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40 Engineering Chemistry and Environmental Studies
SOLUTIONS
2.1 DEFINITION OF SOLUTION, SOLVENT AND SOLUTE
When a small amount of sugar (solute) is mixed with water, sugar uniformally dissolves in water and a sugar solution is obtained. In this solution, sugar molecules are uniformly dispersed in molecules of water. Similarly, a common salt (NaCl) when dissolve in water, if uniformly disperse in water and salt solutions is obtained so a solution of salt in water consist of ions of salt (Na+, Cl –^ ) dispersed in water. Solutions are homogenous mixtures in which one substance is said to have been dissolved in the other. The dissolved substance may be present as individual molecules or ions throughout the other substance. Since both the components of a solution are present in the molecular or ionic state, it constitutes a perfectly uniform and transparent system.
Components of Solution
In the study of solution, it is customary to designate the components in solution. The components are:
Solvent The component present in larger proportion is known as solvent.
Solute The component present in smaller proportion is known as solute. Solution = Solvent + Solute Ex. Sugar solution = Sugar (solute) + Water (solvent) Common salt solution = Salt (solute) + Water (solvent)
2.2 TYPES OF SOLUTION
Homogenous Solution
A solution in which two substances are mixed has uniform composition and the components cannot be identified separately.
Ex.
Sugar (solute) Sugar solutions — Two substances Water (solvent)
Heterogenous Solution
A solution in which two or more substances are mixed has non-uniform composition and the components can be identified separately.
(40)
Solutions (^) 41
Ex.
Naphthalene (solute) Naphthalene solution — Two substances Water (solvent)
In naphthalene solution, both water and naphthalene can be identified and separated from one another and non-uniform in their properties like density, concentration and viscosity etc.
Aqueous Solution
Solution containing water as solvent is called aqueous solution.
Ammoniacal Solution
Solution containing ammonia as solvent is called ammoniacal solution.
Non-aqueous Solution
Solution containing solvent other than water is called non-aqueous solution.
Classification of Solutions
Based on the physical state of components, the solutions are classified into gaseous solutions, liquid solutions and solid solutions.
1. Gaseous solutions: In these solutions, gas is solvent and the solute may be solid or liquid or gas. ( a ) Gas in Gas Ex. H 2 and O 2 mixture, air ( b ) Liquid in Gas Ex. Water in air ( c ) Solid in Gas Ex. Camphor in air 2. Liquid solutions: In these solutions, liquid is the solvent and the solute may be solid or liquid or gas. ( a ) Gas in Liquid Ex. Soda water (CO 2 in water) ( b ) Liquid in Liquid Ex. Alcohol in water ( c ) Solid in Liquid Ex. Salt in water 3. Solid solutions: In these solutions, solid is the solvent and the solute may be solid or liquid or gas. ( a ) Gas in Solid Ex. H 2 palladium ( b ) Liquid in Solid Ex. Hg in Zn ( c ) Solid in Solid Ex. Alloys (Zn in Cu)
Solutions (^) 43
∴ 2 gm of H^2 is equal to one mole of H^2. The gram molecular weight of Na 2 CO 3 is 106 gms. ∴ 106 gms of Na 2 CO^3 is equal to 1 mole of Na 2 CO^3. If we know the gram molecular weight of any substance, we can find the number of moles.
Number of moles =
Weight of the substance Gram molecular weight Example 1: Find the number of present in 196 gms of H 2 SO 4. Solution: The gram molecular weight of H 2 SO 4 = 98 Weight of H 2 SO 4 = 196
Number of moles of H 2 SO 4 =
Weight of the substance Gram molecular weight
=
196 98 = 2 196 gms of H 2 SO 4 has 2 moles. Example 2: What is the weight of 0.2 moles of AgNO 3. Solution: Number of moles of AgNO 3 = 0. Gram molecular weight of AgNO 3 = 108
Number of moles of AgNO 3 =
Weight of the substance Gram molecular weight
0.2 = 3
Weight of AgNO 108 ∴ Weight of AgNO 3 = 0.2 × 108 = 21.6 gms
2.4. MOLARITY
It is indicated by ‘M’. ‘It is the number of moles of solute present in one litre of solutions’ Molarity = Number of moles/1 litre of solution
Molarity =
Weight of the substance (^) 1 litre of solution Gram molecular weight
∴ Molarity =^
Weight of the substance (^) × 1000 Gram molecular weight Volume in ml
or M =
W (^) × 1000 gr. mol. wt. V in ml If the molarity is given we can find out the weight of the substance by using the following equations.
