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Types of Chemical Reactions & Solutions

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1 Types of Chemical Reactions & Solutions
Chapter 4

2 What is a Solution? Soluble – a substance that can dissolve in a given solvent Miscible: two liquids that can dissolve in each other Example: water and antifreeze Insoluble – substance cannot dissolve Immiscible: two liquids that cannot dissolve in each other Example: oil & water

3 Why Do Some Substances Dissolve and not Others?
To dissolve, solute particles must dissociate from each other and mix with solvent particles Attractive 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

4 Solvation/Hydration

5 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 water Water’s hydrogen bonds pulls the sugar molecules apart Oil is not a good solute because it has many C-H bonds (not polar) and few or no O-H (polar) bonds

6 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

7 The Nature of Aqueous Solutions
Composition of a solution can vary by changing amount dissolved: Electrical Conductivity Pure Water is a poor conductor Good conductors have strong electrolytes Weak conductors have weak electrolytes Non-conductors contain non-electrolytes

8 The Nature of Solutions
Originally 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.

9 Strong Electrolytes Strong electrolytes COMPLETELY ionize in solution
Example: NaCl  Na+ & Cl- Arrhenius first associated acidity to the presence of H+ ions (acidus = sour) Acids ionize to form H+ ions HCl, HNO3, H2SO4 are strong acids Strong Acids Completely Ionize Strong Bases also completely ionize OH- compounds completely ionize

10 Weak Electrolytes Weak Electrolytes exhibit a small degree of ionization in water Weak Acids, Weak Bases Example: 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 ions Examples: sugar, alcohol Do not conduct electricity in solution

11 Composition of Solutions
The 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

12 Example What 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+ ions Solution contains 1.0 moles NO3- (2*0.50) This is a CRITICAL conceptual idea to remember for future problems

13 How to Make a Solution of Known Concentration
Example: Make 1.00 L of M K2Cr2O7. How do you do this? Determine moles of K2Cr2O7 needed: 1.00L solution x mol K2Cr2O7/L solution = mol K2Cr2O7 Convert moles K2Cr2O7 grams 0.200 mol K2Cr2O7 x 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.

14 Dilution Dilution: 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 solution M1V1 = M2V2 Proper Procedure: Use measuring or volumetric pipettes to accurately measure solute Measuring pipette – has graduated lines Volumetric pipette – has ONE measurement to fill to

15 Dilution Procedure How to make 500mL or 1.00M acetic acid from a 17.4M stock solution Calculate volume of stock solution needed: Figure out moles of acetic acid: 500mL solution x 1L solution/1000mL Solution x 1.00 mole HC2H3O2 = mole HC2H3O2 V = mol HC2H3O2 /17.4mol HC2H3O2 /1L solution = L or 28.7mL of solution 500 mL of 1M solution – 28.7mL HC2H3O2 = 471.3mL H2O measured into flask Add 28.7mL HC2H3O2 to H2O

16 Types of Chemical Reactions
Last Year, you learned: Single replacement Double replacement combustion, Acid/Base No Longer a Sufficient Concept We must expand upon what you know to better understand what is happening in a reaction

17 New Types of Reactions New Categories of Reactions
Precipitation Acid/Base Oxidation-Reduction Virtually all reactions can fit into these classifications

18 Precipitation Reactions
A precipitation reaction forms a solid that falls (precipitates) from the solution. Example: K2CrO4 (aq) + Ba(NO3)2 (aq)  products including a yellow solid Actually looks more like: 2K+ (aq) + CrO42- (aq) + Ba2+ (aq) + 2 NO3 (aq) -- products How can they form a yellow solid?

