Ch. 9: Introduction to Solutions and Aqueous

Reactions

Dr. Namphol Sinkaset

Chem 200: General Chemistry I

I. Chapter Outline

I. Introduction

II. Solution Concentrations

III. Solution Calculations

IV. Aqueous Solutions

V. Precipitation Reactions

VI. Acid/Base Reactions

VII. Gas-Evolution Reactions

VIII. Oxidation-Reduction Reactions

I. Aqueous Chemistry

• Water-based chemistry is the most well studied – why?

• In this chapter, we will focus on reactions that take place in water and look at:

1) Solution stoichiometry

2) Common aqueous reactions

II. Solution Concentration

• There are two parts of a solution. solute: substance present in smaller

amount solvent: substance present in larger

amount• For stoichiometry, the important aspect

of a solution is its concentration.• concentration: amount of solute present

in a certain volume of solution

II. Concentrated vs. Dilute

• Concentrated solutions have a lot of solute relative to solvent.

• Dilute solutions have a little solute relative to solvent.

II. Quantitative Concentrations

• The most common concentration unit is molarity, which is moles solute per L of solution.

solution L

solute molesM

In a solution, solute is evenly dispersed in the solvent!!

II. Solution Preparation

• Preparing solutions requires a series of exacting steps.

III. Solution Calculations• There are 3 main types of solution-based

calculations.1) What is the concentration?

Use definition of molarity.

2) Solution creation/dilution. Use definition of molarity / M1V1 = M2V2

3) Stoichiometry. Use molarity as a conversion factor.

• Start w/ problems using definition of molarity.

III. Sample Problem

• e.g. Calculate the molarity of a solution formed when 24.2 g NaCl is dissolved in water to make 124.1 mL of solution.

III. Sample Problem

• e.g. How many grams of Na2HPO4 are needed to make 1.50 L of a 0.500 M Na2HPO4 solution?

III. Solution Dilution

• A stock solution is a solution of high concentration.

• Lower concentration solutions can be made from the stock via dilution.

III. Sample Problem

• e.g. How many mL of a 2.0 M NaCl solution are needed to make 250 mL of a 0.50 M NaCl solution?

III. Solution Stoichiometry

• In the stoichiometry we’ve done before, amounts were converted between grams and moles.

• In solutions, amounts are converted between volumes and concentrations.

• The key is remembering that molarity is a conversion factor between moles and volume!

III. Sample Problem

• e.g. How many mL of 0.10 M HCl reacts with 0.10 g Al(OH)3 according to the reaction below?

Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

III. Sample Problem

• e.g. How much PbCl2 forms when 267 mL 1.50 M lead(II) acetate reacts with 125 mL 3.40 M sodium chloride according to the reaction below?

Pb(CH3COO)2(aq) + 2NaCl(aq) PbCl2(s) + 2NaCH3COO(aq)

IV. Chemical Reactions

• There are countless reactions, but only a few categories of reactions.

• With experience, it becomes easier to identify what will happen in a reaction.

• First, we take a close look at the solute and solvent in a solution.

IV. Forming a Solution

• Attractive forces hold solute together.

• When solute is added to a solvent, new potentially attractive forces arise.

• Competition between these forces occurs.

IV. Aqueous Solutions

• Water is a particularly “active” solvent.

• As a solvent, water has one important characteristic: it is polar!

IV. Interactions in Aqueous Solutions

• The polar nature of water allows it to interact with charged species in solution.

IV. Water-Ion > Na+Cl-

IV. Electrolytes• Compounds that dissociate in water and lead to

electrical conductivity are called electrolytes.

IV. Strong/Weak Electrolytes• Strong electrolytes dissociate

completely in water.

NaCl(s) Na+(aq) + Cl-(aq)

• Logically, weak electrolytes do not dissociate completely in water.

CH3COOH(aq) H+(aq) + CH3COO-

(aq)

H2O

H2O

IV. Sample Problem

• e.g. How many moles of each ion are in a solution formed by dissolving 354 g of magnesium hydroxide in water?

V. Precipitation Reactions

• The formation of a precipitate (ppt) is a strong driving force for a reaction.

• Precipitate is a fancy word for solid.

• The attractions in these solids are too strong for H2O to break up.

V. Predicting Precipitates (95%)

1) Li+, Na+, K+, NH4+ salts are soluble.

2) NO3-, CH3COO-, ClO4

- salts are soluble.

3) Ag+, Pb2+, Hg22+ salts are insoluble.

4) Cl-, Br-, I- salts are soluble.

5) CO32-, S2-, O2-, OH- salts are insoluble.

6) SO42- salts are soluble except for

CaSO4,SrSO4, and BaSO4.7) If none of these apply, it’s insoluble.

V. Sample Problem

• e.g. Predict the precipitates in the following aqueous reactions.

a) sodium hydroxide + cadmium(II) nitrate

b) magnesium bromide + potassium acetate

c) ammonium sulfate + barium chloride

d) sodium iodide + lead(II) nitrate

V. Writing Reactions

Nospectator

ions

V. Sample Problem

• Write balanced molecular, total ionic, and net ionic equations for the reaction between strontium chloride and lithium phosphate.

