types of chemical reactions & solutions
Post on 30-Dec-2015
28 Views
Preview:
DESCRIPTION
TRANSCRIPT
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
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.
top related