4.1 General Properties of Aqueous Solutions

Chapter 4 Reactions in Aqueous Solutions

• Solution - a homogeneous mixture

– Solute: the component that is dissolved

– Solvent: the component that does the dissolving

Generally, the component present in the greatest quantity is considered to be the solvent. Aqueous solutions are those in which water is the solvent. – Electrolyte: substance that dissolved in water produces a solution that conducts electricity

• Contains ions:

– Nonelectrolyte: substance that dissolved in water and produces a solution that does not conduct electricity

• Does not contain ions:

• Dissociation - ionic compounds separate into constituent ions when dissolved in solution 3 • Ionization - formation of ions by molecular compounds when dissolved

• Strong and weak electrolytes:

– Strong Electrolyte: 100% dissociation

• All water soluble ionic compounds, strong acids and strong bases

– Weak electrolytes: Partially ionized in solution

• Exist mostly as the molecular form in solution

• Weak acids and weak bases 5

• Examples of weak electrolytes:

– Weak acids:

- + HC2H3O2(aq)  C2H3O2 (aq) + H (aq)

– Weak bases:

+ - NH3 (aq) + H2O(l)  NH4 (aq) + OH (aq) (Note: double arrows indicate a reaction that occurs in both directions - a state of dynamic equilibrium exists)

nonelectrolyte weak electrolyte strong electrolyte

CH3OH H2CO3 NaOH

4.2 Precipitation Reactions

• Precipitation (formation of a solid from two aqueous solutions) occurs when product is insoluble

Solubility Rules of salts in water:

- · Most Nitrate (NO3 ) salts are soluble

· Most salts containing the alkali metal (group I) ions (Li+, Na+, K+, Cs+, Rb+) and the ammonium ion + (NH4 ) are soluble

· Most chloride, bromide and iodide salts are soluble,

+ 2+ 2+ except salts containing Ag , Pb and Hg2

2- · Most sulfate (SO4 ) salts are soluble,

2+ 2+ 2+ 2+ except salts containing Ba , Pb , Hg2 and Ca · Most hydroxide salts (OH-) salts are only slightly soluble.

The compounds Ba(OH)2, Sr(OH)2 and Ca(OH)2 are marginally soluble. NaOH and KOH are soluble

2- 2- · Most sulfide (S ), carbonate (CO3 ), chromate 2- 3- (CrO4 ), and phosphate (PO4 ) salts are only slightly soluble

• Hydration: process by which polar water molecules remove and surround individual ions from the solid.

Identify the precipitate:

Pb(NO3)2(aq) + 2NaCI(aq)  2NaNO3(?) + PbCI2 (?)

Which is soluble or insoluble in water : 9

Ba(NO3)2 soluble AgCI insoluble

Mg(OH)2 insoluble • Molecular equation: shows all compounds represented by their chemical formulas

• Ionic equation: shows all strong electrolytes as ions and all other substances (non- electrolytes, weak electrolytes, gases) by their chemical formulas

Net Ionic equation: shows only the reacting species in the chemical equation

– Eliminates spectator ions: 11

Aqueous solutions of silver nitrate and sodium sulfate are mixed. Write the net ionic reaction:

2AgNO3(aq)+Na2SO4(aq)  2NaNO3(?)+Ag2SO4(?)

+ - + 2- 2Ag (aq) + 2NO3 (aq) + 2Na (aq) + SO4 (aq)

+ -  2Na (aq) + 2NO3 (aq) +

Ag2SO4(s)

+ 2- 2Ag (aq) + SO4 (aq)  Ag2SO4(s) 4.3 Acid-Base Reactions:

HCl(aq) + NaOH(aq) ® NaCl(aq) +

H2O(l)

+ - H (aq) + OH (aq) ® H2O(l)

acid base water

All the other acids and bases are weak electrolytes. 13 • Definitions of acids and bases:

+ – Arrhenius acid - produces H in solution

- – Arrhenius base - produces OH in solution

_ Brønsted acid - proton donor _ Brønsted base - proton acceptor

– Examples of a weak base and weak acid

• Ammonia with water:

