Formality (F): Wt.(g) /F.Wt. F = F.Wt = Formula weight V(L) There is no difference between a substance’s molarity and formality if it dissolves without dissociating into ions. Ex: The molar concentration of a solution of glucose is the same as its formality, since it is not dissociated in the solution to other form.

For substances that ionize in solution, such as NaCl, molarity and formality are different. Ex: Dissolving 0.1 mol of NaCl in 1L of water gives a solution containing 0.1mol of Na+ and 0.1 mol of Cl–. The molarity of NaCl, therefore, is zero since there is essentially no undissociated NaCl in solution. While the formality of NaCl, however, is 0.1 F because it represents the total amount of NaCl in solution.

Chemical Stoichiometry The stoichiometry of a reaction is the relationship among the number of moles of reactants and products as shown by a balanced equation. Thus, the equation: 2NaI (aq) + Pb(NO3)2 (aq) PbI2 (s) + 2NaNO3 (aq) Indicates that 2 mol of aqueous sodium iodide combine with 1 mol of aqueous lead nitrate to produce 1 mol of solid lead iodide and 2 mol of aqueous sodium nitrate.

Stoichiometry steps: 1- When the weight (mass) of reactant or product is given, the weight is first converted to the number of moles using the M.Wt 2- The stoichiometric ratio given by the chemical equation for the reaction is then used to find the number of moles of another reactant or product. 3- Finally, the weight of the other reactant or product is computed by using M.Wt.

Ex: Calculate the weight in gram of AgNO3 (M.Wt = 170 g/mol) required to convert 2.33g of Na2CO3 (M.Wt = 106 g/mol) to Ag2CO3 . 2AgNO3 + Na2CO3 Ag2CO3 (s) + 2NaNO3

Ex: How many (V(ml)) of 0. 1M AgNO3 (M Ex: How many (V(ml)) of 0.1M AgNO3 (M.Wt = 170 g/mol) will be needed to convert 0.348g Na2CO3 (M.Wt = 106 g/mol) to Ag2CO3 ? 2AgNO3 + Na2CO3 Ag2CO3 (s) + 2NaNO3

Ex: La(NO3)3 + 3KIO3 La(IO3)3 (s) + 3KNO3 M.Wt of La(NO3)3 = 325 , KIO3 = 213 , and La(IO3)3 = 664 a) What Wt(g) of KIO3 will react with 0.05 mol of La(NO3)3 ? b) How many grams of La(NO3)3 are required to react completely with 2.15 g of KIO3 ? c) How many milligrams of KIO3 are required to react completely with 2 g of La(NO3)3 ? d) How many grams of La(IO3)3(s) are formed when 3 g of KIO3 are mixed with excess La(NO3)3 ?

Equilibrium constant: Are algebraic equation that relate the concentration of reactants and products in a chemical reaction to numerical quantity called equilibrium constant. Equilibrium constant expression is important because they permit the chemist direction and completeness of chemical reaction. Ke is temperature dependence.

The following table shows that dependence of Kw on temperature. Temperature, ˚C Kw 0.114 × 10-14 25 1 × 10-14 50 5.47 × 10-14 100 49 × 10-14 For convenience Kw written as 1 x 10-14 at 25o C.

Ex: Calculate the hydronium and hydroxide ion concentration of pure water at 25˚C and 100 ˚C?

Ex: Calculate the hydronium and hydroxide ion conc., pOH and pH in 0.2 M aqueous NaOH at 25oC.

Ex: Calculate the hydronium and hydroxide ion conc., pH and pOH in 0.05 M aqueous HCl at 25oC.

Ex: If Ksp for Hg2Cl2 is 1.3 x 10-8 calculate concentration of [Hg22+] in the solution.

Ex: How many grams of Ba(IO3)2 [M.Wt = 487 g/mole] can be dissolved in 500mL of water? Ksp of Ba(IO3)2 = 1.57 × 10-9.

The effect of a common ion on the solubility of a precipitate Generally the solubility of precipitate is decreased in the solution which contains the same common ion with precipitate ions. AgCl is given as an example. Present in silver chloride AgCl are silver ions (Ag+) and chloride ions (Cl¯). Silver nitrate AgNO3 (which is soluble) has silver ion in common with silver chloride. Sodium chloride NaCl (also soluble) has chloride ion in common with silver chloride. The AgCl precipitate solubility was decreased in solution which contains an excess of silver and chloride ions because the soluble part of AgCl make an equilibrium state with the solid part of the salt.

Ex: When added pb(IO3)2 for 0.005 M solution of KIO3 , calculate conc. [pb2+] in the solution at equilibrium. (Ksp for pb(IO3)2 = 2.6 x 10-13 ).

Ex: When added Ba(IO3)2 for 0.02 M solution of Ba(NO3)2 , calculate the solubility of Ba(IO3)2 . (Ksp for Ba(IO3)2 = 1.57 x 10-9 ).