SOIL CHEMISTRY
Soil – may be defined as the unconsolidated mineral on the top layer of the earth’s crust that serves as a natural medium for the growth of land plants.
SOIL COLLOIDS:
Definition: Soil Colloids – are very small organic and inorganic parties present in the soil which are responsible for potential fertility of the soil determine the physical and chemical properties of the soil.
1. How soil colloids are formed
· As soil is formed during weathering processes, some minerals and organic matter are broken down to extremely small particles.
· Chemicals changes further reduce these particles until they cannot be seen with the naked eyes.
· The very smallest particles are what we call soil colloids.
2.
Types of soil colloids
Types of soil colloids
Soil collards may be
a) Inorganic colloids
– Clay minerals (layer silicate clay).
– Iron and Aluminium oxide clays.
– Allophane and amorphous clays.
b) Organic colloids
– Includes highly decomposed organize matter called humus.
– Organic colloids are more reactive chemically and generally have greater influence on soil properties.
Therefore the four major colloids present in the soil are,
i) Layer silicate clay
-These are most important silicate known as phyllosilicates (life-like).
– They comprised of two kinds of horizontal sheets, one dominated by silicon and other by aluminium magnesium.
ii) Iron and Aluminium
These are remnant material which remain after extensive teaching due to its low solubility these are sesquioxides
Sesquioxides are either Al oxide or iron (iii) oxide contaminated with Al(OH)3 or Fe(OH)3
iii) Allophane and other armophous minerals
– Mainly these silicates are mixture of silica and alumina.
– They are amorphous in nature.
iv) Humus
-Humus is armophous, dark-brown to black nearly insoluble in water but more soluble in dil. Alkali e.g NaOH, KOH solution.
– It consists of various chain loops of linked carbon atoms.
– Humus is a temporary intermediate product left after considerable decomposition of plant and animal remains.
-Humus contains partially dissociated enolic, carboxyl and phenolic groups.
PROPERTIES OF SOIL COLLOIDS
-Includes
i. Surface area
ii. Electric change (surface charge )
iii. Ion exchange (adsorption of cation)
i) Surface area.
Because of their small size, all soil colloids expose a large external surface area per unit mass.
· The external surface area of 1g of colloidal clay is at least 1000 times of 1 g of colloidal clay is at least 1000 times that of 1g of coarse sand.
· The external surface area of 1g of colloidal clay is at least 1000 times of 1 g of colloidal clay is at least 1000 times that of 1g of coarse sand.
ii) Electric charge
· Soil colloids surface, both external and internal characteristically carry +ve and or -ve charge.
· Most soil colloids the +ve charge predominate.
· Both organic and inorganic soil colloids when suspended in water carry a negative charge.
Where the negative charge on colloidal particles comes from?
· Negative charge on clays comes from
i) Ionizable hydrogen ions.
ii) Isomorphous substitution.
i) Ionizable hydrogen ions are hydrogen ions from hydroxyl group on clay surfaces
· The O-H bond from Al –OH or Si – O –H portion of clay heterolytically breaks and ionizes to give H+ leaving unneutralized negative charge on oxygen.
· Presence of strong alkaline solution activates the clearage of O-H bond yielding H+ which will combine with OH– from strong alkaline solution in the neutralization reaction.
Hence soil colloids becomes more negative in more alkaline solutions.The same applies to organic colloids which also contain OH in enolic, phenolic or enol.
ii) Isomorphous substitution.
This is due to the substitution of one ion for another of similar size often with lower change
iii) Ion exchange (Adsorption of cations)
As soil colloids posses negative charge, they attract the ions of an opposite charge i.e positively charge ions to the colloidal surfaces. The attraction of cations such as H+, Ca2+, and Mg2+to colloid surface leads to formations of an ionic double layer. The outer layer is made up of a swarm of rather loosely held cations attracted to the negative charged colloidal surface.
ION EXCHANGE IN A SOIL
Colloids are primarily responsible for chemical reactivity in soil
Each colloid has net negative charge thus making possible for the colloid to attract and hold positively charged particles.(cations) like Na+, H+, Ca2+ and Mg2+
– Thus there are cations attached to colloids and in the soil.
– When one of the cations in the soil solution replaces one of the cations on the soil colloids cation exchange is said to take place.
