Solubility and Solutions

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In today’s Chemistry class, We will be learning about Solubility and Solutions. We hope you enjoy the class!

 

CONTENT

• Definition of Terms.
• Calculations based on Solubility.
• Solubility Curves.
• Uses of Solubility.

 

 

SOLUTIONS

A solution is a uniform or homogenous mixture of two or more substances.

Solution = Solvent + Solute

A solute is a dissolved substance which may be a solid, liquid or gas.

A solvent is a substance (usually liquid) which dissolves a solute.

 

TYPES OF SOLUTIONS

1. Aqueous Solution: This is formed when a solute is dissolved in water.
2. Chemical Solution: This is the apparent solution of a solute in a solvent accompanied by a chemical change. For example, magnesium appears to dissolve in dilute hydrochloric acid, what actually happens is that the magnesium attacks the acid to form magnesium chloride, which dissolves in water present.

 

TRUE SOLUTION AND COLLOIDAL SOLUTION

A true solution is formed when solute particles dissolve such that they are able to get in between the solvent particles. Example of a true solution is an aqueous solution of sodium chloride and copper (II) tetraoxosulphate (VI).

A False or Colloidal solution is one in which the individual particles are larger than the particles of a true solution, but not large enough to be seen by the naked eye. Examples of colloids are starch and albumen.

 

TYPES OF COLLOIDS

1. Sols and Gels: These are colloids where solid particles are dispersed in a liquid medium. Example: starch, glue, jelly, etc.
2. Aerosols: In aerosols, liquid particles are dispersed in a gas. Fog, smoke, spray of insecticide is examples of aerosol.
3. Emulsion: For emulsions, a liquid is dispersed in another liquid. Examples of emulsions are milk, hair cream; cleaning action of detergents is due to their ability to form an emulsion.

 

EVALUATION

1. Define the term ‘Solution’.
2. State THREE differences between True solution and False solution.

 

 

SOLUBILITY

The solubility of a solute (substance) in a solvent at a particular temperature is the maximum amount of solute in moles or grams that will dissolve in 1 dm³ of the solvent at that temperature.

The concentration in moldm^-3 of a saturated solution is termed the solubility of the substance i.e. Solubility (moldm³) = Concentration in gdm³

Molar mass

Solubility in mol/dm³ can also be expressed as = $\frac{mass}{Molar mass} \times \frac{1000}{volume}$

 

Solubility in g/dm³ = $\frac{mass}{volume} \times \frac{1000}{1}$

 

The solubility of a solid solute in a solvent increases with a rise in temperature while the solubility of gases decreases with rising temperature.

 

DEFINITION OF TERMS

1. Saturated Solution: A saturated solution at a particular temperature is one which contains as much solute as it can dissolve at that temperature in the presence of undissolved solute particles.
2. Unsaturated solution: This is a solution which contains less of the solute than it can dissolve at a particular temperature.
3. Supersaturated solution: This is a solution which contains more of the solute than it can dissolve at a particular temperature.

 

EVALUATION

1. Define Solubility
2. Differentiate between Saturated solution and Unsaturated solution

 

DETERMINATION OF SOLUBILITY

Solute: KCl,    Solvent: water

Method

1. A saturated solution of KCl is prepared by dissolving excess of the solid in water in a beaker
2. Allow the solution in the beaker to settle down to obtain a clear saturated solution
3. Decant a portion of clear solution into another beaker and measures its temperature
4. Transfer the solution into a weighed evaporation dish and record the mass of the solution
5. Evaporate the solution to complete dryness in a water bath
6. Allow the resulting solid to cool and reweigh the basin with content
7. Obtain mass of the dissolved salt and calculate the mass of the salt that would dissolve in 1dm³ of water at that temperature.

 

CALCULATION

Mass of basin                                  =          xg

Mass of basin + solution               =          yg

Mass of basin  + salt                       =          zg

Mass of solution                              =          (y-x)g

Mass of salt                                      =          (z-x)g

Mass of water used                       =          (y-z)g

:. (y – z)g H₂O dissolves (z – x)g salt

$\mathbf{:}\mathbf{.}\mathbf{ }100g H_{2}0 dissolves \frac{\left(z–x\right)}{\left(y–z\right)} \times 100g salt$

 

[Density of water   =  1gcm³]

:.  No of moles of salt                     =          100(z – x)

(y-z) x M.M

:.  Moles of salt dissolves in 1 dm³ water = 100(z-x)

(y-z) x M.M

 

FACTORS THAT AFFECT SOLUBILITY

1. Nature of solvent and solute
2. Temperature
3. Pressure (often neglected)

 

SOLUBILITY CURVES

These are the graphs of solubility against temperature.  The graph provides a useful source of information.

