# FORM SIX: CHEMISTRY STUDY NOTES-TOPIC 5: SOLUBILITY, SOLUBILITY PRODUCT AND IONIC PRODUCT

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176 TOPIC 5: SOLUBILITY, SOLUBILITY PRODUCT AND IONIC PRODUCT

PHYSICAL CHEMISTRY 1.5-SOLUBILITY

What is solubility ?
Is amount of substances dissolves in water completely so as give free ions.
Since the amount can be in gram (g)  or in moles (mol) then the S .I unit for solubility is g/L or g/dm3
Also solubility can be expressed in mol / dm3 or mol/L .  When solubility of substances is expressed in mol / dm3 and that is called molar solubility.

What is  molar solubility ?
It is amount of solute in moles dissolve in a given dm3 of of solvent to give free ions.
When solubility is expressed in it’s S.I unit  , that is the same as concentration in g /L or g /dmand when it is expressed as molar solubility , that is the same as molarity.
example. a) Define the following :
i) Solubility
ii) Molarity solubility
b) Calculate the following solubility in g /L 0.0004 M Na0H :-
i) Solubility is defined as the amount of a substance dissolve in water completely to give free ions.
ii) Molar solubility is  the amount of solute dissolves in moles in a given  dm3 of to give free ions.

Solution :-
i)Na0H Solubility = solubility in mol / dm3 x Mr
= 0.0004 x 40
= 0.016 g/L
The solubility of Na0H is 0.016 g /L

Solution:
ii) 2.7 x 10-3 mol /dm3 Ca(0H)2
Solubility  = molar  x Mr .
= 2.7 x 10 -3 x 74
= 0.1998 g/ L
The Solubility of Ca(0H)2 is  0.998 g/ L .

iii) 3.24 g of sodium chloride .
Solution:
Solubility  = Molar Solubility x Mr
= 3.24 g/L x  58.5
58.5
= 3.24 g/L
These solubility of sodium chloride = 3.24 g/L

SOLUBILITY PRODUCT OF SPARINGLY SOLUBLE SALTS
Many salts which are referred to as insoluble do infact dissolve to a small/limited extent. They are called sparingly or slightly soluble salts.

In a saturated solution, equilibrium exists between the ions and undissolved salt.

NOTE;
There is a limited number of ions that can exist together in water and this cannot be increased by adding more salts.

In a saturated solution of AgCl in contact with the ions, the equilibrium law can be applied

The concentration of solid is taken as constant at constant temperature

Kc + K = [Ag+][Cl]

Ksp = [Ag+][Cl]

Ksp = Solubility product constant.

By definition, Ksp is the product of maximum concentration of ions of sparingly soluble salt that can exist together in a solution at a given temperature.

OR
It is the product of concentration of all the ions in a saturated solution of sparingly soluble salts.
OR
Solubility product is the product of ions concentration in mol/dm3 of a certain solution raised to their stochiometric coefficient

Solubility product is denoted by Ksp

The unit for Ksp depends on the stochiometric coefficients of the respective ions of such particular substance

How to write Ksp expressions

In writing Ksp expressions one should look on

Ionization equation should be written correctly and balanced.
Stochiometric coefficients become powers of the respective ion.

Generally, for sparingly soluble salts

AxBy  xAy+ + yBx-

Ksp = [Ay+]x[Bx-]y

E.g. Write Ksp expression for the following equilibrium

(i)     Al(OH)3    Al3++ 3OH

(ii)   Ag2CrO4  2Ag+ CrO42-

Solutions:

(i)Ksp = [Al3+] [OH]3

(ii)Ksp = [Ag+]2 [CrO4 2-]

(iii)(Ca3(PO4)   3Ca2+  +  2PO43-

Ksp = [Ca2+ ] [PO4 3-]2

Significance of Ksp
(i) Ksp value is used in the prediction of occurrence of precipitates if ions in the solution are mixed.

