Some other concepts
Basicity of an acid –The number of replaceable hydrogen atoms present in one molecule of an acid is called basicity of the acid. For example, HCl, H2SO4,H3PO4 have one, two and three respectively. They are said to be monobasic, dibasic and tri-basic acids respectively. It should be noted that acetic acid CH3COOH ,is a monobasic acid because it has only one replaceable hydrogen atom which is attached to oxygen atom. The H atoms from CH3 group are not replaceable.
Acidity of a base –The number of replaceable hydroxyl group present in one molecule of an alkali (or base ) is called the acidity of the base. For example, NaOH, Ca(OH)2 and Al(OH)3 have acidity one, two and three respectively. They are said to be monoacidic, diacidic and triacidic bases respectively.
Molecular weight (or Mass ): While expressing the concentration of a solution, molecular weight or mass of a substance is required. Molecular weight (or mass ) of a substance is defined as the average relative weight ( or mass ) of one molecule of the substance as compared to the weight ( or mass ) of one atom of carbon which is taken to be 12 µ. Gram molecular weight or Gram Mole sometimes. (For convenience of calculations, another term called ’gram mole’ is also used). The quantity of a substance expressed in grams numerically equal to the molecular mass is called gram mole or mole of a substance.
Equivalent weight (or mass ): In order to express the concentrations of acids, bases, oxidizing and reducing agents and to carry out the stoichiometric calculations in their reactions, a term other than atomic weight and molecular weight is introduced. This term is called equivalent weight. The weight of a substance which combines with or replaces from a chemical combination 1.008 parts by weight of hydrogen or 35.5 parts by weight of chlorine is called the equivalent weight (or mass ) of a substance.
Equivalent weight is also called a chemical equivalent of the substance. Equivalent weight is an experimental quantity and it changes according to the experimental conditions. Gram equivalent weight or Gram Equivalent Sometimes, for convenience of calculations, another term called ’gram equivalent’ is also used.
The quantity of a substance expressed in grams numerically equal to the equivalent weight (or mass) is called gram equivalent or chemical equivalent of a substance. Relation between molecular weight and equivalent weight of an acid and base For an acid, Equivalent weight = Molecular weight / Basicity of the acid For example, Eq. Wt. of HCl = 36.5 / 1 µ Eq. Wt. of H2SO4 = 98 / 2 µ Eq. Wt. of H3PO4 = 98 / 3 µ For a base, Equivalent weight = Molecular weight / acidity of the base For example, Eq. Wt. of KOH = 56 / 1 µ Eq. Wt. of Ca(OH)2 = 74 / 2 µ Eq. Wt. of Al(OH)3 = 78 / 3 µ Methods of expressing concentration of a solution Following methods are used to express the concentration of a solution.
1. Mass percentage –In this method, the amount of the solute is expressed in grams per 100 grams of the solution. Thus, a 10 % solution of NaCl in water means 10 grams of NaCl is dissolved in 90 grams of water, resulting in a solution of mass 100 grams. Then the solution is 10 % by mass.
2. Parts per million –When a solute is present in very small amount, it is convenient to express its concentration in parts per million, abbreviated as ppm. It is defined as the mass of the solute present in one million (106) parts by mass of the solution. Parts per million (ppm) = (Mass of solute × 106 Mass of the solution)
3. M = (No of moles of solute/ Volume of the solution in liters)
= (Mass of solute/ Molar mass of solute X volume of solution in liters)
4. Molality (m) - The no of moles of solute dissolved in 1 kg (1000 gm) of the solvent is called the molality of the solution, denoted by m.
Molality = (No of moles of solute X 1000)/ (Molar mass of solute X mass of solvent in kg)
= (Mass of solute X 1000)/ (Molar mass of solute X Mass of solvent in kg)
Since mass is independent of temperature, molality is a temperature independent unit.
5. Normality (N) - The number of gram equivalents of the solute per liter ( dm3 ) of solution is called the normality of the solution. It is denoted by N.
Normality = (Number of equivalents of solute Volume of solution in liters) = (Mass in grams of solute Gram Eq. Mass of solute x Volume of solution in liters).
6. Standard solution –A solution in which a known amount of solute is dissolved in a known amount of solution is called a standard solution. In other words, a solution of known concentration is a standard solution. A 1N solution or 1M solution is also a standard solution.
Relation between Normality and Molarity
For an acid, Normality = (Basicity × Molarity of monobasic like HCl or HNO3) i.e. Normality = 1 × Molarity; for Molarity of diabasic like H2SO4, Normality = 2 × Molarity; for tribasic like H3PO4, Normality = 3 × Molarity.
For a base Normality = Acidity × Molarity monoacidic like KOH Normality = 1 × Molarity; for Molarity of diacidic like Ca(OH)2, Normality = 2 × Molarity; for triacidic like Al(OH)3, Normality = 3 × Molarity
Deduction of normality equation
Concentrations of solutions expressed in terms of normality are commonly used in analytical work involving titrations. For the analytical work not involving acid –base reactions or titrations, concentration of solution is usually expressed in molarity. In neutralization reactions, volumes of acids and bases are inversely proportional to their normalities. Let us derive an equation showing the relationship between normality and volume of acid and base solutions.
Let N1 and V1 be the normality and volume of the acid solution and N2 and V2 be the normality and volume of the base solution. Then
V1 ∝ 1 / N1 So V1 = k / N1 —- (1) where k is a proportionality constant or N1V1 = k
V2 ∝ 1 /N2 So V2 = k / N2—- (2) where k is a proportionality constant or N2V2 = k
Dividing equation (1) by (2) ,
we get V1 /V2 = N2 / N1 or N1V1 = N2V2
Normality x Volume (For an acid) = Normality x Volume (For a base)
N1V1 = N2V2
Thus the product of normality and volume of the reacting acid and base is equal. This equation is called the normality equation.
In a titration experiment, the volumes of reacting acid and base solutions are known. If the normality of one solution is given, the normality of the other solution can be found out
Another relation which is useful in finding out the strength of solution in a titration experiment is
Normality × Equivalent Weight = Strength in grams per liter