After completing Unit 3, you know the importance of the " mole" concept in chemistry for calculations involving "weighed" chemical substances. If the mass in grams and the molar mass of a substance is known, then it is a simple operation to convert the mass to moles and to continue this type of stoichiometric calculation. Unit 4 introduces another way to quantitatively measure how much substance in terms of moles is present for chemical calculations. Many chemical and biochemical reactions take place in solution, involving reactants, solutes, which are dissolved in some solvent. Obviously, the most common solvent for biologically important reactions is water, so it will be the common solvent medium in Unit 4 . In Unit 2 , you became familiar with the "Solubility Guidelines " used to predict the relative solubility of different ionic solids in water. If a salt was soluble, it dissociated into "aqueous" ions or ions that were "solvated " by water. Thus, the solute, an ionic salt in this case, interacted with the solvent, water to form a solution of aqueous ions. Among the questions Unit 4 will answer is, what is the concentration of the dissolved ionic species in water defined by the quantitive ratio, mols of solute per liter of solution. This ratio introduced on page 228 of the text, is called molarity and given the symbol, M. A 1.00 M solution of NaCl contains 1.00 mol of NaCl in 1.00 liter of solution, where sodium chloride is the solute and water is the solvent.

A balanced chemical equation is a necessary condition to begin any quantitative calculation using solutions, just as it was for the weight relationships discussed in Unit3. Solution stoichiometry is just small extension of the principles introduced in "Stoichiometry I". Actually, calculations performed by chemists everyday use the "weight" and solution, stoichiometric methods for quantitative problem solving.

Unit 4 begins with Section 5.5, "Working with Solutions", page 228 and ends with "Chapter Highlights" on page 244. Molarity as a concentration factor relating [mols of solute/ liter of solution] is defined in Section 5.5 and the details for preparing solutions of know molar concentrations is discussed. Since the molar concentration, molarity is given as "mols"/ liter, you must constantly remind yourself, this means "mols of solute/ liter of solution" Recognize that molarity is a "Unit Factor". It is expressing the equivalency between a given number of mols of solute and a given volume of solution , 1.00 liter. So, a 0.500 molar solution of sucrose, common sugar, can be represented by 0.5 M C12H22 O11 (sucrose) or 0.5 mols sucrose/ 1 liter of solution. The concept of a unit factor for 0.500 M sucrose is the idea that the concentration is also equal to the expression, 1 liter of solution/05 mols of sucrose the "inverse" of 0.50 mols/ liter. For problem solving using solution, you must learn to use the given molarity like a unit factor similar to math operations with the molar mass in gm/mol.

Section 5.6 "Stoichiometry of Reactions in Aqueous Solution", beginning on page 235, outlines the application of calculations using reactants given in volume and molar concentration terms. Titration is a term used in chemistry to encompass quantitative measurements associated with reactions involving solutions as one or both of the reactants. Chemical glassware used in titrations include volumetric flasks and pipets, and the "unforgettable" for many students, the buret. All of this glassware mentioned, has some "mark" or "scale" that allows you to accurately determine the volume, usually in milliliters, mls. Therefore, many solution problems have the volume of solution given in milliliters not liters and you must change milliliters to liters before doing any calculation with "molarity", otherwise the units "mls" and "liters" don't match. (See FAQ's for more on this comment).

Reading Assignment:

Chapter 5, "Stoichiometry" Sections 5.5 to 5.6, pages 228 to 244. Note: Omit Section 5.4, pages 221 to 228.

Unit 4 Objectives:

1. Using a balanced chemical equation, calculate the relationship between reactants and/or products given factors in mols, mass or molarity and express results in mols mass or molarity. (Review Unit 3, Objective 1 and Screens 5.9, 5.10, 5.10 SB and 5.13)

2. a)Given the formula of the solute, the molarity and volume of the solution, calculate the number of mols or mass of solute present.

b) Given a mass and formula of a solute and the volume of solution to be prepared, calculate the molarity of the resulting solution. (Screens 5.11 and 5.12)

c) Given the formula of the solute, its mass and the molarity of the solution, calculate the volume of solution of the known concentration which contains the given mass of solute. (Screens 5.11 and 5.12).

3. Discuss the concept of titration as an analytic tool in chemistry. Given the appropriate data, calculate the amount of substance present in some sample based on a titration. (Screens 5.14 and 5.15)

e-Mail Activity: There will be no e-mail activity with this unit. Instead a more lengthy quiz worth 50 points will act as both a quiz for Unit 4 and a brief review of Units 1, 2 and 3 in order to prepare for Exam 1 ( 11:00 to 12:15 on Friday, February, 13 "unfortunately" at CSUB, WSL Room 16 ) Read "Course News"for information.)

Workbook Activities: Chapter 5, continuing from Unit 3:

1). View Screens 5.9 and 5.10. Do Exercise 5.5 and check answer by looking at soution to Exercise 5.8 page 132 in the text, and so on for the assigned "exercises" from the workbook.

2) Check out the "Molarity Calculator Tool" Screen 5.10

3) View Screens 5.11 and 5.12 and do Exercise 5.6 and 5.7 .Exercise 5.6 = (Exercise 5.10 on page 232 of the text) and Exercise 5.7 = ( Exercises 5.11 and 5.12 on page 234 of the text)

4) .View Screen 5.13 and do Exercise 5.8 = (Exercise 5.13 on page 236 of the text)

5) View Screens 5.14 and 5.15, the "Titration Simulator" Is this an interseting activity? What do you think ? Comments welcome.

Finally, consider the following "Study Questions" Workbook page 5-12 and 5-13 ( Text # for identical problem)--- 11 (27), 15 (35), 16 (37), 17 (39), 18 (41), 19 (43), 20 (45), 21 (47), 23 (51) and 26 (61).

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