If concentrations of ions are enough to reach Ksp value, salt precipitation occurs

Determination of Solubility product from solubility measurements

The Ksp value of the salt can be determined from its solubility in moles per litre (mol/L)

When concentrations are given in any other units such as g/L, they must be converted to mol/L

Example 1
The solubility of AgI is 1.22 x 10-8 mol/L. Calculate the Ksp for AgI

Solution

AgI(s)         Ag+(aq) + I(aq)
Each 1 mole of AgI that dissolves gives 1 mole of Ag+ and 1 mole of I in solution, concentration of each ion solution is 1.22 x 10-8 mol/L.

Hence

AgI(s)       Ag+     +        Cl

1.22 x 10-8          1.22 x 10-8

Ksp = [Ag+] [Cl]

= (1.22 x 10-8)2

Ksp = 1.4884 x 10-16 mol2L-2

Example 2
PbCl2 dissolves to a slightly extent in water according to the equation

PbCl2(s)    Pb2+(aq)  + 2Cl(aq)

Calculate the Ksp for PbCl2 if (Pb2+) has been found to be 1.62 x 10 -2 mol l-1.

Solution

PbCl2          Pb2+                    +              2Cl

1.62 x 10 -2      1.62 x 10 -2          (2x 1.62 x 10-2 )

Ksp = [Pb2+] [Cl] 2

= 1.62 x 10 -2 molL-1 x 1.0497 × 10-3 mol2L-2
Ksp = 1.7005 x 10 -5 mol3L-3  .

Example 3
The solubility of Pb(CrO4)is 4.3 x 10 -5 gl-1. Calculate the Ksp of Pb(CrO4)

(Pb =207, Cr = 52, O = 16)

Solution

PbCrO       Pb2+     +     CrO4 2-

4.3 x 10 -5gl-1     4.3 x 10 -5 gl-1    4.3 x 10 -5gl-1

To calculate the molar mass of (PbCrO4)

Pb(207) +Cr(52) +O(16)4=323gmol-1

323g     → 1mole

4.3×10-5 → x

x=1.33×10-7mol

PbCrO       Pb2+     +     CrO4 2-

1.33 x 10 -7   1.33 x 10 -7   1.33 x 10 -7

Ksp = [Pb2+] [CrO4 2-]

= (1.33 x 10-7)2

Ksp = 1.76 x 10 -14 M2

Example 4
100 ml sample is removed from water solution saturated with MgF2 at 18oC. The water is completely evaporated from the sample and 7.6mg of MgF2 is obtained. What is the Ksp value for MgF2 at 18oC

Solution

V =100 ml
m= 0.076g.

62g     →       1 mole

0.076g  →       x

x = 1.22 x 10 -3 molel-1

MgF2          Mg2+         +          2F

1.22 x 10 -3    1.22 x 10 -3         (2 ×1.22 x 10 -3 )

Ksp = [Mg2+] [F] 2

= (1.22 x 10 -3) (2.44 x 10-3)

Ksp = 7.33 x 10 -9 moll-1

DETERMINATION OF MOLAR SOLUBILITY FROM Ksp VALUE

If the Ksp value is known, the molar solubility can be obtained since Ksp shows the maximum concentration of ions which exists together in a solution.

Example 1

Calculate the molar solubility of Ag2CrO4 in water at 25oC if its Ksp is 2.4 x 10-12 M3.

Ag2CrO4(s)          2Ag+ (aq)     +    CrO42-(aq)

Let the solubility be S

Ag2CrO4(s)    2 Ag+ (aq) + CrO42-(aq)

S                   2S               S

From Ksp = [Ag+] 2 [CrO42+]

2.4 x 10 -12    = (2S) 2

S = 8.434 x 10 -5 mol L-1

Example 2
Calculate the solubility of CaF2 in water at 25oC if its solubility product is 1.7 x 10 -10 M3

Solution
CaF2     Ca 2+    +    2F

S              S              2S

Ksp = [Ca2+] [F] 2

=S (2S) 2

1.7 x 10 -10    =   4S3

S = 3.489 x 10-4 mol L

SOLUBILITY AND COMMON ION EFFECT

The solubility of sparingly soluble salts is lowered by the presence of second solute that furnishes (produce) common ions. Since the concentration of the common ion is higher than the equilibrium concentration, some ions will combine to restore the equilibrium (Le Chatelier’s principle)

Example 1
In solubility equilibrium of CaF2, adding either Calcium ions or F ions will shift the equilibrium to the left reducing the solubility of CaF2.

