UNIT FOUR - SOLUBILITY EQUILIBRIUM

 

Rules for Determining Ionic and Molecular Compounds

 

An ionic compound is made up of

·        a metal and a non-metal

·        Or a polyatomic cation (NH4) and a non metal

·        Or a metal and a polyatomic anion such as SO4-

·        Usually an Electrolyte - a substance that dissolves to give an electrically conducting solution containing ions

A molecular compound is made up of

·        two non metals

·        Or an organic compound.

·        Usually a Non electrolyte - a substance that dissolves to give a non conducting solution containing only neutral molecules.

 

 

 

Examples

 

Compound

Dissociation Reaction

Electrolyte/Non Electrolyte

NaCl(s)

NaCl(g) à Na+(aq) + Cl-(aq)

Electrolyte

HCl(g)

HCl(g) à H+(aq) + Cl-(aq)

Electrolyte

CH4(g)

CH4(g) à CH4(aq)

Non-Electrolyte

ClO2(l)

ClO2(l)à ClO2(aq)

Non-Electrolyte

FeCl3(s)

FeCl3(s) à Fe+3(aq) + 3Cl-(aq)

Electrolyte

NH4Cr2O7(s)

NH4Cr2O7(s) à NH4+(aq) + Cr2O7-(aq)

Electrolyte

 

Have students guess a number of solutions and demonstrate with an electrolyte apparatus:

 

Do questions Questions 1-2 page 74 of Hebden.



 
Solubility of a substance - The maximum amount of a substance which can dissolve in a given amount of solvent at a given temperature.  The equilibrium concentration of the subtance in solution at a given temperature.

 

Molar solubility – when the concentration is expressed in moles/litre.

 

Saturated solution - A solution that has dissolved the ‘maximum amount’ of a substance.  The dissolved substance is in equilibrium with some of the undissolved substance.

 

Saturation Exists when :

·        Some undissolved material is present and

·        Equilibrium exists between the dissolved and the undissolved material

 

 

A saturated solution is shown by writing equilibrium arrows between the substance and its aqueous products

 

FeCl3(s) « Fe+3(aq) + 3Cl-(aq)

 

You can write it as a dissolving reaction (which is what we will do through out this unit) or a crystallization reaction :

 

NaCl(g) « Na+(aq) + Cl-(aq)                          Dissolving Reaction

Na+(aq) + Cl-(aq) « NaCl(g)                               Crystallization Reaction

 

 

Do Questions 3-7 page 76 of Hebden.

 

Calculating Solubilty

 

It is easy to calculate the solubility of a substance once the mass of a substance present in 1 litre of solution has been measured.

 

Example :

1.0   L of satuated AgBrO3(aq) contains 1.96 g of AgBrO3.  What is the molar solubility in moles/liter?

 

1.96 g x   1 mole   =  8.3 x 10-3 M

               235.8 g

Example :

 

          The molar solubility of PbI2 is 1.37 x 10-3 M.  Express this in g/L

 

          1.37 x 10-3 mol   x    461.0 g    =  0.632 g/L

                             L           1 mole                     

 

Do questions 8 and 10 on page 77 and 15-16 on page 78 of Hebden

 

Do questions 17 together with the class.

 



Dissociation of Ionic Compounds

  • Ionic compounds are made of cations and anions held together by electrostatic attraction
  • if the solvent, usually water, can 'pull apart' the ions making up the crystal, then the substance is said to be soluble.
  • Substances that are soluble, will dissociate into their individual ions.

K3PO4 (s) ® 3K+ (aq) + PO4-3 (aq)

Go over PAGE 84 GENERALIZATIONS

  • Do questions 21 - 24 pages 83-84 Hebden

Finding Molarity (Review)

The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.

Example - What is the molarity when 0.75 mol is dissolved in 2.50 L of solution?

(ANS. = 0.300 M.)

Example - Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?

(ANS. = 0.500 M).



Dilution Calculations (Review)

Remember that you can use the following formula for dilution calculations :

C1V1 = C2V2

C1 = concentration of original solution

V1 = volume of original solution

C2 = concentration of diluted solution

V2 = total volume of diluted solution

Concentration of Individual Ions :

  • when an electrolyte such as NaCl dissolves in water it will ionize to form Na+ ions and Cl- ions.

NaCl (s) ® Na+ (aq) + Cl- (aq)
  • If we dissolved 1 mole of NaCl in 1 liter of water the solution would be 1 molar in both Na+ ions and Cl- ions.
  • if we dissolved 1 mole of K2SO4 in 1 liter of water it would ionize to form 2 moles of K+ ions and 1mole of SO42- ion.

K2SO4 ® 2K+ (aq) + SO42-(aq)
  • The resultant solution would be 2 molar in K+ ions and 1 molar in SO42- ions.
  • We can use simple mole ratios to calculate ion concentrations

 

For example : 1.5 M of Iron (III) Chloride is dissolved in water, what is the final concentration of all the ions in solution.

