (dm^3)…………(i)
= (No. of moles/Volume (cm^3)) 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 (dm^3)…………(iii)
= (Reacting mass/Volume (cm^3)) 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 (dm^3)……(vi)
= (Reacting mass/Molar mass) x 1000/Volume (cm^3)……(vii)
Equations (vi) and (vii) are used when the molar mass with the reacting mass is given.
Solubility Curve
The solubility of a given varies with changes in temperature, therefore, a plot of solubility against on the y-axis against temperature on the x-axis is known as a solubility curve. (Please refer to your textbooks for standard solubility curves)
A study of the solubility curves for some solutes in a given solvent over a range of temperatures shows that the solubility of:
a) KNO3 increases very rapidly over a short range of temperature.
b) NaCl is independent of changes in temperature.
c) KClO3 also increases very rapidly like KNO3, but over a wider range of temperature.
d) KCl is linearly proportional to temperature.
d) Ca(OH)2 is relatively independent of temperature.
Applications of Solubility Curves
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.
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.
c) It is applied in fractional crystallization.
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
Answer
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 cm^3
Therefore, solubility of salt Z in:
i) g dm^-3 = (Mass/Volume (cm^3)) x 1000
= (4.28/21.38) x 1000
= 200.19g dm^-3
ii) mol dm^-3 = (Moles/Volume (cm^3)) 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 dm^3. [Pb = 207, N = 14, O = 16]
Answer
Mass of Pb(NO3)2 = 16.55g
Molar mass of Pb(NO3)2 = (1*Pb) + 2[(1*N) + (3*O)]
= (1 x 207) + 2[(1 x 14) + (3 x 16)]
= 207 + 2(14 + 48)
= 207 + 124
= 331g mol^-1
Volume of water = 100cm^3 (density of water = 1g cm^-3)
Solubility (mol dm^-3) = (Reacting mass/Molar mass) x 1000/Volume (cm^3)
= (16.55/331) x (1000/100)
= 0.05mol dm^-3
Question 3
Potassium trioxochlorate (V) has a solubility of 1.8 mol per dm^3 at 35°C. On cooling this solution to a temperature of 20°C, the solubility was found to be 0.3 mol per dm^3. What mass of KClO3 was crystallized out? [K = 39, Cl = 35.5, O = 16]
Answer
Molar mass of KClO3 = (1*K) + (1*Cl) + (3*O)
= (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 dm^3. 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?
Answer
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 = 250cm^3
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]
Answer
Molar mass of Na3AsO4.12H2O = (3*Na) + (1*As) + (4*O) + 12[(2*H) + (1*O)]
= (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 = 100cm^3 (density of water - 1g cm^-3)
Therefore, solubility of Na3AsO4.12H2O (g dm^-3)
= (Mass/Volume (cm^3 )) 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
40cm^3 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]
Answer
Molar mass of XCl2 = (1*X) + (2*Cl)
= (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 dm^3 of solution contains 8 x 95 = 760g
If, 1000cm^3 of water = 760g of XCl2
Then, 40cm^3 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, 40cm^3 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.
Answer
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 dm^3
Question 2
Calculate the solubility in mol per dm^3 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 20cm^3 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 1dm^3 of distilled water at 25°C. Determine the solubility in mol per dm^3 of 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 50cm^3 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 dm^3 b) mol per dm^3
Question 6
3.06g of a sample of potassium trioxochlorate (V), KClO3, was required to make a saturated solution with 10cm^3 of water at 25°C. What is the solubility of the salt at that temperature?
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Solute
This is a substance that dissolves in a solvent to form a solution. It can be solid, liquid or gas. Examples include common salt, sugar, copper (II) tetraoxosulphate (VI) etc.
Solvent
This is a substance that dissolves a solute to form a solution. It can be liquid or gas. Examples are water, ethanol, benzene etc.
Solution
This is a mixture of solute and solvent, i.e,
Solute + Solvent = Solution
It can be homogeneous with a uniform composition, e.g, an unsaturated solution of a sodium chloride, or heterogeneous with non-uniform composition like an unsaturated solution of the same salt.
Saturated Solution
A saturated solution is one that holds as much solute as it can dissolve in the presence of undissolved solute particles at a given temperature. In a saturated solution, the dissolved and undissolved solutes are in equilibrium at a given temperature.
Unsaturated Solution
An unsaturated solution is one that can still dissolve more solutes at a given temperature until it becomes saturated.
For instance, the solubility limit of sodium chloride is 36g dm^-3, which means that as long as the mass of sodium chloride in 1dm^3 of its solution is below 36g, the solution will remain unsaturated. However, immediately the mass is 36g and above, the solution will be considered to be saturated because any mass in excess of 36g will be precipitated out of solution as undissolved particles. This implies that if the mass of sodium chloride in the solution is 37g; 1g which is the excess will be precipitated out as undissolved particles.