1. Find the molar solubility of CaF2 (Ksp = 3.9 x 10 -11 M3) in a solution containing 0.01M Ca(NO3)2

Solution

Since Ca(NO3)2 is strong electrolyte, it will dissociates completely according to the equation.

Ca (NO3)2 (s)      Ca2+ (aq)      +     2NO3(aq)

0.01                            0.01             2(0.01)

CaF2(s)              Ca2+(aq)           +        2F(aq)

S                        S                        2S

Letting solubility of CaF2 be S

CaF2(s)               Ca2+ (aq)   +      2F(aq)

S                     S             2S

0.01

From Ca (NO3)2

In absence of   Ca(NO3)2

Ksp = [Ca2+] [F]2

3.9 x 10 -11   = S (2S) 2

3.9 x 10 -11   = 4S 3

S 3 = 9.75 x 10 -12

S = 2.136 x 10 -4 molL-1

Assume the concentration of Ca2+ from Ca (NO3)2 does not affect the solubility of CaF2 then concentration of Ca2+ will be

[Ca2+]  = S + 0.01

= (2.136 x 10-4) + 0.01

0.0102 M   ≈ 0.01

Since the Ksp value is very small, the expression 0.01 + S is approximated to 0.01

Ksp = [Ca2+] [F]2

3.9 x 10 -11 = 0.01 (2S) 2

3.9 x 10 -11   = 0.04S2

S2 = 9.75 x 10 -10

S = 3.122 x 10 -5 mol/l
Conclusion
Hence, because of common ions effect, the solubility of CaF2 has reduced from 2.13 x 10 -4 M to 3.122 x 10 -5 M

Example 2
Calculate the mass of PbBr2 which dissolves in 1 litre of 0.1M hydrobromic acid (HBr) at 25oC
(Ksp for PbBr2 at 25oC is 3.9 x 10 -8 M3) (Pb = 207, Br = 80)

Solution

PbBr2        Pb2+    +         2Br

S                  S                  2S

At equilibrium

PbBr2    Pb2+    +  2Br

S                 S       2S + 0.1

2S + 0.1     = 0.1      since Ksp is very small

Ksp   = [Pb2+][Br]2

3.9 x 10 -8   =   S (0.1)2

S = 3.9 x 10 -6 moll-1

m = 1.43 x 10 -3 gl-1 of PbBr2

Example 3
The solubility product of BaSO4 in water is 10 -10 mol2l-2 at 25oC
(a) Calculate the solubility in water in moldm-3

(b) 0.1 M of Na2SO4 solution is added to a saturated solution of BaSO4. What is the solubility of BaSO4 now?

Example 4

Calculate the molar solubility of Mg (OH) 2 in

(a) Pure water

(b) 0.05M MgBr2

(c) 0.17 M KOH

(Ksp for Mg(OH)2 is 7.943 x 10 -12 M3)

(a)Solution

BaSO4 (s)       Ba2+(aq)    +      SO42-(aq)

S                      S                 S

10-10mol2l-2 = [Ba2+]  [SO4-2]

10-10mol2l-2 = S2

S = 1x 10 -5 mol/dm3

(b) Solution

Na2SO4(s)        2Na+(aq)   +         SO42-(aq)

0.1                       2 x 0.1=0.2        0.1

BaSO4 (s)            Ba2+(aq)    +    SO4 2-(aq)

S                            S                   S + 0.1                                     after adding Na2SO4

S + 0.1 0.1 Ksp is very small

Ksp = [Ba2+][SO42-]

1 x 10 -4 mol2/l2   =   S (0.1)

S = 1 x 10 -19 mol l-1

S = 1 x 10 -9 mol l-1

The value of S has reduced after adding NaSO4

1.  (a) Solution

Mg(OH)2(aq)         Mg2+(aq)    +   2OH(aq)