– This exchange only take place when the cations in the soil solution are not in equilibrium to the cations on the soil colloid.
Definition
Ion exchange :
Is reversible reaction which involves an interchange of ions between ion in soil solution and another ion or surface of soil colloid.
It can be anion exchange or cation exchange.
Cation exchange:
Is the interchange between a cation in a solution and another cation on surface of any negatively charged material such as day or organic matter.
In the soil:Cation exchange is the interchange between cation in a soil solution and another cation on the surface of soil colloid.
e.g
Anion exchange
· Soil colloids being negatively charged cannot attract and hold negatively charged particles (like charged repels) like SO4 and NO3
Why? Nitrate is more leached from the soil than ammonium.
· This is because nitrate (NO3) has negative charge like soil colloids. So NO3 is not held by the soil solution to be leached under rainfall conditions.
· NH4 being positively charged is attracted and held by soil colloids and hence it become difficult for NH4 to be leached.
Mechanism of ion exchange in the soil.
· Ion exchange in the soil is well explained by electron –kinetic theory of ion exchange.
According to the theory:
The negative and positive charge associated with soil colloids (clay minerals and organic matter) are balanced by electrostatic attraction of cations and anions respectively.
The balancing ions are turned as EXCHANGEABLE CATIONS OR ANIONS.
· Exchangeable cations and anions form outer sphere complexes with charged surfaces in which water of hydration exist between the charged ion and the oppositely charged colloids.
· Thus the adsorbed cations and anions are said to being state of oscillation forming a diffuse double layer.
· Due to these oscillations, some of the ions move away from the surface of the colloid micelles.
· In presence of the solution of an electrolyte an ion of the soil solution slips in between the inner charged layer and the outer oscillating ion.
· The ion in the soil solution is now adsorbed on the colloid micelles and the surface in remains in solution as an exchange ion and hence the ion exchange occurs.
Factors affecting composition of exchangeable ions in the ion exchange
Explain the factors affecting composition of exchangeable ions in the soil.
i. Strength of adsorption
ii. Relative concentration of the ion in the soil solution.
CATION EXCHANGE CAPACITY (CEC)
– Is the maximum quantity of total cations of any class, that a soil is capable of holding at a given PH value, available for exchange with soil solution.
– It is a measure of the quantity of the negatively charged sites on soil surfaces that can retain positively charged ions by electrostatic force.
Significance of CEC
CEC is useful in the following ways:
It is a measure of soil fertility
-The more cation exchange capacity a soil has the more likely the soil will have fertility.
It is a measure of nutrient retention capacity
-Soil with large value of CEC has large nutrient retention capacity.
It is a measure of the capacity of soil to protect ground water from cation contamination.
-Soil with large CEC has good ability to protect ground water from cation contamination as it exerts more resistance for its cations to be leached away.
Expression of CEC value
It is expressed in terms of number of equivalents (or more specifically as number of millequivale
nts) of cations per 100 grams of dry soil written as:
nts) of cations per 100 grams of dry soil written as:
e.g 100g (meq /100g)
No of equivalents= 
————- (i)
————- (i) Equivalent means the mass of cation that will replace (exchange) 1g (1mole) of H+.
means the mass of cation that will replace (exchange) 1g (1mole) of H+.E.g
23g of Na+ (i.e 1mole of Na+) exchange 1g of 1+ and hence Na+ has equivalent wt of 23g.
4og of Ca2t (I mole of Ca2+) exchange 2g (2moles of H+ which means 20g of Ca2+ exchange 1g of H+
of Ca2+ exchange 1g 
of H+Hence Ca2+ have equivalent wt of 20.
equivalent 
t= 
——— (iii)
= no of moles of the cation number of equivalents =
x Amount of ionic charge Where no of mill moles = No of moles x 1000
No of mill moles = x 1000
x 1000=
Factors affecting value of cation exchange capacity.
Cation exchange capacity is affected by the following factors:
– Amount of clay (soil texture)
– Type of clay
– Soil organic matter
– PH of the soil.
i) Amount of clay in soil
A high silicate- clay soil hold more exchangeable cations than a low silicate clay soil and hence have greater CEC value.
Silicate clay (Si – (OH) easily lose H+ off Hydroxy group in basic medium resulting into net negative charge i.e the greater the –ve charge the greater the CEC
(ii) Clay types
Different clay types have different CEC value due to difference in surface area.
