Normality is a concentration expression. It is defined as the gram equivalent weights of solute in 1 liter of solution. It is calculated using the following equations:
Units of Normality are equivalent/ liter (Eq/ liter). Units of equivalent weight are gram/ equivalent (g/ Eq).
Example 1. Calculate the equivalent weight for sodium chloride (molecular weight = 58.5 g/ mole), and for tribasic potassium phosphate,K3PO4 , (molecular weight = 212 g/ mole).
Answer 1. Both sodium and chloride has a valence of 1 in sodium chloride. Thus, number of equivalents per mole is equal to 1 and the equivalent weight is equal to the molecular weight. However, K3PO4 has three potassium ions each with a valence of 1 or one phosphate ion with a valence of 3. Thus, number of equivalents per mole for the compound is equal to 3 and the equivalent weight = 212/ 3 =70.7 g/ Eq.
Example 2. What is the equivalent weight for Ca3(PO4)2 (molecular weight = 310 g/ mole).
Answer 2. The compound has 3 ions of calcium each with a valence of 2 or it has 2 phosphate ions each with a valence of 3. Thus the total valence in relation to calcium ions or in relation to phosphate ions is 6. Accordingly, equivalent weight = 310/ 6 = 51.7 g/ Eq.
Example 3. Calculate the equivalent weight for monobasic potassium phophate ,KH2PO4 (molecular weight = 136 g/ mole), in relation to potassium, to hydrogen, and to phophate.
Answer 3. The equivalent weight for this compound depends on what ion is considered in the calculation. Thus, it has three different equivalent weights; one in relation to K+, the second in relation to H+ and the third in relation to PO4-3 . In relation to K+ the equivalent weight = 136/ 1 = 136 g/ Eq. In relation to H+ the equivalent weight = 136/ 2 = 68 g/ Eq. In relation to PO4-3 the equivalent weight = 136/ 3 = 45.3 g/ Eq.
Example 4. Calculate the weight of calcium chloride dihydrate ,CaCl2.2H2O , (molecular weight = 147 g/ mole) needed to prepare 1 liter of 0.005 Eq/ liter solution.
Answer 4.
Equivalent weight = 147/ 2 = 73.5 g/ Eq
Equivalents needed to prepare 1 liter of the solution = 0.005 Eq
Weight needed to prepare 1 liter of the solution = 73.5 g/ Eq × 0.005 Eq = 0.3675 g =367.5 mg
Example 5. Calculate the Normality of a potassium chloride solution , KCl, (molecular weight = 74.6 g/ mole) when 1.5 g of the salt are dissolved to prepare 100 ml.
Answer 5.
Equivalent weight = molecular weight = 74.6 g/ Eq
Number of equivalents = 1.5 / 74.6 = 0.02 Eq
Normality = 0.02 Eq/ 0.1 liter = 0.2 Eq/ liter
Note: 100 ml = 0.1 liter
1cc of 0.1N HCl is added to 999cc of aqueous solution of NaCl .the pH of the resulting will be?
ReplyDeleteConcentration 1 × volume 1 = Concentration 2 × volume 2
ReplyDelete0.1 N × 1 ml = Concentration 2 × 1000 ml
Concentration 2 = 0.0001 N (this is the concentration of HCl after dilution which is the same as the hydronium ion concentration)
Assuming that NaCl has no significant effect on activity of hydronium ions in solution then,
pH = - log [H3O+] = -log (0.0001) = 4.0