Chapter 4: Types of Chemical Reactions and Solution Stoichiometry

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Chapter 4: Types of Chemical Reactions and Solution Stoichiometry 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.

8 Stoichiometry of Precipitation Reactions 4.9 Acid-Base Reactions 4.10 Oxidation-Reduction Reactions 4.11 Balancing Oxidation-Reduction Equations (Exam 2) 4.

1 Chapter 4: Types of Chemical Reactions and Solution Stoichiometry 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition of Solutions (MOLARITY!) 4.4 Types of Chemical Reactions 4.5 Precipitation Reactions 4.6 Describing Reactions in Solution 4.7 Selective Precipitation (limited coverage) 4.8 Stoichiometry of Precipitation Reactions 4.9 Acid-Base Reactions 4.10 Oxidation-Reduction Reactions 4.11 Balancing Oxidation-Reduction Equations (Exam 2) 4.12 Simple Oxidation-Reduction Titrations 1 Definitions Solutes, Solvents and Solutions Solute Substance being dissolved, mixed, diluted. Example: compounds extracted from coffee grounds, sugar, milk. Solvent Substance doing the dissolving, mixing, dilution. Example: water Solution Final combination of dissolution, mixing, and dilution. Example: morning coffee 1

Water as a Solvent water is an important solvent dissolves many substances aqueous means a solution in which water is the solvent water is a POLAR molecule Red: more electron density Blue: less

2 Water as a Solvent water is an important solvent dissolves many substances aqueous means a solution in which water is the solvent water is a POLAR molecule Red: more electron density Blue: less electron density 3 Polar and Nonpolar Solutes water dissolves some nonionic substances if they are polar (ethanol-water) ethanol molecules are polar (contain directional O-H bond) nonpolar substances (e.g., octane (C 8 H 18 ), benzene (C 6 H 6 ), fats, oils) will not dissolve in water 4 2

Ionic Solutes polar water molecules dissolve ionic compounds (salts) hydration breaks ionic compounds into anions and cations water dissolves different ionic compounds to different degrees (more in

3 Ionic Solutes polar water molecules dissolve ionic compounds (salts) hydration breaks ionic compounds into anions and cations water dissolves different ionic compounds to different degrees (more in Ch 8) 5 The Role of Water as a Solvent: Dissolution of Ionic Compounds Electrical conductivity: the flow of electrical current in a solution is an indicator of the presence of ions in solution and the solubility of ionic compounds. Electrolyte: a substance that conducts a current when dissolved in water. Soluble, ionic compounds that dissociate completely conduct a large current and are called strong electrolytes. NaCl (s) + H 2 O (l) Na + (aq) + Cl - (aq) Ions become solvated/hydrated are surrounded by water molecules. These ions are labeled aqueous, are free to move throughout the solution, and conduct electricity. 3

Strong Electrolytes Produce ions in aqueous solution and conduct electricity well. Strong electrolytes are soluble salts (NaCl, NH 4 NO 3 ), strong acids, and strong bases.

HCl, HNO 3, and H 2 SO 4 are strong acids: HNO 3 (aq) H + (aq) + NO 3- (aq) Strong bases produce OH - ions when they dissolve in water: NaOH and KOH are strong bases:

4 Strong Electrolytes Produce ions in aqueous solution and conduct electricity well. Strong electrolytes are soluble salts (NaCl, NH 4 NO 3 ), strong acids, and strong bases. Strong acids produce H + ions when they dissolve in water. HCl, HNO 3, and H 2 SO 4 are strong acids: HNO 3 (aq) H + (aq) + NO 3- (aq) Strong bases produce OH - ions when they dissolve in water: NaOH and KOH are strong bases: NaOH(s) Na + (aq) + OH - (aq) All of the above species are ionized nearly 100%. 7 HCl (aq) is completely ionized. Strong acids fully dissociate, forming the anion and a hydrated proton. NaOH (aq) is completely ionized. Strong bases fully dissociate, forming a cation and the hydroxide anion. 4

Weak Electrolytes Produce relatively few ions in aqueous solutions The most common weak electrolytes are weak acids and weak bases: acetic acid is a typical weak acid:

Acetic acid (HC 2 H 3 O 2 ) exists in water mostly as undissociated molecules.

