In the previous section, we saw solubility product constant. We saw some solved examples also. In this section we will see two more solved examples. After that, we will see common ion effect on solubility. We will also see the effect of pH on solubility
Solved example 7.92
Equal volumes of 0.002 M solutions of
sodium iodate and cupric chlorate are mixed together. Will it lead to
precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10-8)
Solution:
1. Let V be the original volume of NaIO3 (sodium iodate)
• In 1 liter, there will be 0.002 moles of NaIO3
♦ So in V liters, there will be 0.002V moles of NaIO3
2. Cu(ClO3)2 (cupric chlorate) will be having the same volume V
• In 1 liter, there will be 0.002 moles of Cu(ClO3)2
♦ So in V liters, there will be 0.002V moles of Cu(ClO3)2
3. The two solutions are mixed together. So the volume of the final mixture will be 2V
• Then molarity of NaIO3 in the final mixture
= No. of moles⁄Volume = 0.002V⁄2V = 0.001
• Similarly, molarity of Cu(ClO3)2 in the final mixture
= No. of moles⁄Volume = 0.002V⁄2V = 0.001
4. Dissociation equations:
• NaIO3 will dissociate as: NaIO3 (s) ⇌ Na+ (aq) + IO3- (aq)
♦ NaIO3 will completely dissociate to give the following concentrations:
✰ [Na+] = 0.001
✰ [IO3-] = 0.001
• Cu(ClO3)2 will dissociate as: Cu(ClO3)2 (s) ⇌ Cu2+ (aq) + 2ClO3- (aq)
♦ Cu(IO3)2 will completely dissociate to give the following concentrations:
✰ [Cu2+] = 0.001
✰ [ClO3-] = (2 × 0.001) = 0.002
5. Cu2+ and IO3- will combine together to give Cu(IO3)2 (cupric iodate):
Cu2+ + 2IO3- ⇌ Cu(IO3)2
6. The solid Cu(IO3)2 thus formed will be in equilibrium with it's two ions:
Cu(IO3)2 (s) ⇌ Cu2+ (aq)+ 2IO3- (aq)
6. Q of this reaction is given by: Q = [Cu2+][IO3-]2 = (0.001)(0.001)2 = 10-9
◼ This Q is less than Ksp of Cu(IO3)2
7. So the system will try to increase Q
• That means, more and more Cu(IO3)2 will dissolve
• The precipitation of Cu(IO3)2 will not occur
Solved example 7.93
What
is the maximum concentration of equimolar solutions of ferrous sulphate
and sodium sulphide so that when mixed in equal volumes, there is no
precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3× 10-18)
Solution:
1. Let V be the original volume of FeSO4 (ferrous sulphate)
• Let x be the molarity of the solution
⇒ In 1 liter, there will be x moles of FeSO4
⇒ In V liters, there will ve Vx moles of FeSO4
2. Let the same V be the original volume of Na2S (sodium sulfide)
• Let the molarity be the same x
⇒ In 1 liter, there will be x moles of Na2S
⇒ In V liters, there will ve Vx moles of Na2S
3. The two solutions are mixed together. So the volume of the final mixture will be 2V
• Then molarity of FeSO4 in the final mixture
= No. of moles⁄Volume = Vx⁄2V = 0.5x
• Similarly, molarity of Na2S in the final mixture
= No. of moles⁄Volume = Vx⁄2V = 0.5x
4. Dissociation equations:
• FeSO4 will dissociate as: FeSO4 (s) ⇌ Fe2+ (aq) + SO42- (aq)
♦ FeSO4 will completely dissociate to give the following concentrations:
✰ [Fe2+] = 0.5x
✰ [SO42-] = 0.5x
• Na2S will dissociate as: Na2S (s) ⇌ 2Na+ + S2-
♦ Cu(IO3)2 will completely dissociate to give the following concentrations:
✰ [Na+] = (2 × 0.5x) = x
✰ [S2-] = 0.5x
5. Fe2+ and S2- will combine together to give FeS (iron sulphide):
Fe2+ + S2- ⇌ FeS
6. The solid FeS thus formed will be in equilibrium with it's two ions:
FeS (s) ⇌ Fe2+ (aq) + S2- (aq)
6. Q of this reaction is given by: Q = [Fe2+][S2-] = (0.5x)(0.5x) = 0.25x2
◼ This Q must be less than or equal to Ksp
If it is greater than Ksp, precipitation willtake place
7. To find the maximum aloowable value, we will equate the two:
Q = Ksp
⇒ 0.25x2 = 6.3 × 10-18
⇒ x = 5.02 × 10-9
Solved example 7.94
The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen
sulphide is 1.0 × 10-19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of
the following: FeSO4, MnCl2, ZnCl2 and CdCl2. In which of these solutions
precipitation will take place?