44 Engineering Chemistry and Environmental Studies
W = M × gr. mol. wt. × V in ml 1000 For dilution, molarity relation M 1 V 1 = M 2 V (^2) where M 1 = Molarity of concentrated solution V 1 = Volume of concentrated solution M 2 = Molarity of dilute solution V 2 = Volume of dilute solution For volumetric analysis, molarity relation 1 1 1
M V n =^
2 2 2
M V n where M 1 = Molarity of Ist solution V 1 = Volume of Ist solution n 1 = Number of moles of Ist solution M 2 = Molarity of IInd solution V 2 = Volume of IInd solution n 2 = Number of moles of IInd solution The unit of molarity is moles/lit. Molarity depends on temperature. If temperature increases, molarity decreases.
2.5 NORMALITY (N)
The number of gram equivalents of the solute present in 1 litre or 1000 ml of solution at a given temperature is called normality (N).
Normality (N) =
No. of gram equivalents of solute Volume of solution in litres
or N = (^) V in litres
x
where x = No. of gram equivalents of solute It is calculated by the following relation
x =
weight of solute in gram (W) gram equivalent weight of solute (GEW)
Then N =
W (^) × 1 GEW V in litres
or N =
W (^) × 1000 GEW V in litres Normality relation for dilution, N 1 V 1 = N 2 V (^2) N 1 = Normality of concentrated solution V 1 = Volume of concentrated solution
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Table 2.3: Equivalent weight of some Bases Name of the base Formula of base Acidity Equivalent weight Sodium hydroxide NaOH 1 40/1 = 40 Potassium hydroxide KOH 1 56/1 = 56 Ammonium hydroxide NH 4 OH 1 35/1= 35 Magnesium hydroxide Mg(OH) 2 2 58/2 = 29 Calcium hydroxide Ca(OH) 2 2 74/2 = 37
3. Equivalent weight of a salt: It is the ratio of molecular weight to the total valency of cations or anions.
ESalt =
Molecular weight Total valency of cations or anions Table 2.4: Equivalent weight of some Salts Name of the salt Formula of salt Total valency of Equivalent weight cations or anions Sodium chloride NaCl 1 58.5/1 = 58. Sodium carbonate Na 2 CO 3 2 106/2 = 53 Magnesium chloride MgCl 2 2 95/2 = 47. Magnesium sulphate MgSO 4 2 120/2 = 60 Calcium carbonate CaCO 3 2 100/2 = 50 Silver nitrate AgNO 3 1 170/1 = 170 Copper sulphate CuSO 4 2 159.5/2 = 79.
4. Equivalent weight of an oxidising agent: It is the ratio of molecular weight to the number of electrons gained
E (^) OA =
Molecular weight No. of electrons gained Table 2.5: Equivalent weight of some oxidising agents Name of the Formula of No. of Equivalent compound compound electrons gained weight Potassium permanganate KMnO 4 5 (in acidic medium) 158/5 = 31. Potassium permanganate KMnO 4 3 (in neutral) 158/3= 52. Potassium dichromate K 2 Cr 2 O 7 6 294/6 = 49
5. Equivalent weight of a reducing agent: It is the ratio of molecular weight to the number of electrons lost.
E (^) RA =
Molecular weight No. of electrons lost
Solutions (^) 47
Table 2.6: Equivalent weight of some reducing agents Name of the Formula of No. of Equivalent compound compound electrons lost weight
Mohr’s salt FeSO 4 (NH 4 ) 2 SO 4 × 6H 2 O 1 392/1 = 392 Hypo Na 2 S 2 O 3 1 158/1 = 158 Oxalic acid H 2 C 2 O 4 × 2H 2 O 2 126/2 = 63 Ferrous sulphate FeSO 4 × 7H 2 O 1 278/1 = 278
6. Equivalent weight of an element: It is the ratio of atomic weight to the valency
Eele =
Atomic weight Valency Table 2.7: Equivalent weight of some elements Element Symbol Valency Equivalent weight Sodium Na 1 23/1 = 23 Magnesium Mg 2 24/2 = 12 Aluminium Al 3 27/3 = 9 Silver Ag 1 108/1 = 108 Ferrous Fe +2^2 56/2 = 28 Ferric Fe +3^3 56/3 = 18. Zinc Zn 2 65.4/2 = 32. Cupric Cu +2^2 63.5/2 = 31. Potassium K 1 39/1 = 39
2.7 NUMERICAL PROBLEMS ON MOLARITY, NORMALITY
Problem 1: 2 moles of a solute is dissolved in 5 litres of solution. What is its molarity? Solution: Number of moles of solute ( n ) = 2 Volume of solution (V) = 5 litres M =?
M = (^) V^ n^ = 25 = 0.4 M
Problem 2: Find the number of moles of solute present in 500 ml of 0.2 M solution. Solution: Number of moles of solute ( n ) =? Volume of solution (V) = 500 ml = 0.5 lit. Molarity of solution (M) = 0.