5 Weak Electrolytes Produce relatively few ions in aqueous solutions The most common weak electrolytes are weak acids and weak bases: acetic acid is a typical weak acid: HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2- (aq) ammonia is a common weak base: NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH - (aq) Both of these species are ionized only ~1% 9 Acetic acid (HC 2 H 3 O 2 ) exists in water mostly as undissociated molecules. Weak acids partially dissociate, forming only a small number of anions and hydrated protons. The reaction of NH 3 in water. Weak bases partially dissociate (or react with water to a limited extent), forming only a small number of cations and hydroxide anions. 5

6 Nonelectrolytes Dissolve in water but produce no ions in solution. Nonelectrolytes do not conduct electricity because when a sample of the compound dissolves, the substance remains intact as whole molecules and no ions are produced. Common nonelectrolytes include: ethanol (CH 3 CH 2 OH) table sugar (sucrose, C 12 H 22 O 11 ) Insoluble ionic compounds are also nonelectrolytes: Ca 3 (PO 4 ) 2, MgCO 3 11 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I How many moles of each ion are formed when the following compounds are dissolved in water: a) 4.0 moles of sodium carbonate b) 46.5 g of rubidium fluoride c) 9.32 x formula units of iron(iii) chloride H 2 O a) Na 2 CO 3 (s) 2 Na + (aq) + CO 3-2 (aq) moles of Na + = 4.0 moles Na 2 CO 3 x 2 mol Na + 1 mol Na 2 CO 3 = 8.0 moles Na + (and 4.0 moles of CO 3-2 ) 6

7 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II H 2 O b) RbF (s) Rb + (aq) + F - (aq) moles of RbF = 46.5 g RbF x thus, mol Rb + and mol F - 1 mol RbF = moles RbF g RbF H 2 O c) FeCl 3 (s) Fe +3 (aq) + 3 Cl - (aq) moles of FeCl 3 = 9.32 x formula units = mol FeCl 3 = mol Fe +3 x 1 mol FeCl x formula units FeCl 3 moles of Cl - = mol FeCl 3 x 3 mol Cl - = mol Cl 1 mol FeCl - 3 Molarity and Stoichiometry When we first met stoichiometry, we talked about combining moles or masses of reactants. But, A LOT of interesting chemistry happens in aqueous phase, where it s more convenient to talk about molarity. The molarity of a solution gives you information about the number of moles of solute available to participate in a reaction. 7

8 Describing Solutions: Molarity Molarity (M) = moles solute volume solution = n V 500. ml water 0.1 mol C 6 H 12 O mol 500. ml OR ml 1 L = 0.2 mol 1 L 0.1 mol = 0.2 mol = 0.2 M L 1 L = 0.2 M [ ] C 6 H 12 O 6 = 0.2 M Using Molarity Molarity (M) = moles solute volume solution = n V 500. ml water How many grams of glucose are in this solution? (MM = g/mol) [ ] C 6 H 12 O 6 = 1.5 M n = MV = (1.5 mol/l)*(0.500 L) = mol? g = 0.75 mol g 1 mol = g = 140 g 8

9 Molarity of Ionic Solutions Recall, ionic compounds separate into their component ions when they dissolve in water. H 2 O(l) MgCl 2 (s) Mg 2+ (aq) + 2Cl - (aq) For every MgCl 2 formula unit that dissolves......you get one Mg 2+ ion, and two Cl - ions in solution. This means... [MgCl 2 ] [Mg 2+ 1 = ] = [Cl - ] 2 Think about the solution in two ways: What is the molarity (M) with regards to the initial composition of the solute? What is the M of each species given the electrolytic nature of the solute? Examples Determine the concentrations of all the ions in each of the following solutions: 1 M FeCl 3 : [Fe 3+ ] = 1 M [Cl - ] = 3 M [ions] = 4 M 0.50 M Co(NO 3 ) 2 : [Co 2+ ] = 0.50 M [NO 3- ] = 1.0 M [ions] = 1.50 M How many moles of NO 3 - ions are there in 350. ml of a solution that is M in Co(NO 3 ) 2? 9