(iii)Soil Organic Matter
(iii)Soil Organic Matter
Organic matter in the Soil, also are negative charged. Therefore
– High – Organic Soil have greater CEC value than a low organic Soil.
– On another hand if an organic matter continue to decay, the CEC tends to decrease with the decomposition.
– They can retain more cations.
Therefore organic matter particles have greater CEC than clay particles.
(iv) PH of the soil.
PH affects CEC of colloids that are charged based on hydroxyl group instead isomorphous substitution.
Under acidic (low PH ) condition.
H+ are in excess resulting to less ionization of the colloids, less negative charge and hence low CEC value.
Under basic (High PH ) condition.
OH– are in excess resulting to more ionization of the colloids , more negative charge on the colloids surface and hence high CEC value.
PERCENTAGE BASE SATURATION
Acid cations and Base cations
Acid cations are exchangeable cations (mainly (H+ and AI3+) which tend to acidify the soil
– They are also known as exchangeable acids
– In very acidic soil, H+ and AI3+ dominates other adsorbed cations.
– The acidity of AI3+ is explained by its cationic hydrolysis according to the following equation:
· Base cations (or exchangeable bases) are exchangeable cations which are capable of neutralizing soil acidity . Common exchangeable bases are Ca2+, Mg2+, K+ and Na+
Base saturation
Base saturation is the fraction of exchangeable cations that are base cations.
-It is expressed as percentage and hence the name percentage base saturation.
Percentage base saturation =
Relationship between percentage base saturation and percentage acid saturation
Percentage base saturation + percentage acid saturation
= x 100%
x 100%but number of base cations + Number of acid cations= Total number Of exchangeable cations = CEC of the soil
= x 100% =100%
x 100% =100% Therefore
Percentage base saturation + percentage acid = 100% Saturation.
Q. 1. (a) Define the following terms as applied to soil:
(i) Cation Exchange Capacity
(ii) Percentage Base saturation
(iii)Salinity
(b) A soil sample has a CEC of 25meq per 100g of 200mg of soil sample were shaken with 40c of 0.1M HCl. After filtering and washing the soil the filtrate and washing were titrated against NaOH solution, 24.0cm3 of 0.1 M NaOH were required for complete neutralization. Calculate the percentage base saturation of the soil sample.
of 0.1M HCl. After filtering and washing the soil the filtrate and washing were titrated against NaOH solution, 24.0cm3 of 0.1 M NaOH were required for complete neutralization. Calculate the percentage base saturation of the soil sample. (c) A soil sample (20g) was analyzed and found to contain 0.0015g of calcium, what is the concentration of Ca in the soil sample in meq/100g of soil.
1eq of Na + 23g = 0.023meq
ANSWERS:
(b) Solution:
Remaining HCl will react with NaOH hence HCl was in excess
Used = 0.004 – 0.0024
= 0.0016moles of H+ (No of moles which attached themselves to the soil colloid)
Now we know 1eq = 1 mole of H+
0.0016 eq was in 200g of soil
? 100g
0.0016 e.g. = 200
? = 100g
8 x 10-4 x 1000 = 0.8meq/100g
P. B.S. = 3.2%
Example 4
A Soil test shows the following:
Nutrient | meq / 100 soil |
Ca2+ | 9.9 |
Mg2+ | 2.1 |
K+ | 2.0 |
Al3+ | 7.6 |
NH4+ | 0.6 |
Na+ | 0.1 |
a) Calculate the CEC of the Soil.
b) Calculate the percent base saturation of the Soil
c) Calculate the percent aluminium saturation of the soil
SOLUTION:
a) CEC of the Soil
= number of exchangeable base cations + number of exchangeable acid cations.
