Types of Chemical Reactions & Solutions

Chapter 4

What is a Solution?Soluble – a substance that can dissolve in a given solventMiscible: two liquids that can dissolve in each

otherExample: water and antifreeze

Insoluble – substance cannot dissolve Immiscible: two liquids that cannot dissolve in

each otherExample: oil & water

Why Do Some Substances Dissolve and not Others?

To dissolve, solute particles must dissociate from each other and mix with solvent particlesAttractive forces between solute and solvent

must be greater than attractive forces within the solute

Process of surrounding solute particles with solvent particles is called SOLVATION In water, it is also called HYDRATION

Solvation/Hydration

Aqueous Solutions of Molecular Compounds

Water is also a good solvent for many molecular compounds (Example: sugar)Sugar has many O-H bonds (polar)When water is added, the O-H bond becomes a

site for hydrogen bonding with waterWater’s hydrogen bonds pulls the sugar

molecules apartOil is not a good solute because it has many C-H

bonds (not polar) and few or no O-H (polar) bonds

Factors that Affect Solvation Rate

Increase Solvation Rate (Dissolve Faster) by:Agitation (stirring) Increase surface area (make particles smaller)Temperature (make it hotter)

All these increase the number of collision between water and the solute

The Nature of Aqueous Solutions

Composition of a solution can vary by changing amount dissolved:

Electrical ConductivityPure Water is a poor conductorGood conductors have strong electrolytesWeak conductors have weak electrolytesNon-conductors contain non-electrolytes

The Nature of SolutionsOriginally identified by Arrhenius

Conductivity arises from the presence of IONS in the solution Ions are charged particles Ionic theory starts to make sense

Arrhenius postulates:The extent to which a solution can conduct

electricity depends DIRECTLY on the number of ions present.

Strong ElectrolytesStrong electrolytes COMPLETELY ionize in solutionExample: NaCl Na+ & Cl-

Arrhenius first associated acidity to the presence of H+ ions (acidus = sour)Acids ionize to form H+ ionsHCl, HNO3, H2SO4 are strong acidsStrong Acids Completely Ionize

Strong Bases also completely ionizeOH- compounds completely ionize

Weak ElectrolytesWeak Electrolytes exhibit a small degree of ionization in waterWeak Acids, Weak BasesExample: Acetic Acid is a weak acid (~1/100

molecules dissociates)Example Ammonia is a weak base (~1/100

molecules dissociates)

Non-Electrolytes dissolve in water, but do not dissociate into ionsExamples: sugar, alcoholDo not conduct electricity in solution

Composition of SolutionsThe nature of the chemical reaction frequently depends on the amounts of chemicals present: Molarity = moles solute/1 liter solution

Concentration is determined BEFORE it dissolves 1.0M NaCl is made by measuring 1.0 Moles of NaCl and

adding enough water to make 1 L of solutions 1.0M does not mean it contains 1.0 mole of NaCl units It contains 1.0 mole of Na+ ions and 1.0 mole of Cl- ions

Finding moles from molarity: Liters solution x molarity = moles of solute

ExampleWhat is the concentration of each type of ion in the solution 0.50 M Co(NO3)2

Solid compound dissolves into Co2+ ions and NO3- ions

Co(NO3)2 (s) - Co2+ (aq) + 2 NO3- (aq)Solution contains 0.50 moles Co2+ ionsSolution contains 1.0 moles NO3- (2*0.50)

This is a CRITICAL conceptual idea to remember for future problems

How to Make a Solution of Known Concentration

Example: Make 1.00 L of 0.200 M K2Cr2O7. How do you do this?Determine moles of K2Cr2O7 needed:

1.00L solution x 0.2000 mol K2Cr2O7/L solution = 0.200 mol K2Cr2O7

Convert moles K2Cr2O7 grams

0.200 mol K2Cr2O7 x 294.20 g K2Cr2O7/mol K2Cr2O7 = 58.8g K2Cr2O7

Measure out 58.8g K2Cr2O7. Transfer it to a 1.00L volumetric flask. Add distilled water to the mark on the flask.