S                          S                     2S

Ksp = [Mg2+][OH-]2

7.943 x 10 -12   =   4S3

S3 = 1.985 x 10 -12

S = 1.25 x 10 -4 moll-1

(b) Solution
KOH(aq)            K+ (aq)    +                OH(aq)

0.17                      0.17                        0.17

Mg(OH)2      Mg2+      +   2OH

S                       S                 2S

2S + 0.17

2S + 0.17 = 0.17 Ksp is very small

Ksp =[Mg2+][OH]2

7.943 x 10 -12 M3   = S(0.17)2

S = 2.748 x 10 -10 moll-1

(c) Solution

MgBr2        Mg2+         +      2Br

0.05                0.05              2 x 0.05 ≈ 0.1

Mg (OH)2           Mg2+          +             2OH

S                           S + 0.05     2S

S + 0.05   0.05    Ksp is very small

Ksp  =  [Mg2+][OH]2

7.943 x 10 -12 M3 = (0.05)(2S)2

S2 = 3.9715 x 10 -11

S = 6.3 x 10 -6 moll

PRECIPITATION OF SPARINGLY SOLUBLE SALTS

If the equilibrium of sparingly soluble salts are approached by starting with ions in solutions and producing pure undissolved solute, then the process involved is precipitation reaction.

By definition, precipitation is the reaction where solid particles are formed by mixing ions in solution.

A substance will start precipitating as the reaction quotient (Q) becomes greater than the solubility product. Therefore, knowing the solubility product of a salt, it is possible to predict whether on mixing the solutions of ions precipitations will occur or not and what concentration of ions are required to begin the precipitation of the salt.

As for Ksp, Qsp are also given by

Qsp = [Ay+]x [Bx-]y

Where [Ay+]  and [Bx-]  are the actual ions concentrations and not necessary those at equilibrium

If Qsp  =  Ksp    the system is at equilibrium.

If Qsp < Ksp The solution is unsaturated and precipitation does not occur.

If Qsp > Ksp the solution is super saturated and precipitations occurs. REASON : So as to maintain Ksp value.

Example 1
The concentration of Ni 2+ ions in a solution is 1.5 x 10 -6 M. If enough Na2CO3 is added to make the solution 6.04 x 10 -4 M in the CO32- ions will the precipitates of Nickel carbonate occur or not?
Ksp for Ni2+      6.6 x 10 -9 M2

Solution
NiCO3          Ni2+   +    CO32-
Qsp = [Ni2+][CO2-]
= (2 x 1.5 x 10 -6)2 (6.04 x 10 -4)
Qsp = 5.4 x 10 -4 M2
Qsp >Ksp (Precipitation will occur)

Example 2
Predict whether there will be any precipitates on mixing 50cm3 of 0.001m NaCl with 50cm3 of 0.01m of AgNO3 solution. (Ksp for AgCl 1.5 x 10 -10M2)

Solution
Concentration of Cl
NaCl           Na+                  +          Cl

0.01                 0.001                     0.001
0.001    →      1000cm3
x          →      50cm3
x = 5 x 10 -5 moles (These are in 100cm3 since we added 50cm3 ofAgNO3)
Now 50 x 10 -5 moles    →      100 cm3

x                →      1000cm3

x = 5 x 10 -4 moll-1

Now for Ag+

AgNO3       Ag+         +    NO3-

0.01m              0.01m         0.01m

0.01 moles →     1000cm3

x               →        50cm3

x =5 x 10 -4 moles

5 x 10 -4moles   →      100cm3

x              →       1000cm3

x = 3 x 10-3 mol l

AgCl     Ag+   +   Cl

Qsp =  [Ag+][Cl]

Qsp = (5 x 10 -3)(5 x 10-4)

Qsp = 2.5 x 10-6 M2

Qsp > Ksp

Precipitation will occur.