VI. Acids/Bases

• There are many definitions of acids and bases, but we will use the Arrhenius definitions for now.

• Acids are molecular compounds that produce H+ ions in aqueous solution.

• Bases are substances that produce OH- ions in aqueous solution.

VI. Aqueous Acids

• What is H+ comprised of?• Water interacts so strongly w/ H+, that it

forms a bond with it. e.g. HCl(g) + H2O(l) (H2O)H+

(aq) + Cl-(aq)

• This is called the hydronium ion, and it’s usually written as H3O+. Note that H+

(aq) = H3O+(aq)

• Polyprotic acids have more than one ionizable H+, e.g. H2SO4.

VI. Common Acids & Bases

VI. Acid Nomenclature

• There are two categories of acids that have different naming rules.

1) Binary acids

2) Oxoacids

• You can easily recognize acids because their formula has H as the first element.

VI. Naming Binary Acids

• Binary acids contain a nonmetal anion. HCl = hydrochloric acid HBr = hydrobromic acid H2Se = hydroselenic acid

HI = hydroiodic acid

VI. Naming Oxoacids

• Set 1 HNO3 = nitric acid

H2SO4 = sulfuric acid

HClO3 = chloric acid

HClO4 = perchloric acid

• Set 2 HNO2 = nitrous acid

HClO2 = chlorous acid

HClO = hypochlorous acid

H2SO3 = sulfurous acid

Oxoacids contain an oxoanion.

VI. Acid/Base Reactions

• The driving force for this reaction is the formation of water; other product is a salt.

• The net ionic equation for acid/base reactions is always the same!! H+

(aq) + OH-(aq) H2O(l)

VI. Acid/Base Titrations

• The concentration of an acid or base can be determined experimentally.

VI. Titration Terminology

• titration: procedure in which one solution of known [ ] is used to determine the [ ] of another solution

• indicator: a substance used to visualize the end of a reaction

• The equivalance point occurs when moles acid = moles base.

• The endpoint occurs when the solution changes color due to the indicator.

VI. Titration Problems

• The first step in solving a titration problem is writing the titration reaction!!

• After identifying the reaction, it becomes a solution stoichiometry problem!

• Again, use unit labels on the numbers to guide your calculation.

VI. Sample Problem

• e.g. To determine the concentration of a solution of H2SO4, you titrate a 50.00 mL sample of it with 0.250 M NaOH. If it takes 22.35 mL of the NaOH solution to reach the endpoint, what’s the concentration of the H2SO4?

VII. Gas-Evolution Reactions

• Many gas-evolution reactions are also acid-base reactions.

• The formation of the gas could be direct or through decomposition of a product. H2CO3(aq) H2O(l) + CO2(g)

VII. It Makes Bubbles

VII. Common Gas Products

VII. Sample Problem

• Write balanced molecular, total ionic, and net ionic equations for the reaction between aqueous solutions of hydrobromic acid and potassium sulfite.

VIII. Reactions Involving e- in Motion

• The movement of e- from one atom to another is another driving force for reactions.

• Oxidation is the loss of e-.• Reduction is the gain of e-.• They are coupled processes; one

cannot occur w/out the other.

VIII. Reduction-Oxidation Reactions

• These are also known as redox reactions.• Formation of NaCl is an example.

Na Na+ + e- ½ Cl2 + e- Cl-

• Cl2 oxidizes Na; Cl2 is the oxidizing agent.

• Na reduces Cl2; Na is the reducing agent.

VIII. NaCl Formation via Redox

VIII. Nonpolar to Polar

• You don’t need complete e- transfer for a redox reaction.

• Can also have just a shift of e- density.

• Consider H2(g) + Cl2(g) 2HCl(g)

VIII. Oxidation States

• Oxidation states (oxidation numbers) allow us to keep track of which atoms are gaining/losing e- in a reaction.

• oxidation state (number): the charge an atom would have if e- are transferred completely and not shared

• Note that in ionic compounds, we consider e- as totally transferred, so the ionic charge is the oxidation state.

VIII. Rules for Assigning O.N.

1) Atoms in elemental form have O.N. = 0.2) Charge on a monatomic ion equals its O.N.3) The sum of all O.N. must equal the total charge.4) For Group 1, O.N. = +1.5) For Group 2, O.N. = +2.6) For H, O.N. = +1 w/ nonmetals, -1 w/ metals and

B.7) For F, O.N. = -1.8) For O, O.N. = -1 in peroxides and -2 in all others.9) For Group 17, typically O.N. = -1.

VIII. Assigning O.N.

• e.g. Determine O.N. for all atoms in the following.

a) KMnO4

b) NH4+

c) IF3

d) ZnCl2

VIII. Sample Problem

• e.g. Determine the substances that are oxidized and reduced in the reaction below.

5CO(g) + I2O5(s) I2(s) + 5CO2(g)