VIEWS: 19 PAGES: 22 POSTED ON: 2/2/2012 Public Domain
Normality Dr. Steve Badger Concentrations of Solutions A brief review of molarity & molality Number of moles of solute Molarity = Liter of solution Number of moles of solute Molality = Kg of solvent Preparing a Solution of Known Molarity Would you use a volumetric flask to prepare a solution of known molality? How would you prepare a solution of known molality? Another way to express solution concentration • Now that you understand molarity… Number of moles of solute Molarity = Liter of solution …let’s consider normality! Number of equivalents of solute Normality = Liter of solution What’s an equivalent ? • An equivalent of a substance is the mass (grams) of that substance that will combine with one mole of another reactant. • In an acid-base reaction, an equivalent is that amount of a substance that reacts with or liberates 1.0 mole of H+. What’s an equivalent ? • An equivalent of a substance is the mass (grams) of that substance that will combine with one mole of another reactant. • In a redox reaction, an equivalent is that amount of a substance that gains or loses 1.0 mole of e–s. Making a 1.00N Solution • Calculate the mass of one equivalent of the substance, then measure that number of grams of the substance. • Put that substance in…. ?? • Add how much solvent? Preparing a Solution of Known Normality AbNormality? • Why don’t many modern chemistry textbooks cover normality? • If we take a dimensional analysis approach to problems solving, normality is a superfluous concept. • Consider the following problem: What volume of a 0.500 M KOH solution is required to titrate 10.0 mL of a 0.20 M H2SO4 solution? H2SO4 + KOH H2O + K2SO4 Could we just use this equation? MaVa = MbVb Here’s what we’d get if we used MaVa = MbVb 0.20 M H2SO4 X 10.0 mL = 0.500 M KOH X ? mL Solving this, we get 4.0 mL of 0.500 M KOH. But is this correct? No, it’s wrong! Why? So let’s see how we solve this correctly. What volume of a 0.500 M KOH solution is required to titrate 10.0 mL of a 0.20 M H2SO4 solution? BALANCE THE CHEMICAL EQUATION! H2SO4 + KOH H2O + K2SO4 M rx M volume acid moles acid moles base volume base acid coef. base What volume of a 0.500 M KOH solution is required to titrate 10.0 mL of a 0.20 M H2SO4 solution? H2SO4 + 2KOH 2H2O + K2SO4 M rx M volume acid moles acid moles base volume base acid coef. base ___ mol H2SO4 __ mol KOH ____ ml soln 10.0 mL x x x = ___ mL ____ mL soln __ mol H2SO4 ____ mol KOH What volume of a 0.500 M KOH solution is required to titrate 10.00 mL of a 0.20 M H2SO4 solution? H2SO4 + 2KOH 2H2O + K2SO4 M rx M volume acid moles acid moles base volume base acid coef. base 0.20 mol H2SO4 2 mol KOH 1000 ml soln 10.0 mL x x x = 8.0 mL 1000 mL soln 1 mol H2SO4 0.500 mol KOH Notice this! M rx M volume acid moles acid moles base volume base acid coef. base 0.20 mol H2SO4 2 mol KOH 1000 ml soln 10.0 mL x x x = 8.0 mL 1000 mL soln 1 mol H2SO4 0.500 mol KOH Notice this! This is what makes normality superfluous! Is the following statement true or false? • Any volume of a base will completely react with (neutralize) that same volume of acid if the two solutions have the same normality (vice versa too). • In other words, is this true: NaVa = NbVb Is the following statement true or false? • Any volume of a reducing agent will completely react with that same volume of an oxidizing agent if the two solutions have the same normality. • In other words, is this true: NoxVox = NredVred Let’s solve a few problems using normality • Look at the handout that has sample problems and work the first one. And another thing… • We also use equivalents and milliequivalents as an amount of a substance • The same way that we use moles (mol) and millimoles (mmol) as an amount of a substance If you need more help: Come by my office and I can give you a few pages of worked examples from a Schaum’s Solved Problems Series. You can download this PowerPoint file at my EU web site: www.evangel.edu/Personal/badgers/Web/ The End Now wasn’t that fascinating?