10 Example A water sample from the Great Salt Lake in Utah contains 83.6 mg of Na + per g of lake water. What is the molarity of sodium ions in the lake? (d lake water = g/ml) Mass of Na + Mass of lake water Moles of Na+ in g lake water Vol of g lake water Moles of Na + L of lake water Molarity of Great Salt Lake Example (cont) 1 g 1 mol Na? mol Na + = 83.6 mg Na mg g Na + = x 10-3 mol Na +? L lake water = g wtr 1 ml wtr 1 L g wtr 1000 ml = x 10-4 L? [Na + ] = x 10-3 mol Na x 10-4 L lake wtr = mol/l 10

11 Solution Preparation: Standard Solutions Solution for which the concentration is accurately known. Often prepared by starting with a solid solute weighed out on a balance with several significant figures. Solution Preparation: Standard Solutions How would you prepare 1.00 L of a M solution of ammonium carbonate? Determine the moles of ammonium carbonate required: mol (NH 1.00 L x 4 ) 2 CO 3 = mol (NH 4 ) 2 CO 3 L solution Convert to grams using the molar mass: g (NH mol (NH 4 ) 2 CO 3 x 4 ) 2 CO 3 = g (NH 4 ) 2 CO 3 mol (NH 4 ) 2 CO 3 To make 1.00 L of solution, 35.3 g of (NH 4 ) 2 CO 3 are weighed out and transferred to a 1.00 L volumetric flask. DI water is added to dissolve the solid and then to dilute the solution to the mark on the neck of the flask. Recalculate molarity based on specific mass

Solution Dilution Dilution is the process of making a solution with a certain molarity from a

Concentrated acetic acid Diluted acetic acid Volumetric glassware has very high precision 500.

12 Solution Dilution Dilution is the process of making a solution with a certain molarity from a solution with a greater molarity. Concentrated acetic acid Diluted acetic acid Volumetric glassware has very high precision ml Dilution The number of moles (n) of solute stays the same only the volume of solution (V) changes. We can formalize this relationship: M 1 V 1 = M 2 V 2 where M 1, V 1 = molarity and volume of concentrated solution M 2, V 2 = molarity and volume of diluted solution Recall: M = n/v n = MV n remains unchanged: M 1 V 1 = n = M 2 V 2 12

Using M 1 V 1 = M 2 V 2 What volume of 5.00 M calcium nitrate solution is needed to prepare 500.0 ml of a 0.250 M Ca(NO 3 ) 2 solution? V 1 =? M 1 = 5.00 M M 1 V 1 = M 2 V 2 V 1 = V 2 = 500.

13 Using M 1 V 1 = M 2 V 2 What volume of 5.00 M calcium nitrate solution is needed to prepare ml of a M Ca(NO 3 ) 2 solution? V 1 =? M 1 = 5.00 M M 1 V 1 = M 2 V 2 V 1 = V 2 = ml M 2 = M M 2 V 2 (0.250 mol/l)*( L) = M mol/l = L = 25.0 ml Stoichiometry Flowchart Vol of known soln of known M Vol of desired soln of known M Convert to moles Convert to volume 13

Quick Review. - Chemical equations - Types of chemical reactions - Balancing chemical equations - Stoichiometry - Limiting reactant/reagent

Quick Review. - Chemical equations - Types of chemical reactions - Balancing chemical equations - Stoichiometry - Limiting reactant/reagent Quick Review - Chemical equations - Types of chemical reactions - Balancing chemical equations - Stoichiometry - Limiting reactant/reagent Water H 2 O Is water an ionic or a covalent compound? Covalent,