DilutionDilution: Adding water to a prepared (or stock) solution in order to achieve a desired molarity.Key: Moles of solute after dilution = moles of

solute before solutionM1V1 = M2V2

Proper Procedure:Use measuring or volumetric pipettes to

accurately measure soluteMeasuring pipette – has graduated linesVolumetric pipette – has ONE measurement to fill to

Dilution ProcedureHow to make 500mL or 1.00M acetic acid from a 17.4M stock solution

1. Calculate volume of stock solution needed:Figure out moles of acetic acid:500mL solution x 1L solution/1000mL Solution x 1.00 mole HC2H3O2 = 0.500 mole HC2H3O2

V = 0.500 mol HC2H3O2 /17.4mol HC2H3O2 /1L solution = 0.0287L or 28.7mL of solution

2. 500 mL of 1M solution – 28.7mL HC2H3O2 = 471.3mL H2O measured into flask

3. Add 28.7mL HC2H3O2 to H2O

Types of Chemical ReactionsLast Year, you learned:Single replacementDouble replacementcombustion,Acid/Base

No Longer a Sufficient ConceptWe must expand upon what you know to better

understand what is happening in a reaction

New Types of ReactionsNew Categories of ReactionsPrecipitationAcid/BaseOxidation-Reduction

Virtually all reactions can fit into these classifications

Precipitation ReactionsA precipitation reaction forms a solid that falls (precipitates) from the solution.

Example:K2CrO4 (aq) + Ba(NO3)2 (aq) products

including a yellow solidActually looks more like:2K+ (aq) + CrO42- (aq) + Ba2+ (aq) + 2 NO3 (aq) --

productsHow can they form a yellow solid?

Precipitation ReactionsPredicting products is very hard

Actual reaction products must be confirmed experimentally before you can really conclude the reaction

Predicting from what we know: Compound must be electrically neutral Must contain both cations and anions What possible combinations exist? K2CrO4, KNO3, BaCrO4, Ba(NO3)2

Can’t be K2CrO4 or Ba(NO3)2 – these are reactants

KNO3 will always be soluble so precipitate must be BaCrO4

Precipitation ReactionsHow Do We Know That?Based on Simple Solubility Rules

Terms:Soluble – the salt will dissolve in water to a great

extentSlightly Soluble = Insoluble – only a tiny,

insignificant portion of the salt dissolves in water

Simple Solubility RulesText page 150 – Memorize them!

1. Most nitrates (NO3) are soluble2. Most salts of alkali metals and ammonium ions are

soluble3. Most Chloride, Bromide, and Iodide salts are soluble,

EXCEPT Silver, Lead, Mercury4. Most Sulfates are soluble, EXCEPT Barium, Lead,

Mercury, and Calcium5. Most Hydroxide salts are slightly soluble, EXCEPT

Sodium and Potassium which are highly soluble. Barium, Tin, and Calcium are marginally soluble

6. Most Sulfide, carbonates, chromates, and phosphates are only slightly soluble

Describing Solution ReactionsConvert the Formula Equation to Complete Ionic EquationList all the ions on both sidesSolids (precipitates) are not ionsAll strong electrolytes are shown as ions in (aq)This will reveal that some ions do not participate

in the reaction and are spectator ionsBe Able to Identify spectator Ions

Create a Net Ionic EquationRe-write the complete ionic equation, but remove

the spectator ions from both sides

Example

K2CrO4 (aq) + Ba(NO3)2 (aq) BaCrO4 (s) + 2KNO3 (aq)

2K+ (aq) + CrO42- (aq) + Ba2+ (aq) + 2 NO3 (aq) -- BaCrO4 (s) + 2K+ (aq) + 2NO3

- (aq)

Step 1: Formula Equation

Step 2: Complete Ionic Equation

2K+ (aq) + CrO42- (aq) + Ba2+ (aq) + 2 NO3 (aq) -- BaCrO4 (s) + 2K+ (aq) + 2NO3

- (aq)

Step 3: Eliminate Spectator Ions

CrO42- (aq) + Ba2+ (aq) -- BaCrO4 (s)

Step 4: Net Ionic Equation

Predict Products of Reactions:CaCl2(aq) + 2Ag2SO4(aq) ??

H2SO4 + Na2CO3 ??

Na2CrO4 + AgNO3 ??

Write the total ionic equation for the reaction of hydrofluoric acid with potassium hydroxide.

When aqueous solutions of iron(III) sulfate (Fe2(SO4)3) and sodium hydroxide were mixed, a precipitate formed. What is the precipitate?

Stoichiometry for Solution Reactions

1. Identify all the species (ions or compounds) present in the reaction and determine what reaction occurs

2. Write the balance NET IONIC Equation

3. Calculate Moles of Reactants

4. Determine Limiting Reactant

5. Calculate Moles of Product or products

6. Convert to grams or other units

Acid/Base ReactionsArrhenius Acids:H+ ions = AcidOH- ions = base

Refinement of Concept by Bronsted and Lowry:Acid is a proton donorBase is a proton acceptor

Acid/Base ReactionsWhat’s the Difference?What does a H+ ion look like?A bare protonBut you can have bases that are not OH-

Example:KOH (aq) + HC2H3O2 (aq) ??