= (9.9 + 2.1 + 2.0 + 7.6 + 0.6 + 1.0) meq = 22.3 meq
= (9.9 + 2.1 + 2.0 + 7.6 + 0.6 + 1.0) meq = 22.3 meq
Hence CEC of the soil is 22.3 meq
b) From a given cations, are acid cations
= (22.3 – (0.6 + 7.6)) = 14.1
(Recall: CEC = Number of exchangeable base cation + Number of exchangeable acid cations, and therefore number of exchangeable base cation = CEC – Number of exchangeable acid cations)
(Recall: CEC = Number of exchangeable base cation + Number of exchangeable acid cations, and therefore number of exchangeable base cation = CEC – Number of exchangeable acid cations)
Using; Percent base saturation
Hence percent base saturation is 63.23
Hence percent base saturation is 63.23
Percent aluminium saturation
Hence aluminium base saturation is 34.1%
Hence aluminium base saturation is 34.1%
SOIL REACTION – It is the acidity or alkalinity of the soil. The soil reaction can be acidic, neutral or alkaline due to the soil solution. It is
measured in PH using electrometric methods (PH – meter). All soil PH range from 4 to 8. Soils with PH
4 generally contain sulphuric acids while those with PH
8 contain high percentage of Na+ and thus are alkaline. Importance of soil PH
- Most crop plant prefers PH to be between 6 and 7.5 i.e slightly acidic up to slightly alkaline.
- The availability of nutrients to plant depends on soil PH. All plant nutrients are reasonably available between PH 6 and 7.5 which is the optimum PH. The availability of N, P, K, S,Ca, Mg decreases with the increase in soil acidity. Thus acidic soils are often deficient of these nutrients.
E.g. Nutrient phosphorus is found as phosphate (v) in the soil. At PH 5, the soluble iron, aluminium and manganese react with phosphates (v) to form insoluble complexes and then fix them (makes them unavailable for plants).
5, the soluble iron, aluminium and manganese react with phosphates (v) to form insoluble complexes and then fix them (makes them unavailable for plants). Fe, Mn and Cu precipitate at high PH. Thus deficiency of these nutrients often limits plants growth in alkaline soils. Manganese and Iron become plant toxic at low PH.
The problem with heavy metals toxic to humans arises near industrial area due to acid rain.
E.g. Ladmium and lead, below PH 4 get dissolved and may enter plant and make crops unfit for human consumption
Conclusion:
Determination of PH of agricultural soil is very important for the selection of suitable crops. Below the PH 4.8 a soil may need liming in order to avoid aluminum toxicity.
CAUSES OF SOIL ACIDITY
. Soil acidity is caused when heavy rains leach bases like Ca2+, Mg2+, K+ and Na+ from the soil to the ground water table leave surplus H+ in the soil.
. Soil acidity is caused when heavy rains leach bases like Ca2+, Mg2+, K+ and Na+ from the soil to the ground water table leave surplus H+ in the soil.
. Also industrial regions may bring sulphuric (vi) acid and nitric (v) acid to the soil which increases its acidity.
. Acidic mineral fertilizers like ammonium sulphate (vi) and ammonium chloride make the soil more acidic due to hydrolysis.
. Also the nitrification of ammonium ions by bacteria produces H+(aq)
NOTE: Organic acids are produced during the decomposition of organic matter. Due to this, most soils in humid tropics are Acidic.
Liming & liming materials
Discuss:
– Meaning and significance of liming as treatment to soil PH.
– The efficiency of liming materials i.e. neutralizing values of carbonates, oxides, hydroxides and silicates.
– Beneficial effects of liming
– Detrimental effects of over-liming
Liming:
It refers to the process of adding basic compounds of calcium and magnesium to acid soil (PH 5) in order to raise the PH of the soil to the required level.
The compound of calcium and magnesium (oxides, hydroxides, carbonates and silicates) are called agricultural limes. Such compounds are contained in limestone, dolomite, building (slaked) lime, oyster shells and blast furnace slag which can be used for liming soil if finely grind.
5) in order to raise the PH of the soil to the required level. The compound of calcium and magnesium (oxides, hydroxides, carbonates and silicates) are called agricultural limes. Such compounds are contained in limestone, dolomite, building (slaked) lime, oyster shells and blast furnace slag which can be used for liming soil if finely grind.
Significance of Liming:
- To raise the soil PH
Q. 1. Explain similarities and differences between industrial fertilizers and manure
Q. 2. To mention the advantages of straight and mixed fertilizers of N, P and K
Q. 3. Mention examples of straight and mixed fertilizers and classify them
Q. 4. Compare the relative advantages and disadvantages of manures as compare industrial fertilizers.
How to maintain and improve soil fertility:
A fertile soil provides all essential plant nutrients in amounts which are suitable for the growth of most plants.