• Hydrofluoric acid with water: 15

• Types of acids:

– Monoprotic: one ionizable hydrogen

+ - HCl + H2O  H3O + Cl

– Diprotic: two ionizable hydrogens

+ - H2SO4 + H2O  H3O + HSO4

- + 2- HSO4 + H2O  H3O + SO4

– Triprotic: three ionizable hydrogens:

+ - H3PO4 + H2O  H3O + H2PO4

- + 2- H2PO4 + H2O  H3O + HPO4

2- + 3- HPO4 + H2O  H3O + PO4 Polyprotic: generic term meaning more than one ionizable hydrogen 17

• Types of bases:

- – Monobasic: One OH group KOH  K+ + OH-

- – Dibasic: Two OH groups

2+ - Ba(OH)2  Ba + 2OH

• Neutralization: Reaction between an acid and a base Acid + Base  Salt + Water Molecular equation:

HCl(aq) + NaOH(aq)  NaCl(aq) +

H2O(l)

Ionic equation:

+ - + - H (aq)+ Cl (aq) + Na (aq) + OH (aq)

+ -  Na (aq) + Cl (aq) + H2O(l) Net ionic equation: (formation of water)

+ - H (aq) + OH (aq)  H2O(l)

4.4 Oxidation-Reduction Reactions

• Often called “redox” reactions

• Electrons are transferred between the reactants

– One substance is oxidized, loses electrons Reducing agent

– Another substance is reduced, gains electrons

• Oxidizing agent

• Oxidation numbers change during the reaction 19

Example:

Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)

2+ 2+ Zn(s) + Cu (aq)  Zn (aq) + Cu(s)

– Zinc is losing 2 electrons and oxidized.

• Reducing agent

2+ - • Zn(s)  Zn (aq) + 2e

– Copper ions are gaining the 2 electrons.

• Oxidizing agent

2+ - • Cu (aq) + 2e  Cu(s) • Rules for assigning oxidation numbers

· Oxidation state of an atom in an element is zero

e.g: Na(s), O2(g), O3(g) and Hg(l)

· Oxidation state of an ion is its charge: e.g: Na + and Cl -

· Fluorine in its compounds is always assigned an oxidation state of -1

· Oxygen is usually assigned an oxidation state of -2, except in: 1. Peroxides is assigned an oxidation state of -1

(e.g. H2O2)

2. OF2 is assigned an oxidation state of +2 21 · Hydrogen in its covalent compounds (with non - metals) is assigned an oxidation state of +1 and in its hydrides (with metals) is assigned an oxidation state of -1.

· The sum of the oxidation states for a neutral compound must be zero.

Oxidation numbers must sum to the overall charge of the species:

2- SO4 = -2 (O is usually -2)

X + 4(-2) = -2

X - 8 = -2 \ X = +6 (S is +6) 23

Assign oxidation numbers for all elements in each species:

MgBr2 (Mg +2, Br -1)

- ClO2 (Cl +1 , O -2)

Al2 (SO4) 3 (Al +3, SO4 -2]

KMnO4 (K +1, MnO4 -1]

Displacement reactions:

– A common reaction: active metal replaces (displaces) a metal ion from a solution:

Mg(s) + CuCl2(aq)  Cu(s) +

MgCl2(aq) 25

Activity series:

• Balancing redox reactions:

– Electrons (charge) must be balanced as well as number and types of atoms

– Consider this net ionic reaction:

2+ 3+ Al(s) + Ni (aq)  Al (aq) + Ni(s)

– Divide reaction into two half-reactions:

3+ - Al(s)  Al (aq) + 3e (oxidation)

2+ - Ni (aq) + 2e  Ni(s)

(reduction)

– Make electrons gained equal electrons lost:

3+ - 2 [Al(s)  Al (aq) + 3e ] 2+ - 3 [Ni (aq) + 2e  Ni(s) ] 29

– Cancel electrons and write balanced net ionic reaction:

3+ 2Al(s)  2Al - (aq) + 6e

2+ - 3Ni (aq) + 6e  3Ni(s)

- Sum the two half-reactions:

2+ 3+ 2Al(s) + 3Ni (aq)  2Al (aq) + 3Ni(s) • Combination Reactions:

– Many combination reactions may also be classified as redox reactions

– Consider the reaction between Hydrogen gas reacts and oxygen gas:

2H2(g) + O2(g)  2H2O(l) Identify the substance oxidized and the Substance reduced.