 

USES OF SOLUBILITY CURVES

1. It provides useful information about suitable solvent and temperature for solvent extraction from natural sources
2. It provides useful information about temperature for fractional crystallization of a mixture of soluble salts.
3. The curves enable pharmacists to determine the number of solid drugs that must be dissolved in a given quantity of solvent to give a prescribed drug mixture.

 

EVALUATION

1. Define a super-saturated solution
2. State two applications of solubility curves

 

CALCULATION ON SOLUBILITY

1. If 12.2g of Pb(NO₃)₂ were dissolved in 21cm³ of distilled water at 20^oC. Calculate the solubility of the solute in moldm^-3

 

Solution:

Molar mass of Pb(NO₃)₂ = 331g

No of moles of Pb(NO₃)₂  =  12.2/331 = 0.037moles

If 21cm³ of water at 20⁰C dissolved 0.037mole salt

:. 1000cm³ of water at 20⁰C dissolves 0.037 x 1000/21

= 176moles Pb(NO₃) per dm³ H₂O

2. 1.0dm³ of an aqueous solution at 90^oC contains 404g of KNO₃ and 245g of KClO₃. Determine
   1. • which of the two salts will separate out when the solution is cooled to 60^oC
      • mass of salt that will separate out at 60^oC

(Solubility of KNO₃ in H₂O at 60^oC = 5.14moldm^-3, solubility of KClO₃ in H₂O at 60^oC = 1.61moldm^-3)

 

Solution:

No of moles of KNO₃ = 404/101 = 4.0moles dm^-3

No of moles of KClO₃ = 245/122.5 = 2.0 moldm^-3

The solubility of KClO₃ at 60^oC (5.14 moldm^-3) is higher than the amount in solution (4.0 moldm^-3), then KNO₃ will remain in solution while KClO₃ will crystallize out at 60^oC since the solubility at 60^oC is lower than the amount in solution.

B.Mass of salt that will separate out at 60^oC = 2.0 – 1.61 = 0.39mole

Mass of salt = Number of moles x Molar mass

= 0.39 x 122.5 = 47.78g

 

 

3. The solubility of KNO₃ is exactly 1800g per 1000g water at 83^oC and 700g per 1000g water at 40^oC. Calculate the mass of KNO₃ that will crystallize out of solution if 155g of the saturated solution at 83^oC is cooled to 40^oC.

 

Solution:

Saturated solution of KNO₃ at 83^oC = 1000 + 1800 = 2800g

Saturated solution of KNO₃ at 40^oC = 1000 + 700 = 1700g

Mass of solute deposited = 2800 – 1700 = 1100g

From 83^oC to 40^oC, 2800 of saturated solution deposited 1100g of solute

155g of saturated solution will deposit 1100 x 155/2800 = 60.80g of salt.

EVALUATION

1. Define the following terms: Solubility, Saturated solution, Unsaturated solution.
2. 33 dm³ of water at 70^oC are saturated by 2.25 moles of lead (II) trioxonitrate (V) and 1.33 dm³ of water at 18oc are saturated by 0.53 mole of the same salt. If 4.50dm³ of the saturated solution are cooled from 70^oC to 18^oC, calculate the amount of solute that will be deposited in (a) moles (b) grams.

 

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O.Y. Ababio (6^thedition) pages 303-310

 

WEEKEND ASSIGNMENT

SECTION A: Write the correct option ONLY

1. A saturated solution is a solution a. in which the solute is in equilibrium with the solvent b. in which the solute saturates the solution c. the solvent can still accept more solute except when the temperature is lowered d. whose solvent has low solubility at a given temperature
2. A graph of solubility against temperature is called a. sigmoid curve b. supernant curve c. solubility curve d. dispersion curve
3. On heating 25g of a saturated solution to dryness at 60^oC, 4g of anhydrous salt was recovered. Calculate its solubility in g/dm³.  A. 160    B. 180      C. 200       D. 220
4. The solubility of alcohols in water is due to their A. covalent nature   B. hydrogen bonding    C. their low boiling point D. their ionic character
5. A common solvent of sulphur is   A.  water   B. carbon(IV)sulphide    C. alcohol     D. ethanoic acid

 

SECTION B

1. Define the following:

(a) Solubility (b) Saturated solution (c) Unsaturated solution

2. If the solubility of KNO₃ at 0^oC is 1.33mol/dm³, determine whether a solution containing 30.3g/dm3 at 0^oC is saturated or unsaturated.

 

 

 

We have come to the end of this class. We do hope you enjoyed the class?

Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.

In our next class, we will be learning about Mass/Volume Relationship. We are very much eager to meet you there.

 

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