FeCl3 ® Fe+3 (aq) + 3 Cl- (aq)

 

1.5 M FeCl3 x 1 Fe+3 = 1.5 M Fe3+

1 FeCl3

 

1.5 M FeCl3 x 3 Cl- = 4.5 M Cl-

1 FeCl3

 

Do questions 18 -20 (every other letter) page 81 Hebden

 

Describing Reactions in Solution

 

A. The "formula" equation lists the whole formula for reactants and products, not taking into account the form of the solute in solution. For example, NaCl doesn't exist as NaCl in water solution; it exists as sodium and chloride ions.

 

Example :

2AgNO3 (aq) + Na2S (aq) ® Ag2S (s) + 2NaNO3 (aq)

 

B. The Complete Ionic equation shows all strong electrolytes, reactants and products, as separated ions. Thus, NaCl would be shown as Na+ and Cl- ions in an ionic equation. Remember to carry your balancing throughout.

 

Example

2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + S2-(aq) ® Ag2S(s) + 2Na+(aq) + 2NO3-(aq)

 

C. Net Ionic: This gives information on those species that undergo a chemical change. Ions that appear in the same form on both sides of the complete ionic equation are called spectator ions, and are not included in the net ionic equation.

2Ag+ (aq) + S2- (aq) ® Ag2S (s)



For example : The reaction of a solution of BaCl2 with a solution of Na2SO4 to form the insoluble solid BaSO4

Formula :

BaCl2(aq) + Na2SO4(aq) ® BaSO4(s) + 2NaCl(aq)

Total Ionic :

2Na+(aq) + SO42-(aq) + Ba2+(aq) + 2Cl-(aq) ® 2Na+(aq) + 2Cl-(aq) + BaSO4(s)

Net Ionic :

Ba2+(aq) + SO42-(aq) ® BaSO4(s)

Try question 25 on page 87 of Hebden (do every other letter)

 

Separating Mixtures of Ions By Precipitation Methods

We can use experimental procedures to find which ions are present in a mixture, and/or to separate specific ions out of a mixture.

Example :

  • Assume a solution contains one or both of the ions Ag+ and Sr+2
  • Try to find an anion that will form a precipitate with only one of the two cations
  • If a precipitate is formed, you can assume that the ion being looked for is present
  • If no precipitate, the ion is absent

 

Cl-

SO4-2

S-2

OH-

PO43-

Ag+

ppt

ppt

ppt

ppt

ppt

Sr+2

-

ppt

-

-

ppt

  • We see that adding either Cl-, S-2, or OH- will precipitate out the silver ion if it is present.

  • USE THE POINTERS ON PAGE 84 OF HEBDEN TO DECIDE WHICH COMPOUNDS TO ADD AS A SOURCE OF THE ION YOU DESIRE.
  • For example add NaCl or Na2S, or NaOH to ppt the Ag+
  • Once we add one of those ions we can add SO4-2 or PO43- , in the form of Na2SO4 or Na3PO4 to precipitate out the strontium ion.

 

Example:

A solution contains one or more of Ag+, Ba+2, and Ni2+. What ions could be added, and in what order, to determine which of these cations are present.

 

 

Cl-

SO4-2

S-2

OH-

PO43-

Ag+

ppt

ppt

ppt

ppt

ppt

Ba+2

-

ppt

-

ppt

ppt

Ni+2

-

-

ppt

ppt

ppt

 

 

I) use NaCl to ppt out the Ag+, and filter out the AgCl

II) now either use

Na2SO4 to ppt out the Ba+2, and filter out the BaSO4

or Na2S to ppt out the Ni+2 , and filter out the NiS

III) now do which ever step you did not do in part II

 

Do Questions 26 - 36 (odds) and do 37, 38 and 39 page 90-91 of Hebden.

Solubility Product

  • When a salt undergoes a dissociation reaction, if the solution is saturated, an equilibrium is established.

CaF2(s) « Ca+2(aq) + 2F-(aq)
  • The equilibrium expression describing the dissociation of a salt is called the "Solubility Product Expression" and is given the symbol Ksp (solubility product constant)

Ksp = [Ca+2][F-]2
  • Ksp is the same as keq, except we are describing the equilibrium of a dissociation reaction
  • The Ksp values for many salts is given in the table of "Solubility Product Constants at 25oC"
  • Please note that the larger the Ksp value, the more soluble a salt is, and the smaller the Ksp value, the less soluble a salt is.