Solution:
1. Finding the concentration of S2- (sulphide ions):
(i) The original concentration of S2- is 1.0 × 10-19 M
⇒ 1 liter will contain 1.0 × 10-19 moles of S2-
⇒ 1 mL will contain (1.0 × 10-19 ÷ 1000) = 1.0 × 10-22 moles of S2-
⇒ 10 mL will contain (1.0 × 10-22 × 10) = 1.0 × 10-21 moles of S2- ions
(ii) This 10 mL is added to 5 mL each of matallic solutions
⇒ The final volumes of each metallic solution is (10 + 5) = 15 mL
⇒ 1.0 × 10-21 moles of S2- is present in 15 mL of solution
⇒ No. of moles of S2- in 1 mL = (1.0 × 10-21 ÷ 15) = 6.67 × 10-23
⇒ Concentration of S2-
= No. of moles in 1 liter = (6.67 × 10-23 × 1000) = 6.67 × 10-20 moles
⇒ Concentration of S2- in the 15 mL solution = [S2-] = 6.67 × 10-20 M
2. FeSO4 will completely dissociate in aqueous solution
• So 0.04 M FeSO4 will give 0.04 M Fe2+ ions
• Let us find the concentration of Fe2+:
(i) The original concentration of Fe2+ is 0.04 M
⇒ 1 liter will contain 0.04 moles of Fe2+
⇒ 1 mL will contain (0.04 ÷ 1000) = 4.0 × 10-5 moles of Fe2+
⇒ 5 mL will contain (4.0 × 10-5 × 5) = 20.0 × 10-5 moles of Fe2+ ions
(ii) This 5 mL is added to 10 mL of sulphide solutions
⇒ The final volumes of each metallic solution is (10 + 5) = 15 mL
⇒ 20.0 × 10-5 moles of Fe2+ is present in 15 mL of solution
⇒ No. of moles of Fe2+ in 1 mL = (20.0 × 10-5 ÷ 15) = 1.33 × 10-5
⇒ Concentration of Fe2+ = No. of moles in 1 liter
= (1.33 × 10-5 × 1000) = 1.33 × 10-2 moles
⇒ Concentration of Fe2+ in the 15 mL solution = [Fe2+] = 1.33 × 10-2 M
3. In a similar way, MnCl2, ZnCl2 and CdCl2 will also completely dissociate into ions. So we get:
[Mn2+] = [Zn2+] = [Cd2+] = 1.33 × 10-2 M
4. Consider the solution of FeSO4
• The Fe2+ will combine with S2- to form FeS
• The Q of this reaction = [Fe2+][S2-] = (1.33 × 10-2 × 6.67 × 10-20) = 8.87 × 10-22
◼ From the data book, we have: Ksp of FeS = 6.3 × 10-18
♦ So Q has to increse. That means, the concentration of ions must increase
♦ So more and more FeS will dissolve. It will not precipitate
5. Consider the solution of MnCl2
• The Mn2+ will combine with S2- to form MnS
• The Q of this reaction = [Mn2+][S2-] = (1.33 × 10-2 × 6.67 × 10-20) = 8.87 × 10-22
◼ From the data book, we have: Ksp of MnS = 2.5 × 10-13
♦ So Q has to increse. That means, the concentration of ions must increase
♦ So more and more MnS will dissolve. It will not precipitate
6. Consider the solution of ZnCl2
• The Zn2+ will combine with S2- to form ZnS
• The Q of this reaction = [Zn2+][S2-] = (1.33 × 10-2 × 6.67 × 10-20) = 8.87 × 10-22
◼ From the data book, we have: Ksp of ZnS = 1.6 × 10-24
♦ So Q has to decrease. That means, the concentration of ions must decrease
♦ So more and more solid ZnS has to form. It will precipitate
7. Consider the solution of CdCl2
• The Cd2+ will combine with S2- to form CdS
• The Q of this reaction = [Cd2+][S2-] = (1.33 × 10-2 × 6.67 × 10-20) = 8.87 × 10-22
◼ From the data book, we have: Ksp of CdS = 8.0 × 10-27
♦ So Q has to decrease. That means, the concentration of ions must
decrease
♦ So more and more solid CdS has to form. It will
precipitate
Common ion effect on solubility
Some basics can be written in 3 steps:
1. Consider the general form of an ionic equilibrium:
Mxp+Xyq- ⇌ Mxp+ + Xyq-
♦ On the left side of the equation, we have the salt
♦ On the right side of the equation, we have the ions
• The solid salt will be in equilibrium with the ions
2. If we increase the concentration of one of the ions, the equilibrium is disturbed
• Then according to Le Chatelier’s principle,the system will try to decrease the effect of those extra ions
• For that, some of the extra ions will combine with the opposite ions to give the salt
• This leads to precipitation of the salt
3. If we decrease the concentration of one of the ions, then also, equilibrium is disturbed
• Then according to Le Chatelier’s principle,the system will try to decrease the effect of those depleted ions
• For that, some of the solid will dissociate into the solution
• This leads to a decrease in the quantity of solids in the solution