Supersaturated Solution
A solution which contains more of the solute than it can normally hold at a given temperature is said to be supersaturated. The supersaturated solution is unstable at a given temperature and the solute particles can crystallize out of the solution at the slightest touch or addition of a dry crystal of the solute. At times, dust particles can cause the crystallization of the solute particles. Also, merely scratching the inside of the glass vessel holding the solution with a glass rod or spatula can cause the solute particles to crystallize out. This is known as seeding, and is employed in crystallization as a separation technique.
Solubility
The solubility of a solute in a given solvent is the amount of a solute in moles or grams required to saturate 1 dm^3 of the solvent at a definite temperature. Its units are moles per dm^3 (mol dm^-3), g per dm^3 (g dm^-3) or g per 100g of solvent.
It is a way of comparing the level to which different solutes dissolve in a given solvent at a particular temperature.
Generally, the solubility of some substances increases with increase in temperature, especially most solids. This is why your hot water dissolves more sugar or beverages than cold water. However, the solubility of gases decreases with increase in temperature and increases with increase in pressure. Gases like hydrogen chloride and ammonia are highly soluble in water and do not follow this rule.
Calculations on Solubility
Mathematically, solubility can be expressed as:
Solubility (mol dm^3) = No. of moles/Volume (dm^3)…………(i)
= (No. of moles/Volume (cm^3)) 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 (dm^3)…………(iii)
= (Reacting mass/Volume (cm^3)) 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 (dm^3)……(vi)
= (Reacting mass/Molar mass) x 1000/Volume (cm^3)……(vii)
Equations (vi) and (vii) are used when the molar mass with the reacting mass is given.
Solubility Curve
The solubility of a given varies with changes in temperature, therefore, a plot of solubility against on the y-axis against temperature on the x-axis is known as a solubility curve. (Please refer to your textbooks for standard solubility curves)
A study of the solubility curves for some solutes in a given solvent over a range of temperatures shows that the solubility of:
a) KNO3 increases very rapidly over a short range of temperature.
b) NaCl is independent of changes in temperature.
c) KClO3 also increases very rapidly like KNO3, but over a wider range of temperature.
d) KCl is linearly proportional to temperature.
d) Ca(OH)2 is relatively independent of temperature.
Applications of Solubility Curves
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.
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.
c) It is applied in fractional crystallization.
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
Answer
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 cm^3
Therefore, solubility of salt Z in:
i) g dm^-3 = (Mass/Volume (cm^3)) x 1000
= (4.28/21.38) x 1000
= 200.19g dm^-3
ii) mol dm^-3 = (Moles/Volume (cm^3)) 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 dm^3. [Pb = 207, N = 14, O = 16]
Answer
Mass of Pb(NO3)2 = 16.55g
Molar mass of Pb(NO3)2 = (1*Pb) + 2[(1*N) + (3*O)]
= (1 x 207) + 2[(1 x 14) + (3 x 16)]
= 207 + 2(14 + 48)
= 207 + 124
= 331g mol^-1
Volume of water = 100cm^3 (density of water = 1g cm^-3)
Solubility (mol dm^-3) = (Reacting mass/Molar mass) x 1000/Volume (cm^3)
= (16.55/331) x (1000/100)
= 0.05mol dm^-3
Question 3
Potassium trioxochlorate (V) has a solubility of 1.8 mol per dm^3 at 35°C. On cooling this solution to a temperature of 20°C, the solubility was found to be 0.3 mol per dm^3. What mass of KClO3 was crystallized out? [K = 39, Cl = 35.5, O = 16]
Answer
Molar mass of KClO3 = (1*K) + (1*Cl) + (3*O)
= (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 dm^3. 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?
Answer
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 = 250cm^3
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]
Answer
Molar mass of Na3AsO4.12H2O = (3*Na) + (1*As) + (4*O) + 12[(2*H) + (1*O)]
= (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 = 100cm^3 (density of water - 1g cm^-3)
Therefore, solubility of Na3AsO4.12H2O (g dm^-3)
= (Mass/Volume (cm^3 )) 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
40cm^3 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]
Answer
Molar mass of XCl2 = (1*X) + (2*Cl)
= (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 dm^3 of solution contains 8 x 95 = 760g
If, 1000cm^3 of water = 760g of XCl2
Then, 40cm^3 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, 40cm^3 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.
Answer
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 dm^3
Question 2
Calculate the solubility in mol per dm^3 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 20cm^3 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 1dm^3 of distilled water at 25°C. Determine the solubility in mol per dm^3 of 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 50cm^3 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 dm^3 b) mol per dm^3
Question 6
3.06g of a sample of potassium trioxochlorate (V), KClO3, was required to make a saturated solution with 10cm^3 of water at 25°C. What is the solubility of the salt at that temperature?
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