1. Normality Dr. Steve Badger

2. Concentrations of Solutions A brief review of molarity & molality Molarity = Number of moles of solute Liter of solution Molality = Number of moles of solute Kg of solvent

3. Preparing a Solution of Known Molarity

4. Would you use a volumetric flask to prepare a solution of known molality? How would you prepare a solution of known molality?

5. Another way to express solution concentration Now that you understand molarity… Normality = Number of equivalents of solute Liter of solution …let’s consider normality! Molarity = Number of moles of solute Liter of solution

6. 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 +.

7. 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.

8. 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?

9. Preparing a Solution of Known Normality

10. 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:

11. H 2 SO 4 + KOH H 2 O + K 2 SO 4 What volume of a 0.500 M KOH solution is required to titrate 10.0 mL of a 0.20 M H 2 SO 4 solution? M a V a = M b V b Could we just use this equation?

12. Here’s what we’d get if we used M a V a = M b V b 0.20 M H 2 SO 4 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.

13. volume acidmoles acidmoles basevolume base H 2 SO 4 + KOH H 2 O + K 2 SO 4 M acid rx coef. M base What volume of a 0.500 M KOH solution is required to titrate 10.0 mL of a 0.20 M H 2 SO 4 solution? BALANCE THE CHEMICAL EQUATION!

14. What volume of a 0.500 M KOH solution is required to titrate 10.0 mL of a 0.20 M H 2 SO 4 solution? volume acidmoles acidmoles basevolume base ___ mol H 2 SO 4 ____ mL soln x __ mol KOH __ mol H 2 SO 4 x ____ ml soln ____ mol KOH x 10.0 mL = ___ mL M acid rx coef. M base H 2 SO 4 + 2KOH 2H 2 O + K 2 SO 4

15. What volume of a 0.500 M KOH solution is required to titrate 10.00 mL of a 0.20 M H 2 SO 4 solution? volume acidmoles acidmoles basevolume base H 2 SO 4 + 2KOH 2H 2 O + K 2 SO 4 0.20 mol H 2 SO 4 1000 mL soln x 2 mol KOH 1 mol H 2 SO 4 x 1000 ml soln 0.500 mol KOH x 10.0 mL = 8.0 mL M acid rx coef. M base

16. volume acidmoles acidmoles basevolume base 0.20 mol H 2 SO 4 1000 mL soln x 2 mol KOH 1 mol H 2 SO 4 x 1000 ml soln 0.500 mol KOH x 10.0 mL = 8.0 mL M acid rx coef. M base Notice this! This is what makes normality superfluous!

17. 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: N a V a = N b V b

18. 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: N ox V ox = N red V red

19. Let’s solve a few problems using normality Look at the handout that has sample problems and work the first one.

20. 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

21. 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/

22. The End Now wasn’t that fascinating?