Let us see an example. It can be written in 7 steps:
1. A vessel contains Ag+ (silver ions) in a solution
• We want to know how much silver ions are present in that solution
2. For that, take about 10 mL of that solution
• To that 10 mL, add enough Cl- ions
3. The equilibrium is disturbed due to the excess Cl- ions
• In order to reduce the effect of excess Cl- ions, they combine with Ag+ ions to form AgCl
• The reaction is: Ag+ + Cl- ⇌ AgCl
♦ AgCl is insoluble in water. It will precipitate
4. If there is enough Cl- ions, all the Ag+ in that 10 mL will precipitate as solid AgCl
• Solid AgCl is sparingly soluble in water. So it can be easily separated form the solution
5. The mass of that solid AgCl is measured
◼ From that mass, we get an important information:
The quantity of Ag+ ions in the 10 mL solution
6. If quantity in the 10 mL solution is known, we can easily calculate:
The quantity of Ag+ ions in the original solution
◼ This method of finding the quantity of a substance is called gravimetric estimation
7. Let us see two more examples:
• We can calculate the quantity of ferric ions
♦ For that, ferric ions are made to precipitate as ferric hydroxide
♦ Ferric hydroxide is sparingly soluble in water
• The quantity of barium ions
♦ For that, barium ions are made to precipitate as barium sulphate
♦ Barium sulphate is sparingly soluble in water
Solved example 7.95
Calculate the molar solubility of Ni(OH)2 in 0.10 M NaOH. The ionic product of
Ni(OH)2 is 2.0 × 10-15
Solution:
1. The dissolution of Ni(OH)2 will be according to the equation:
Ni(OH)2 (s) ⇌ Nip+ (aq) + 2OHq- (aq)
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of Ni(OH)2 dissolve in water,
♦ 1 mole of Nip+ is produced
♦ 2 moles of OHq- are produced
3. So when S moles of Ni(OH)2 dissolve in water,
♦ S moles of Nip+ are produced
♦ 2S moles of OHq- are produced
4. But in our present case, 0.10 moles of OHq- is already present
♦ This extra OHq- ions is due to the dissociation of 0.10 M NaOH
• Then we can write:
Ksp = [Nip+]1[OHq-]2 = (S)1(2S+0.10)2
• Here, 2S will be very small (when compared to 0.10) because, the solubility of Ni(OH)2 is very small
♦ So we can approximate (2S+0.10) as 0.10
5. So we can write:
Ksp = [Nip+]1[OHq-]2 = (S)1(2S+0.10)2 = (S)1(0.10)2
⇒ 2.0 × 10-15 = 0.01 × S1
⇒ S = 2.0 × 10-13
Effect of pH on solubility
This can be explained with the help of an example. It can be written in 11 steps:
1. Consider the sparingly soluble salt CaCO3
• It’s dissociation can be written as:
CaCO3 (s) ⇌ Ca2+ (aq) + CO32- (aq)
2. CO32- attracts one proton from the water molecule to become HCO3-
♦ When the water molecule loses one proton, it becomes OH-
3. So there are two reactions taking place:
♦ CaCO3 (s) ⇌ Ca2+ + CO32-
♦ CO32- + H2O ⇌ HCO3- + OH-
4. The net reaction can be obtained by adding the two. We get:
CaCO3 (s) + H2O ⇌ Ca2+ + HCO3- + OH-
5. The presence of OH- ions, make the solution basic
♦ So it’s pH will be greater than 7
6. If we add more OH- ions, the system will try to reduce the effect by favoring the backward reaction
• So more and more solid CaCO3 will be produced
• In effect, the CaCO3 will become more and more insoluble
7. Remember that:
When we add more OH- ions, we are increasing the pH
◼ So we can write:
When we increase the pH, CaCO3 which is originally sparingly soluble becomes even more insoluble
8. If in step (6), instead of OH- ions, we add H3O+ ions, those extra H3O+ ions will react with OH- ions to form water
• The reaction is: H3O+ + OH- ⇌ 2H2O
9. Thus the OH- ions are removed
• The system will try to reduce the effect of this OH- depletion by favoring the forward reaction
• So more and more solid CaCO3 dissolves
10. Remember that:
When we add more H3O+ ions, we are decreasing the pH
◼ So we can write:
When we decrease the pH, CaCO3 which is originally sparingly soluble, becomes more soluble
11. This is the reason for the damages suffered by limestone (CaCO3) statues due to acid rains
•
In the next chapter, we will see redox reactions
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