KNOW THE DIFFERENCE :

Solubility - the amount of substance required to make a saturated solution

Molar Solubility - the molar concentration of a saturated solution

Solubility Product - Ksp value

 

More Ksp Notes

 

There are two ways of getting a solution that is saturated :

 

1.     add a salt into a solution until it becomes saturated.  If you do this, the concentration of each ion can be determined easily if we know the concentration of the dissolved salt using stoichiometry.

CaF2(s) à  Ca+2 + 2F-

 

2.     add the ions that make up the salt into a solution until it becomes saturated.  You must be aware that if you do this, the ksp will still be the same as if you use method one, but the ion concentrations can NOT be determined from the dissociation reaction.   In other words, you must be given separate values for the concentrations of the ions, and can NOT determine them from the molarity of the dissolved salt, as you could in case one.

 

Solubility questions are easier than previous equilm questions since the salt is a solid and is not entered into the Ksp expression.

 

For each question write the following information down

·        The equilibrium equation showing the dissociation of the salt

·        The ksp expression

 

Example

 

A solution in equilibrium with a ppt of BaF2 has 4.59 x 10-2 M Ba+2 and 2.00 x 10-3 M F-.  What is the Ksp for BaF2?

 

          BaF2(s) « Ba+2 (aq)  +  2F-(aq)

 

          Ksp = [Ba+2][F-]2 = (4.59 x 10-2 M)( 2.00 x 10-3)2 = 1.84 x 10-7

 

 

Example

 

A satuated solution of BaF2 conatins 3.58 x 10-3 mol of BaF2 in 1.00 L of solution.  What is Ksp for BaF2?

 

          BaF2(s)               «     Ba+2 (aq)               +         2F-(aq)

          3.58 x 10-3 M               3.58 x 10-3M                7.16 x 10-3 M

 

          Ksp = [Ba+2][F-]2 = (3.58 x 10-3M)( 7.16 x 10-3 M)2 = 1.84 x 10-7

 

Example

 

a) What is the [Mg+2] in a saturated solution of Mg(OH)2

 

Since no values are given, you must use an x.  Also you find Ksp in your table.

 

Mg(OH)2(s)   «   Mg+2(aq) + OH-(aq)

        x                   x            2x

 

 

Ksp =  [Mg+2][ OH-]2 = (x)(2x)2 = 5.6 x 10-12

 

                   4x3 = 5.6 x 10-12

                  

                   x3 = 1.4 x 10-12

 

                   x = 1.12 x 10-4


[Mg+2] = x = 1.12 x 10-4 M

 

b)  What mass of Mg(OH)2 will dissolve in 250 mL of water.

 

          C=n/V  n = CV = (1.12 x 10-4 M)(0.25L) = 2.8 x 10-5 mol

                   2.8 x 10-5 mol   x   58.3 g   =   1.6 x 10-3

                                              1 mole

 

Do questions 42, 44, 46, 48, 50, 52, 55 on page 95 of Hebden.

Trial Ksp

 

Predicting Whether a PPT will form

 

If two ions are mixed, we can predict whether a ppt will form

 

if there is sufficient concentration of both ions to establish an equilibrium, then we can calculate a trial Ksp (or Q) and compare it to the real Ksp.

 

Q is what we have and Ksp is what we need.

 

If   AgCl(s)  «  Ag+(aq)  + Cl-(aq)

 

                   Ksp = [Ag+][Cl-]

 

 

If Q < Ksp we have less than we need to form a saturated solution, no ppt is formed

 

If Q > Ksp we have more than we need to form a satuated solution, a ppt is formed

 

If Q = Ksp we have just enough to form a ppt, the minimum possible amount of ppt is formed

 

 

Example

 

Will a ppt form when 5.0 mL of 6.0 x 10-5 M Ag+ mixes with 10.0 mL of 4.2 x 10-6 M Cl-?

 

AgCl(s)  «  Ag+(aq)  + Cl-(aq)                 ksp = 1.8 x 10-10  (from table)

 

                   Ksp = [Ag+][Cl-]

 

C1V1 = C2V2

 

C2 =   C1V1  = (6.0 x 10-5)(0.0050L) = 2.0 x 10-5 M

                   V2             (0.0150L)

C2 =   C1V1 =   (4.2 x 10-6)(0.0100L) = 2.8 x 10-6 M

                   V2             (0.0150L)

 

                   Q = [Ag+][Cl-] = (2.0 x 10-5)( 2.8 x 10-6) = 5.6 x 10-11

 

 

5.6 x 10-11 < 1.8 x 10-10  so a ppt will NOT form

 

 

Example : 

 

If 25.0 mL of 4.50 x 10-3 M Pb(NO3)2 is mixed with 35.0 mL of 2.80 x 10-3 M MgI2 will a ppt form?

 

1)     calculate the concentrations of Pb(NO3)2 and MgI2 using C1V1 = C2V2

2)     write dissociation equations and calculate the concentrations of the individual ions using stoichiometry

3)     calculate Q

4)     compare it to Ksp 

5)     a ppt will form!