Example 3
If a solution contains 0.001M  CrO42- , what concentration of Ag+ must be exceeded by adding AgNO3 to the solution to start precipitation Ag2CrO4? Neglect any increase in volume due to additional of AgNO3 (Ksp of Ag2CrO4    9.0 x 10-12M2)

Solution

Ag2CrO4       2Ag+   +   CrO42-

Ksp = [Ag+][CrO42-]

9 x 10 -12 = [Ag+]2  0.001

[Ag+]2 9 x 10 -9

[Ag+] = 9.48 x 10-5 M

Example 4
The Ksp value of AgCl at 18°C is 1 x 10 -10 mol2l-2, What mass of AgCl will precipitate if 0.585g of NaCl is dissolved in 1l of saturated solution of AgCl.

(Ag = 108, Na = 23, Cl = 35.5)

Solution

AgC  l   Ag+   +   Cl

Ksp = [Ag+][Cl]

1 x 10 -10   =   S2

S=1 x 10-5M

n  = 0.01moles

AgCl     Ag+      +       Cl-

S                  S                0.01 + S

0.01                                             S ≈ 0.01

Ksp = [Ag+][Cl]

1×10 -10 = S(0.01)

S = 1 x 10 -8 moll-1 (Solubility has decreased)

To find the amount which has precipitated,

S = 10-5    –    10-8

S = 9.99 x 10 -6M

9.99 x 10 -6 mole   →      1l

143.5g                   →   1 mole

9.99 x 10 -6

m = 1.43 x 10 -3g    of   AgCl

Question 1
Should precipitation of PbCl2 be formed when 155cm3 of 0.016M KCl is added to 245cm3 of 0.175M  Pb(NO3)2?  Ksp is 3.9 x 10 -5

Question 2
If concentration of Zn2+ in 10cm3 of pure water is 1.6  x 10 -4 M. Will precipitation of Zn(OH)2 occur when 4mg of NaOH is added.(Ksp for Zn(OH)2  is 1.2 x 10 -17)

Question 3
A cloth washed in water with manganese concentration exceeding 1.8 x 10 -6 may be stained as Mn(OH)2 (Ksp  4.5 x 10 -14) . At what pH will Mn2+ ions concentration be equal to 1.8 x 10 -6M

Solution 1
Mn(OH)2   Mn2+  +  2OH

Ksp = [Mn2+][OH]2

4.5 x 10 -14 = 1.8 x 10 -6  [OH]2

[OH-]2 = 2.5 x 10 -8

[OH-] = 1.58 x 10 -4 M

pOH  = -log[OH]

=  -log(1.58 x 10-4)

p(OH) = 3.801

pH + pOH  =14

pH = 14 – pOH

=14 – 3.801

pH = 10.199

Solution 2
We find concentration of OH- in NaOH

Qsp  >  Ksp  Hence precipitation will occurs

PRECIPITATION REACTION IN QUALITATIVE ANALYSIS (ION SEPARATION)
Qualitative analysis refers to a set of laboratory procedures  that can be used to separate and test for presence of ions in solutions. This can be done by precipitations with different reagents or by selective precipitation.

Consider an aqueous solution which contains the following metals ions Ag+, Pb2+,Cd2+  and Ni2+ which have to be separated. All the ions form very insoluble sulphides (Ag2S, PbS, CdS, NiS). Therefore sulphide is a precipitating reagent of all the above metal ions.  However, only two of them form insoluble chlorides (AgCl and PbCl) i.e aqueous HCl can be used to precipitate them while the other two ions remain in solution.

The separation of PbCl2 from AgCl is not difficult since PbCl2 dissolves in hot water while AgCl remains insoluble

The separation of Cd 2+ and Ni2+ can be done by selective precipitation with sulphide ions by considering the Ksp values of the two compounds.

Example Ksp (CdS) = 3.6 x 10 -29  and Ksp(NiS) = 3.0 x 10 -21

The Compound that precipitate first is the one whose Ksp is exceeded first (one with smaller Ksp). Suppose the solution contains 0.02M in both Cd2+ and Ni2+, the sulphide ions concentration necessary to satisfy the solubility product expression for each metal sulphide is given by

Concentration of S2- can exist in which the solution without precipitation (above which precipitation occurs)

The much smaller S2- concentration is needed to precipitate (CdS than to begin forming NiS thus CdS precipitate first before NiS.

