Title BIOCHEM 231 Practical Manual.pdf ✅

Exercise No. : 01 Preparation of Solutions, pH and Buffers Standard solution means a solution which the concentration is known, irrespective of the manner in which its concentration is expressed. The concentration of solutions may be expressed in a number of ways, the general methods are : 1. The per cent solutions 2. Molarity and or molar solutions. * Atomic weight of element is proportional to the actual weight of an element, hence it is customary to express in grams. * The atomic weight of an element expressed in gram, is called Gram atomic weight or simply gram atom. * Similarly, Molecular weight of any given substance is simply the Gram molecular weight or gram-mole. Gram mole = Mole Molarity (M) : The conc. of solute in a solution expressed as the number of moles of solute per liter of solution. Moles of solute Molarity = Litre of solution Molality (m) : The conc. of solute in a solution expressed as the number of moles of solute per kilogram of solvent. Moles of solute Molality = Kilogram of solvent Equivalent weight : Equivalent weight of an element may be defined as that quantity of the element in grams which reacts with or displace 1 gram atom of hydrogen or 1 gram atom of oxygen or 1 gram of chlorine. Molecular weight Gm Equivalent weight = Replaceable H+ or OH- or Cl- Solution : A solution consists of a Solute (one substance) dissolved in Solvent (another substance).
Concentration : The concentration of a solution is the amount of the solute dissolved per unit volume or weight of the solvent. The concentration of solution can be expressed in a number of ways : (a) Gram of solute per 100 gm of the solvent (w/w) per cent (b) Gram of solute per 100 ml of the solvent (w/v) per cent (c) Gram of solute per 1000 ml (litre) of the solvent (w/v) Molar (d) Gram of solute per 1000 ml (litre) of the solvent (w/w) Molal (e) Milli litre (ml) per 1000 ml (litre) of the solvent (v/v) Normal solution (N) : A Normal solution (N) is one which is prepared by dissolving 1 gram equivalent of the solute per 1000 ml (litre) of the solution. The unit symbol N is used to denote "mol/L". Sometimes "Ëq/L" also used.. Molecular weight Equivalent weight = Valency Amount of substance gm/L of solution Normality = Eq. weight of substance. Chemical substance Equivalent weight 1 N = Eq. wt. x1 2 N = Eq. wt. x 2 0.5 N = Eq. wt.x 0.5(N/2) = Eq. wt /2 1 N = Eq. wt. x 0.1 (N/10) = Eq. wt. /10 HCl 36.45 36.45 72.90 18.23 3.65 H2SO4 49.01 49.01 98.02 24.51 4.90 H3PO4 32.64 32.64 65.29 16.32 3.26 NaOH 40.00 40.00 80.00 19.99 4.00 Ca(OH) 2 37.03 37.03 74.05 18.51 3.70 Na2CO3 52.97 52.97 105.94 26.49 5.30 Equivalent weight of acids : The equivalent weight of an acid is that weight of acid which contains one replaceable H atom. Thus the molecular weight itself is the equivalent weight for monobasic acids, for dibasic acids it is mole weight divided by two, for tri basic molecular weight divided by three --------
Acid Mol. weight Valency Eq. wt. HCl 36.45 1 36.45 H2SO4 98.02 2 49.01 H3PO4 97.93 3 32.64 Equivalent weight of bases : The equivalent weight of an base is that weight of acid which contains one replaceable OH group. 17.008 g OH are equivalent to 1.008 grams of H. Thus, equivalent weight of NaOH, KOH, NH4OH are equal to its molecular weight, whereas equivalent weight of Ca(OH)2, Ba(OH) 2 the molecular weight divided by two. Base Mol. weight Valency Eq. wt. NaOH 40.00 1 40.00 Ca(OH)2 74.05 2 37.03 Na2CO3 105.94 2 52.97 Equivalent weight of acids : The equivalent weight of an acid is that weight of it which contains one gram equivalent weight of replaceable hydrogen (all acids contains replaceable H atoms). Example 1 : One mole of HCl contains one replaceable hydrogen atom, hence the gram equivalent weight of HCl (1:1) i.e. (1 + 35.5 = 36.5). 36.5 Equivalent weight of HCl =--------------------- = 36.5 gm 1 Example 2 : One mole of H2SO4 contains two replaceable hydrogen atom, hence the equivalent weight of H2SO4 (2:1:4) i.e. (2 + 32.6 + 63.96 = 98.56). 98.565 Equivalent weight of H2SO4 =------------------= 49.28 gm 2 Equivalent weight of Bases : The gram equivalent weight of base is that weight of the base, which contains one gram equivalent weight of the hydroxyl ion (all acids contains replaceable H atoms). Example 1 : One mole of NaOH contains one replaceable hydroxyl ion, hence the gram equivalent weight of NaOH (1:1:1) i.e. (23 + 16 + 1 = 40). 40.00
Equivalent weight of NaOH = -------------------- = 40.00 gm 1 Example 2 : One mole of Ca(OH)2 contains two replaceable hydroxyl ion, hence the equivalent weight of Ca(OH)2 (1:2:2) i.e. (40.02 + 31.98 + 2 = 74). 74 Equivalent weight of Ca(OH)2 =-------------= 37 gm 2 Dilution of acids : Example 1 : Prepare 500 ml, 1 N solution of Sulphuric acid (H2SO4) from concentrated acid (purity 99 %, Specific gravity 1.84 and Molecular weight 98.08). Eq wt. of H2SO4 = 49.04 (98.08 / 2 i.e. replaceable H) Eq. wt. of acid x V2 x Normality x 100 Required vol. of conc. acid = - (VI) for the solution 1000 x Sp. Gravity x Purity (%) 49.04 x 500 x 1 x 100 VI = 1000 x 1.84 x 99 = 13.46 ml Answer : 13.46 ml. conc. H2SO4 is to be diluted in 500 ml distilled water in a volume flask to get 1 N solution of H2SO4 . Example 2 : Prepare 750 ml, 0.1 N solution of Orthophosphoric acid (H3PO4) from conc. acid (purity 93 %, Sp. Gr. 1.75, and Mol. Wt. 98.00). Eq wt. of H3PO4 = 32.66 (98.00 / 3 i.e. replaceable H) Eq. wt. of acid x V2 x Normality x 100 Required vol. of conc. acid = - (VI) for the solution 1000 x Sp. Gravity x Purity (%) 32.66 x 750 x 0.1 x 100 VI = 1000 x 1.75 x 98 = 1.43 ml Answer : 1.43 ml. conc. H3PO4 is to be diluted in 750 ml distilled water in a volume flask to get 0.1 N (N/10) solution of H3PO4 . Parts per million (ppm) concept : ppm solutions are those solutions in which a substance is present in a very small quantity.