How can a soil be made more fertile
1. Good cropping system
2. Adding manure to the soil
3. Adding industrial fertilizers to the soil
3. Adding industrial fertilizers to the soil
1.GOOD CROPPING SYSTEM:
Crop rotation should be practiced. Every season, another crop should be planted on a given field. Legumes should be rotated with cereal
s, shallow rooted plants with deep rooted ones .
s, shallow rooted plants with deep rooted ones .
2.ADDING MANURE:
Manures are organic materials that can be added to the soil to increase soil fertility.
E.g. Kraal manure – from cattle kraal
Farmyard manure – wastes from animals
Compost manure-mixture of soil and decomposing organic matter
Bio gas manure – from effluent of bio gas plants
3.ADDING INDUSTRIAL FERTILIZERS:
– Fertilizers are most inorganic compounds which contain one or more plant nutrients in a concentrated form.
E.g. Nitrogen fertilizers, phosphorus fertilizers, potassium fertilizers.
Similarities between manure and industrial fertilizers.
– They both increase humus to the soil and promote good soil structure
– Both manure and industrial fertilizers contain minerals Nitrogen, phosphorus and potassium which are essential for plant growth.
Differences between manure and industrial fertilizers
Manure | |
1. It is organic material | 1. It is inorganic compound |
2. It is composition varies depending on the type of manure | 2. It is composition is fixed |
3. It is less toxic | 3. It is toxic if used in large amounts |
4. It has low contents of nutrients | 4. It has high content of nutrients |
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 /dm3 and 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.
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 /dm3 and 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
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.
Stochiometric coefficients become powers of the respective ion.
Generally, for sparingly soluble salts
AxBy xAy+ + yBx-
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)2 3Ca2+ + 2PO43-
Ksp = [Ca2+ ]3 [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
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.
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–
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 dissolves to a slightly extent in water according to the equation
PbCl2(s) Pb2+(aq) + 2Cl–(aq)
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–
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 .
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)
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
PbCrO4 Pb2+ + CrO4 2-
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
PbCrO4 Pb2+ + CrO4 2-
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
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.
m= 0.076g.
62g → 1 mole
0.076g → x
x = 1.22 x 10 -3 molel-1
MgF2 Mg2+ + 2F–
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)
2Ag+ (aq) + CrO42-(aq)Let the solubility be S
Ag2CrO4(s) 2 Ag+ (aq) + CrO42-(aq)
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
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
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 –
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.
In solubility equilibrium of CaF2, adding either Calcium ions or F– ions will shift the equilibrium to the left reducing the solubility of CaF2.
i. 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)
Ca2+ (aq) + 2NO3(aq)0.01 0.01 2(0.01)
CaF2(s) Ca2+(aq) + 2F–(aq)
Ca2+(aq) + 2F–(aq) S S 2S
Letting solubility of CaF2 be S
CaF2(s) Ca2+ (aq) + 2F–(aq)
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 expres
sion 0.01 + S is approximated to 0.01
sion 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
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)
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—
Pb2+ + 2Br— S S 2S
0.1 After adding HBr
At equilibrium
PbBr2 Pb2+ + 2Br
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
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)
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)
2Na+(aq) + SO42-(aq)0.1 2 x 0.1=0.2 0.1
BaSO4 (s) Ba2+(aq) + SO4 2-(aq)
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
4. (a) Solution
Mg(OH)2(aq) Mg2+(aq) + 2OH–(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)
KOH(aq)
K+ (aq) + OH–(aq)0.17 0.17 0.17
Mg(OH)2 Mg2+ + 2OH–
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–
Mg2+ + 2Br–0.05 0.05 2 x 0.05 ≈ 0.1
Mg (OH)2 Mg2+ + 2OH–
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
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)
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)
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–
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
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-
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–
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)
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-
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.
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–
Ag+ + Cl–Ksp = [Ag+][Cl–]
1 x 10 -10 = S2
S=1 x 10-5M
n = 0.01moles
AgCl Ag+ + Cl-
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
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)
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
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–
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
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?
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.
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)
ANSWERS
Question 2 solution
Given [Ag+] = 0.01 M [Ba2+] = 0.02M
For Ag2CrO4 to begin precipitating
ii. 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-
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–
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.