M = (^) V^ n n = M × V = 0.2 × 0.5 = 0.
Solutions (^) 49
M =
W (^) × 1 GMW V in litres W = M × GMW × V = 0.2 × 60 × 2 = 24 g Problem 7: Find the volume of water required to prepare 0.1 M H 2 SO 4 from 200 ml of 0.5 M solution.
Solution: M 1 = 0. V 1 = 200 ml M 2 = 0. V 2 =? M 1 V 1 = M 2 V (^2) 0.5 × 200 = 0.1 × V 2 V 2 = 1000 ml Volume of water required V 2 – V 1 = 1000 – 200 = 800 ml Problem 8: 300 ml of water is added to 200 ml of 0.5 M HCl solution. Calculate the molarity of dilute solution.
Solution: M 1 = 0. V 1 = 200 ml V 2 = 300 + 200 = 500 ml M 2 =? M 1 V 1 = M 2 V (^2)
M 2 = 1 1 2
M V
V =^
0.5 × 200 500 = 0. Problem 9: 9.8 grams of H 2 SO 4 dissolved in 2 litres of water calculate the normality of solution. Solution: Wt. of solute (W) = 9.8 g Volume of solution (V) = 2 lit. GMW of H 2 SO 4 = 98 g Basicity of H 2 SO 4 = 2
GEW of H 2 SO 4 =
98 2 = 49 g Normality (N) =?
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N =
W (^) × 1 GEW V in litres
=
9.8 (^) × 1 49 2 =
1 10 or^ 0.1 N Problem 10: Find the normality of solution prepared by dissolving 10 grams of NaOH in 500 ml of water.
Solution: W = 10 g V = 500 ml GMW of NaOH = 40 g Acidity of NaOH = 1
GEW of NaOH =
40 1 = 40 g N =?
N =
W (^) × 1000 GEW V in ml
=
(^10) × 1000 40 500 =^
1 2 or^ 0.5 N Problem 11: Calculate the weight of Na 2 CO 3 present in 100 ml of 0.5 N solution. Solution: W =? V = 100 ml N = 0. GMW of Na 2 CO 3 = 106 g
GEW of Na 2 CO 3 =
106 2 = 53 g
N =
W (^) × 1000 GEW V in ml
W =
N × GEW × V 1000 =^
0.5 × 53 × 100 1000 = 2.65 g Problem 12: Calculate the weight of oxalic acid (H 2 C 2 O 4 × 2H 2 O) required to prepare 0.05 normal solution in 2 litres.
Solution: W =? N = 0. V = 2 litres
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Q. 1. Explain the terms solvent and solute with suitable examples. Ans. The larger component of solution is called solvent and the smaller component is called solute. Example: If 5 grams of a salt dissolved in 100 ml of water, salt solution is formed. In this solution, salt is solute and water is solvent. Q. 2. What are the saturated, unsaturated and supersaturated solutions? Ans. A solution containing maximum amount of the dissolved solute at a given temperature is called saturated solution. A solution in which the amount of dissolved solute is less than that required for saturation is called unsaturated solution. A solution containing more amount of dissolved solute than that required for saturation is called supersaturated solution. Q. 3. What is molarity? Calculate molarity of a solution prepared by dissolving 5.85 g of NaCl in 500 ml of water. Ans. The number of moles of solute present in 1 litre of solution at a given temperature is called molarity. W = 5.85 g V = 500 ml GMW = 58.5 g M =?
M =
W (^) × 1000 GMW V in ml
=
5.85 (^) × 1000 58.5 500 = 0. Q. 4. What is normality? Find the normality of a solution prepared by dissolving 1.58 g of KMnO 4 is 200 ml of water. (GMW of KMnO 4 is 158 g) Ans. The number of gram equivalents of solute dissolved in 1 litre of solution is called normality W = 1.58 g V = 200 ml GMW = 158 g
GEW =
158 5 =^ 31.6 g N =? (KMnO 4 is an oxidising agent. It takes five electrons)
N =
W (^) × 1000 GEW V in ml
=
1.58 (^) × 1000 31.6 200 =^
1 4 or^ 0.
Questions and Answers
Solutions (^) 53
Q. 5. Calculate molarity and normality of a solution prepared by dissolving 10.6 g of Na 2 CO 3 in 2 litres of water. Ans. W = 10.6 g GMW = 106 g V = 2 lit.
GEW =
106 2 =^ 53 g
M =
W (^) × 1 GMW V in litres
=
10.6 (^) × 1 106 2
=
1 20 or^ 0.05 M
N =
W (^) × 1 GEW V in litres
=
10.6 (^) × 1 53 2
=
1 10 or^ 0.1 N
