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Equivalent Weights and Normality

Equivalent Weights and Normality. Normality - the number of equivalents (equivalent weights) of solute per liter of solution. Normality ( N ) = # eq. / L of solution Molarity ( M ) = # moles/ L of solution

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Equivalent Weights and Normality

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  1. Equivalent Weights and Normality • Normality - the number of equivalents (equivalent weights) of solute per liter of solution. • Normality (N) = # eq. / L of solution • Molarity (M) = # moles/ L of solution • Equivalent weight of an acid/Base - the mass of the acid (g) that could furnish 6.022*1023 (1 mole) protons (H+)/hydroxide ions (OH-) • hydrochloric acid - 1 mol HCl provides 1 mol of protons • Therefore, a 1 M HCl solution is a 1 N solution • sulfuric acid - 1 mol H2SO4 provides 2 moles of protons • Therefore, a 1 M H2SO4 solution is a 2 N H2SO4 solution • phosphoric acid - 1 mol H3PO4 provides 3 moles of protons • Therefore, a 1 M H3PO4 solution is a 3 N H3PO4 solution

  2. Sodium hydroxide - 1mol NaOH provides 1 mol OH- • Therefore, a 1M NaOH solution is a 1N NaOH solution • Barium hydroxide - 1mol Ba(OH)2 provides 2 mol OH- • Therefore, a 1M Ba(OH)2 solution is a 2N Ba(OH)2 solution • Sodium Carbonate - 1mol Na2CO3 provides 2 mol OH- • Therefore, a 1M Na2CO3 solution is a 2N Na2CO3 solution • ? N = M * (# eq./mol) • What is the normality of a 0.6589 M diprotic tartaric acid, (C2H4O2)(COOH)2? • N = 0.6589 mol/L * (2 eq./mol) = 1.318 N tartaric acid

  3. Why normality? • One mole of an acid doesn’t necessarily neutralize one mole of a base • NaOH + H2SO4  NaHSO4 + H2O • One equivalent of an acid always neutralizes one equivalent of a base • 1 mol protons + 1 mol hydroxide ions  1 mol H2O • Therefore, when acids and bases are given in normality we can use: • Nacid * Vacid = Nbase * Vbase • in acid base neutralization reactions. • What is the normality of a sulfuric acid solution if 30.00mL of it are necessary to neutralize 24.00mL of a 0.2000 N NaOH soln? • Nacid * 30.00mL = 0.2000 N * 24.00mL • N = 0.1600 N H2SO4

  4. Equivalent Weights • Equivalent weight of an acid/Base - the mass of the acid (g) that could furnish 6.022*1023 (1 mole) protons (H+)/hydroxide ions (OH-) • Equivalent weight = molecular/formula weight * (moles/eq.) • What is the equivalent weight of NaOH? • Eq. weight = 40.00 g/mol NaOH*(1mol NaOH/1 eq.) = 40.00g/eq • What is the equivalent weight of diprotic oxalic acid (COOH)2? • Eq. weight = 90.04g/mol (COOH)2*(1mol/2eq)= 45.02g/eq. • What is the normality of a solution prepared with 35.25 g of oxalic acid in 2.000L of solution? • N = (35.25g/2.000L) * (1 eq./45.02g) = 0.3915 N oxalic acid

  5. Balancing oxidation-reduction reactions • 1/2 reaction method: • 1)Write as much of the overall unbalanced equation as possible. • 2) Construct unbalanced oxidation and reduction 1/2 rxns • 3) Balance all atoms except H and O. Then balance the H and O. • 4) Balance the charge on each 1/2 rxn by adding electrons • 5) Balance the electron transfer • 6) add the resulting 1/2 rxns and eliminate any common terms

  6. Balancing oxidation-reduction reactions • Change in oxidation state method: • 1) Write as much of the overall unbalanced equation as possible. • 2) Assign oxidation numbers to the elements undergoing oxidation and reduction. • 3) Draw a bracket to connect atoms of the elements being oxidized and reduced. • 4) Balance the atoms being oxidized and reduced & then balance the electron transfer • 5) Balance remaining atoms and charge by inspection

  7. 1) Nitrite ions oxidize metallic copper to copper(II) ions and are reduced to nitrogen oxide in acidic soln. • 2) balance the follow redox equation in acidic solution. • MnO4- + H2S  S8 + Mn2+ • 3) Hydrogen peroxide can be oxidized to oxygen gas by AuCl4- in acidic soln. Metallic gold and chloride ions form in the process. • 4) In basic solution, chlorate ions can oxidize dinitrogen tetroxide to nitrate ions forming chloride ions as the other product.

  8. 1) Nitrate ions oxidize metallic copper to copper(II) ions and are reduced to nitrogen oxide in acidic soln. • 2NO3- + 8H+ + 3Cu  2NO + 3Cu2+ + 4H2O • 2) balance the follow redox equation in acidic solution. • MnO4- + H2S  S8 + Mn2 • 16MnO4- + 48H+ + 40H2S  5S8 + 16Mn2+ +64H2O • 3) Hydrogen peroxide can be oxidized to oxygen gas by AuCl4- in acidic soln. Metallic gold and chloride ions form in the process. • 3H2O2 + 2AuCl4- 3O2 + 8Cl- + 2Au + 6H+ • 4) In basic solution, chlorate ions can oxidize dinitrogen tetroxide to nitrate ions forming chloride ions as the other product. • ClO3- + 3N2O4 + 6OH-6NO3- + Cl- + 3H2O

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