Just before NiS begins to precipitate, how many Cd2+ remains in the solution?

Concentration of S2- needs to be slightly in excess of 1.5 x 10 -19 M for NiS to begin precipitation. The [Cd2+] that can exist in solution when the concentration of S2- ions is 1.5 x 10 -19 is given by

To find  % of Cd2+ which has precipitated.

= 99.99%

This means that we can separate Cd2+ and Ni2+ ions in aqueous solution by careful controlling concentration of S2- ions.

Question 1

1.The Ksp of AgX are (AgCl) = 1.7 x 10 -10

(AgBr) = 5.0 x 10 -13

(AgI)  = 8.5 x 10 -17

A solution contains 0.01M of each of Cl, Br, and I-. AgNO3 is gradually added to the solution. Assume the addition of AgNO3 does not change the   volume.

(a)  Calculate the concentration of Ag+ required starting precipitation of all three ions.

(b)Which will precipitate first

(c)  What will be the concentrations of  this ion when the second ion start precipitating

(d)  What will be the concentration of both ions when the third ion starts precipitation

Solution

(b)  AgI will precipitate first because the Ag concentration is very small

(c) When second  ion starts to precipitate ie AgBr start to precipitate, concentration of Ag will be

(b)For [I] when AgBr starts to precipitate

For [Br-] when AgBr starts to precipitate

Question 2
A solution contains 0.01M of Ag+  and 0.02M of  Ba2+. A 0.01M solution of Na2CrO4 is added gradually to it with a constant stirring.
(a) At what concentration of Na2CrO4will precipitation of Ag+ ions and Ba2+ starts?

(b) What will precipitate first?

(c) What will be the concentration of the first precipitated species when the precipitation of the second species starts?

(Ksp (Ag2CrO4) 2 x 10 -12 M3 , Ksp(BaCrO4)  8.0 x 10 -11 M2)

Question 3
To precipitate calcium and magnesium ions, ammonium oxalate (NH4)2 C2O4 is added to a solution i.e 0.02M in both metal ions. If the concentration of the oxalate ions is adjusted properly, the metal oxalate can be precipitated separately.

(a) What concentration of oxalate ions  (C2O42-) will precipitate the maximum amount of Ca2+ ions without precipitating Mg2+ ions.

(b) What concentration of Ca2+ ions remain when Mg2+ ions just begin precipitation.

(c) The Ksp of two slightly soluble salts, AB3 and PQ2 are each equal to 4.0 x 10 -18. Which salt is more soluble?

(d) What is the minimum volume of water required to dissolve 3g of CaSO4 at 298K

(Ksp (CaSO4) = 9.1 x 10 -6 M2)

Question 2 solution

Given [Ag+] = 0.01 M  [Ba2+] = 0.02M

For Ag2CrO4 to begin precipitating

1. Ag2CrO4 will start to precipitate since of the lower concentration of CrO42-  needed.

iii.   When BaCrO4 start to precipitate

Question 3 solution
CaSO4    Ca2+ +   SO42-

S                   S             S

Ksp = [Ca2+][SO42-]

9.1 x 10 -6 = S2

S = 3.02 x 10 -3 moll-1

Molecular  mass of CaSO4

CaSO4 = 40 + 32+ 64

=136gmol-1

136g  →  1 mol

3g  →       x

x = 0.022moles

3.02 x 10 -3 moles   → 1 dm3

0.022moles → x

x = 7.3 dm3

Question 4 solution

AB3  A3+   +   3B

S             S            3S

Ksp = [A3+][B]3

4 x 10 -18 = S(3S)3

4 x 10 -18 = 27S4

S4 = 1.48 x 10 -19

S = 1.96 x 10 -5 M   for AB3

PQ2 = P2+   + 2Q

S          S        2S

Ksp = [P2+][Q]2

4 x 10 -18 = S(2S)2

S3 = 1 x 10 -18

S = 1x 10 -6 M for PQ2

Therefore AB3 will be more soluble.