 

Do questions 57, 59, 61 page 98 of Hebden.

Removing Pollution and Hardness from Water by Precipitation Methods

 

A. Removing Metal Ion Pollutants by Precipitation Methods

 

·        heavy metals such as Cu+2, Hg+2 and Pb+2 interfere with biological systems

·        the higher the concentration, the greater the toxicity

 

Example :  Waste water in the ‘tailings’ pond of a mining operation had [Cd+2] = 0.005 M.  The legal [Cd+2] is 1.0 x 10-5 M.  What [OH-]  would be required to bring [Cd+2] to acceptable values?  Ksp = 5.3 x 10-15 for Cd(OH)2

 

Cd(OH)2 ß> Cd+2(aq) + 2OH-(aq)

 

Ksp = [Cd+2][OH-] = 5.3 x 10-15

 

[OH-] =      Ksp    =  5.3 x 10-15   = 5.3 x 10-10

              [Cd+2]      1.0 x 10-10 

 

        [OH-] = 2.3 x 10-5 M

 

B. Hardness in Water

·        bitter

·        leaves white deposits (can clog pipes)

·        inhibits cleaning action of soaps by forming calcium & magnesium stearate

·        (could get rid of the ions, but is expensive)

 

i)   origins of hardness in water

·        results from Ca+2, and/or Mg+2

·        often a result of acid rain reacting with deposits of limestone

 

        CaCO3(s) + 2 H+ ß> Ca+2(aq) + CO2(g) + H2O(l) + heat

 

·        also from a reaction of atmospheric carbon dioxide dissolving in water and producing carbonic acid

 

        CaCO3(s) + CO2(g) + H2O(l) ß> Ca+2(aq) + 2HCO3-(aq) + heat

 

·        some MgCO3 also reacts (but much less)

·        rain water can travel down several hundred meters and eat away the limestone forming caverns.

 

ii)  getting rid of hardness in water

·        add washing soda  Na2CO3 10H2O  forms CaCO3 and MgCO3 which have low solubility in water, so they precipitate the unwanted ions out.

OR

·        if the water also contains HCO3-, the HC03- will react with the Ca+2 ions with heat (boiling the water)

 

        Ca+2(aq) + 2HCO3- + heat ß> CaCO3(s) + CO2(g) + H2O(l)

 

(when no HCO3- is present, it is permanently hard - need precipitation method, if it can boil away, it is temporarily hard water)

The Common Ion Effect

 

·        the solubility of an ionic compound in pure water does not change

·        if pure water is not used, it is possible to increase or decrease the solubility (but not the solubility constant) by applying Le Chatelier’s Principle

 

dissolving rxn

        AgCl(s) à Ag+(aq) + Cl-(aq)

crystallization rxn

        Ag+(aq) + Cl-(aq) à AgCl(s)

 

·        If we can increase the rate of dissolving more than the rate of crystallization, then more AgCl(s) will leave the solid phase (Solubility of AgCl(s) increases)

·        If we can increase the rate of crystallizing more than the rate of dissolving, then more Ag+ and Cl- will leave the aqueous phase and enter the solid phase (Solubility of AgCl(s) decreases)

 

A.  Decreasing the solubility of a salt

 

·        if we increase the concentration on ONE of the ions in solution, then by Le Chatelier’s Principle, he equilibrium will shift to use up some of the added ions and cause more solid to form

 

Example :

                        AgCl(s) ß> Ag+(aq) + Cl-(aq)

 

increase he [Ag+] by adding a soluble salt containing Ag+, such as AgNO3

 

or increase the [Cl-] by adding a soluble salt containing Cl-, such as NaCl

 

 

·        the lowering of the solubility of a salt by adding a second salt which has one ion “in common” with the first salt is called the common ion effect

·         used to prevent a particular salt from dissolving to any great extent or to force a particular dissolved ion to leave a solution.

 

 

B.  Increasing the Solubility of a Salt

 

·        if we decrease the concentration of one of the ions in solution, the equilibrium will shift to bring the conc of ions back up again.

 

Example : In the equilibrium AgCl(s) ß> Ag+(aq) +  Cl-(aq) we can

 

decrease the [Ag+] by adding some ion which precipitates the Ag+ present.  Look at the solubility, SO4-2, S-2, OH-, or CO3-2 will work.  Just pick one - S-2 and add Na2S.

       

                AgCl(s) ß> Ag+(aq) + Cl-(aq)

                                   +

                                S-2

 


                                Ag2S(s)

 

or, decrease the [Cl-] by adding some ion which precipitates the Cl-.  Pb+2 or Cu+ will.

 

                        AgCl(s) ß> Ag+(aq) + Cl-(aq)

                                               +

                                            Pb+2

 


                                           PbCl2(s)