• Decomposition reactions:

– Many decomposition reactions may also be classified as redox reactions

– Consider strong heating of Potassium chlorate:

2KClO3(s)  2KCl(s) + 3O2(g) Identify substances oxidized and reduced. 31

• Disproportionation reactions:

– One element undergoes both oxidation and reduction

– Consider:

• Combustion reactions:

– Common example, hydrocarbon fuel reacts with oxygen to produce carbon dioxide and water

– Consider:

33 4.5 Concentration of Solutions

• Concentration is the amount of solute dissolved in a given amount of solvent.

• Qualitative expressions of concentration

– Concentrated – higher ratio of solute to solvent

– Dilute - smaller ratio of solute to solvent 35

• Quantitative concentration term:

– Molarity is the ratio of moles solute per liter of solution:

– Symbols: M or [ ]

Calculate the molarity of a solution prepared by dissolving 45.00 grams of KI into a total volume of 500.0 mL. How many milliliters of 3.50 M NaOH can be prepared from 75.00 grams of the solid?

• Dilution:

– Process of preparing a less concentrated solution from a more concentrated one. 37 For the next experiment the class will need 250. mL of 0.10

M CuCl2. There is a bottle of 2.0 M CuCl2. Describe how to prepare this solution. How much of the 2.0 M solution do we need?

McLc = MdLd

(2.0 M) (Lc) = (0.10 M) (250.mL)

Lc = 12.5 mL

• Solution Stoichiometry:

– Soluble ionic compounds dissociate completely in solution.

– Using mole ratios we can calculate the concentration of all species in solution.

NaCl dissociates into Na+ and Cl-

+ Na2SO4 dissociates into 2Na and 2- SO4 39

3+ - AlCl3 dissociates into Al and 3Cl

Find the concentration of all species in a 0.25 M

solution of MgCl2

2+ - MgCl2  Mg + 2Cl

Given: MgCl2 = 0.25 M

[Mg2+ ] = 0.25 M (1:1 ratio) [Cl- ] = 0.50 M (1:2 ratio)

Using the square bracket notation, express the molar concentration for all species in the following solutions:

0.42 M Ba(OH)2

[Ba2+ ] = 0.42 M (1:1 ratio) [OH- ] = 0.84 M (2:1 ratio)

1.2 M NH4Cl

+ [NH4 ] = 1.2 M (1:1 ratio) [Cl- ] = 1.2 M (1:1 ratio) 4.6 Aqueous Reactions and Chemical Analysis

• Types of quantitative analysis:

– Gravimetric analysis: (mass analysis)

Example: precipitation reaction

– Volumetric analysis: (volume analysis)

Example: titration 41

• Gravimetric Analysis:

One form: isolation of a precipitate

A 0.825 g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess silver nitrate. If 1.725 g of AgCl precipitate forms, what is the percent by mass of Cl in the original sample?

• Volumetric analysis :

– Commonly accomplished by titration

• Addition of a solution of known concentration (standard solution) to another solution of unknown concentration.

– Standardization is the determination of the exact concentration of a solution.

– Equivalence point represents completion of the reaction. – Endpoint is where the titration is stopped.

nacid = nbase

– An indicator is used to signal the endpoint. 43 Calculate the molarity of 25.0 mL of a monoprotic acid if it took 45.50 mL of 0.25 M KOH to neutralize the acid.

nacid = nbase For monoprotic acid and base:

(MV)acid = (MV)base

Calculate the concentration of the acid or base remaining in solution when 10.7 mL of 0.211M HNO3 is added to 16.3 mL of 0.258 M NaOH.

A) 7.21 x10-2 M NaOH