Lesson Notes By Weeks and Term - Senior Secondary School 2

SOLUBILITY AND SOLUTIONS

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SUBJECT: CHEMISTRY

CLASS:  SS 2

DATE:

TERM: 3rd TERM

REFERENCE MATERIALS

  • New School Chemistry for Senior Secondary Schools by O. Y. Ababio
  • New System Chemistry for Senior Secondary Schools by T. Y. Toon et al
  • S.S.C.E  Past Questions and Answers on Chemistry 
  • U.T.M.E Past Questions and Answers on Chemistry

 

 
WEEK TWO

TOPIC: SOLUBILITY AND SOLUTIONS

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 true solution is 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 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 dispersedin another liquid. Examples of emulsions are milk, hair cream; cleaning action of detergents is due to their ability to form 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 = mass x      1000

                        Molar mass      volume

Solubility in g/dm3 = mass   x    1000

        volume    1

Solubility of a solid solute in a solvent increases with rise in temperature while solubility of gases decreases with rise in temperature.   

 

DEFINITION OF TERMS

  1. Saturated Solution: A saturated solution at a particular temperature is one which contains as much solute as it can dissolves 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. Super saturated 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 a 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

:.  100g H2O dissolves (z – x)/(y – z) x 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 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 amount of solid drugs that must be dissolved in a given quantity of solvent to give a prescribed drug mixture.

 

EVALUATION

  1. Define 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 20oC.  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

 

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

(Solubility of KNO3 in H2O at 60oC = 5.14moldm-3, solubility of KClO3 in H2O at 60oC = 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 60oC (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 60oC since the solubility at 60oC is lower than the amount in solution.

 

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

     Mass of salt = Number of moles x Molar mass

    = 0.39 x 122.5 = 47.78g

 

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

 

Solution: 

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

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

Mass of solute deposited = 2800 – 1700 = 1100g

From 83oC to 40oC, 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. 1.33 dm3 of water at 70oC 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 70oC to 18oC, calculate the mount of solute that will be deposited in (a) moles (b) grams.

 

GENERAL EVALUATION/REVISION

  1. Calculate the solubility of KCl in g/dm3 if 5g of the salt was dissolved in 50cm3 of water at 40oC
  2. If 50cm3 of a saturated solution of potassium chloride at 30oC yielded 18.62g of dry salt, calculate the solubility of the salt in mol/dm3 at 30oC
  3. Define solubility
  4. A certain mass of a gas occupies 300cm3 at 35oC. At what temperature will it have its volume reduced by half, assuming its pressure remains constant?
  5. A certain mass of hydrogen gas collected over water at 10oc and 760mm Hg pressure has a volume of 37cm3. Calculate the volume when it is dry at s.t.p. (Saturated vapour pressure of water at 10oC = 9.2mmHg)



READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O.Y.Ababio (6thedition) 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 
  1. supernant curve c. solubility curve d. dispersion curve
  1. On heating 25g of a saturated solution to dryness at 60oC, 4g of anhydrous salt was recovered. Calculate its solubility in g/dm3. a. 160 b. 180 c. 200 d. 220
  2. The solubility of alcohols in water is due to a. their covalent nature b. hydrogen bonding c. their low boiling point d. their ionic character
  3. 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

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





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