Lecture 4: Aqueous solution chemistry Lecture 4 Topics Brown, chapter 4 1. Solutes & solvents 4.1 Electrolytes & non-electrolytes Dissociation 2. Solution concentration & stoichiometry 4.5 – 4.6 Molarity & interconversion Dilution Types of aqueous chemial reactions 3. Precipitation reactions 4.2 Complete ionic equations 4. Neutralization reactions 4.3 Acids & bases Neutralization reactions Non-hydroxide bases produce gases Titration 4.6 Summary of complete ionic equations 5. Reduction & oxidation reactions 4.4 Oxidation numbers Oxidation of metals by acids & salts Activity series

Solution concentration & stoichiometry Molarity (moles/L) measures concentration. Molarity relates mass to volume. Dilution decreases concentration. Solution stoichiometry: volume & moles The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

Solution concentration: molarity (M) Remember what a solution is? A homogeneous mixture of some amount of solute dissolved in a large(r) volume of solvent. Molarity (M) a measure of solution concentration moles 1 M = 1 mole/L L of solution What’s the molarity of a solution made by dissolving 23.4 g of sodium sulfate in enough water to make a final volume of 125 mL? MW of Na2SO4 ~ 142 g/mole 23.4 g 1 mole = 1.32 M 142 g 0.125 L What is the molarity of a solution made by dissolving 5.00 g glucose (C6H12O6) to make 100 mL of solution? MW ~ 180 g/mol 5.00 g 1 mole 1000 mL = 0.278 M 180 g 100 mL 1 L What are the ionic concentrations of this strong electrolyte, 0.74 M MgCl2? MgCl2 --> Mg+2 + 2Cl-1 0.74 0.74 (2x0.74) = 1.48 M Cl-1 p.142-6

Interconversion: molarity, moles & volume The ‘conversion flower’ works well for this. Or, remember that: moles = (M)(V) grams moles liters MW M (mol/L) What volume of 1.50 M HCl would we need to get 0.250 moles HCl? The key is starting with a single value and ‘operating’ on it with the conversion factor. 0.250 moles 1 L = 0.167 L = 167 mL 1.50 moles How many grams of Na2SO4 are needed to make 0.350 L of 0.500 M? MW ~142 0.350 L 0.500 mol 142 g = 24.9 g Na2SO4 1 L 1 mol How many grams of Na2SO4 are there in 15 mL of 0.500 M? O.015 L 0.500 mol 142 g = 1.1 g Na2SO4 1 L 1 mol p.144-6

Solution stoichiometry “mall map” atoms X molecules A subsc Av# mass A moles A volume A MW M Stoic mass B moles B volume B MW M Av# atoms Y subsc molecules B p.149-50

Dilution Dilution is what happens to solutions of molecular compounds whose volume is increased. Mix chocolate syrup int0 a cold glass of milk; syrup is diluted. In labs, solutions that are used often are often made up in a concentrated form so that they can be quickly diluted for use. This saves time & work. Think canned OJ concentrate or margarita mix. How do we know how much dilution will be needed to get to the desired concentration from our concentrated stock solution? MiVi = MfVf where ‘i’ is initial (stock) & ‘f’ is final (diluted sol’n) Units of volume don’t matter but must be same on both sides. What’s the final concentration produced by diluting 25.0 mL of 4.50 M NaCl to a volume of 500.o mL? (4.5o M)(25.0 mL) = (Mf)(500.0 mL)  Mf = 0.225 M How many mL of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M? (3.0 M)(Vi) = (0.10 M)(450 mL)  Vi = 15.0 mL p.146-8

Examples: solution stoichiometry How many grams of Ca(OH)2 are needed to react with 25.0 mL of 0.100 M H(NO3)? Ca(OH)2 + 2H(NO3)  Ca(NO3)2 + 2H(OH) 25.0 mL 1 L 0.100 mole acid 1 mole base 74.12 g Ca(OH)2 103 mL 1 L 2 mole acid 1 mole base = 0.0926 g Ca(OH)2 How many mL of 2.5 M Ba3(PO4)2 are needed to react with 25.0 g of MgCl2? MgCl2 = 95.21 g/mol Ba3(PO4)2 + 3MgCl2  3BaCl2 + Mg3(PO4)2 25.0 g 1 mol 1 mol Ba3(PO4)2 1 L 103 mL 95.21 g 3 mol MgCl2 2.5 mole 1 L = 35.0 mL Ba3(PO4)2 p.149-50