K+ (aq) + OH- (aq) + HC2H3O2 (aq) are the species present before any reaction occurs

A precipitation reaction could occur between K+ and OH- but KOH is soluble

Or is there another possible proton donor to OH-?

Weak Acid – does not ionize

Acid/Base ReactionsYES: HC2H3O2 molecules

Hydroxide ion is such a strong base that for purposes of stoichiometric equations, it can be assumed to react completely with any weak acid encountered

Actual net ionic equation is:OH- + HC2H3O2 H2O + C2H3O2

-

Acid/Base reactions are called neutralization reactions

Acid/Base Reaction Calculations

List species present in combined solution BEFORE any reaction occursWrite a balance NET ionic equationCalculate moles of reactants using volumes and molaritiesDetermine limiting reactant where appropriateCalculate moles of required reactant or productConvert to grams or volume as required

Acid/Base TitrationsTitration is a volumetric analysis:Uses a buretUses volume of a KNOWN solution (titrant)Delivered into an unknown solution (analyte)

When titrant added is exactly reacted with analyte you have the equivalence point or stoichiometric pointEquivalence point is marked with an indicatorWhen the indicator changes color, you have

reached the endpoint of the titration

Acid/Base TitrationsOnce you have reached the ENDPOINT, it is a stoichiometry problem.How much standard solution (titrant) was used?How many molesWhat was the reaction?How many moles of analyte was neutralized?What volume of analyte was neutralized?Calculate molarity of the analyte.

Review Sample Exercise 4.15

ReDox in Acidic Solutions1. Write separate equations for the half-reactions

2. For each half-reaction:a. Balance all elements except H and O

b. Balance O with water

c. Balance H using H+

d. Balance charge using electrons

3. Multiply half-reaction by integer to equalize electron totals

4. Add half-reactions and cancel identical species

5. Check that elements & Charges are balanced

ReDox in Basic Solutions1. Write separate equations for the half-reactions2. For each half-reaction:

a. Balance all elements except H and Ob. Balance O with waterc. Balance H using H+

d. Balance charge using electrons3. To both sides of the equation, add OH- ions to

equal H+ ionsa. Form H2O and eliminate from both sides

4. Multiply half-reaction by integer to equalize electron totals

5. Add half-reactions and cancel identical species6. Check that elements & Charges are balanced

ReDox in Basic SolutionsBook: Page 177, Example 4.20As(s) + CN-(aq) + O2(g) Ag(CN)-

2(aq)

Balance oxidation ½ reaction first Balance as if H+ ions were present. Balance C

and N first2CN-(aq) + Ag(s) Ag(CN)-

2(aq)

Balance the charge

2CN-(aq) + Ag(s) Ag(CN)-2(aq) + e-

Balance the reduction ½ reaction (O2)

O2(g) 2H2O(l)

ReDox in Basic SolutionsO2(g) 2H2O(l)

Balance the Hydrogen with H+

O2(g) + 4H+(aq) 2H2O(l)

Balance the Charge4e- + O2(g) + 4H+(aq) 2H2O(l)

Multiply Balanced oxidation ½ reaction by 4

4(2CN-(aq) + Ag(s) Ag(CN)-2(aq) + e-)

8CN-(aq) + 4Ag(s) 4Ag(CN)-2(aq) + 4e-

ReDox in Basic SolutionsAdd the ½ reactions and cancel identical species:8CN-(aq) + 4Ag(s) 4Ag(CN)-

2(aq) + 4e-

4e- + O2(g) + 4H+(aq) 2H2O(l)

4e- + O2(g) + 4H+(aq) 8CN-(aq) + 4Ag(s)

4Ag(CN)-2(aq) + 4e- + 2H2O(l)

ReDox in Basic SolutionsAdd OH- to both sides to cancel the H+ ions:

O2(g) + 4H+(aq) 8CN-(aq) + 4Ag(s) 4Ag(CN)-2(aq) + 2H2O(l)

+ 4OH- +4OH-

Eliminate water molecules formed:O2(g) + 4H2O(l) 8CN-(aq) + 4Ag(s) 4Ag(CN)-

2(aq) + 2H2O(l) +4OH-

-2H2O -2H2O

Final Balanced ReDox Reaction:

O2(g) + 2H2O(l) 8CN-(aq) + 4Ag(s) 4Ag(CN)-2(aq) +4OH-

Double-Check to see that elements balance, charges balance.