19 Precipitation Reactions
Predicting 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

20 Precipitation Reactions
How Do We Know That? Based on Simple Solubility Rules Terms: Soluble – the salt will dissolve in water to a great extent Slightly Soluble = Insoluble – only a tiny, insignificant portion of the salt dissolves in water

21 Simple Solubility Rules
Text page 150 – Memorize them! Most nitrates (NO3) are soluble Most salts of alkali metals and ammonium ions are soluble Most Chloride, Bromide, and Iodide salts are soluble, EXCEPT Silver, Lead, Mercury Most Sulfates are soluble, EXCEPT Barium, Lead, Mercury, and Calcium Most Hydroxide salts are slightly soluble, EXCEPT Sodium and Potassium which are highly soluble. Barium, Tin, and Calcium are marginally soluble Most Sulfide, carbonates, chromates, and phosphates are only slightly soluble

22 Describing Solution Reactions
Convert the Formula Equation to Complete Ionic Equation List all the ions on both sides Solids (precipitates) are not ions All strong electrolytes are shown as ions in (aq) This will reveal that some ions do not participate in the reaction and are spectator ions Be Able to Identify spectator Ions Create a Net Ionic Equation Re-write the complete ionic equation, but remove the spectator ions from both sides

23 Example Step 1: Formula Equation
K2CrO4 (aq) + Ba(NO3)2 (aq)  BaCrO4 (s) + 2KNO3 (aq) 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 2K+ (aq) + CrO42- (aq) + Ba2+ (aq) + 2 NO3 (aq) -- BaCrO4 (s) + 2K+ (aq) + 2NO3- (aq) Step 4: Net Ionic Equation CrO42- (aq) + Ba2+ (aq) -- BaCrO4 (s)

24 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?

25 Stoichiometry for Solution Reactions
Identify all the species (ions or compounds) present in the reaction and determine what reaction occurs Write the balance NET IONIC Equation Calculate Moles of Reactants Determine Limiting Reactant Calculate Moles of Product or products Convert to grams or other units

26 Acid/Base Reactions Arrhenius Acids:
H+ ions = Acid OH- ions = base Refinement of Concept by Bronsted and Lowry: Acid is a proton donor Base is a proton acceptor

27 Acid/Base Reactions What’s the Difference? Example:
What does a H+ ion look like? A bare proton But 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

28 Acid/Base Reactions YES: 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

29 Acid/Base Reaction Calculations
List species present in combined solution BEFORE any reaction occurs Write a balance NET ionic equation Calculate moles of reactants using volumes and molarities Determine limiting reactant where appropriate Calculate moles of required reactant or product Convert to grams or volume as required

30 Acid/Base Titrations Titration is a volumetric analysis:
Uses a buret Uses 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 point Equivalence point is marked with an indicator When the indicator changes color, you have reached the endpoint of the titration

31 Acid/Base Titrations Once you have reached the ENDPOINT, it is a stoichiometry problem. How much standard solution (titrant) was used? How many moles What 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

32 ReDox in Acidic Solutions
Write separate equations for the half-reactions For each half-reaction: Balance all elements except H and O Balance O with water Balance H using H+ Balance charge using electrons Multiply half-reaction by integer to equalize electron totals Add half-reactions and cancel identical species Check that elements & Charges are balanced

33 ReDox in Basic Solutions
Write separate equations for the half-reactions For each half-reaction: Balance all elements except H and O Balance O with water Balance H using H+ Balance charge using electrons To both sides of the equation, add OH- ions to equal H+ ions Form H2O and eliminate from both sides Multiply half-reaction by integer to equalize electron totals Add half-reactions and cancel identical species Check that elements & Charges are balanced

34 ReDox in Basic Solutions
Book: Page 177, Example 4.20 As(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 first 2CN-(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)

35 ReDox in Basic Solutions
O2(g)  2H2O(l) Balance the Hydrogen with H+ O2(g) + 4H+(aq)  2H2O(l) Balance the Charge 4e- + 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-

36 ReDox in Basic Solutions
Add 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)

37 ReDox in Basic Solutions
Add OH- to both sides to cancel the H+ ions: O2(g) + 4H+(aq) 8CN-(aq) + 4Ag(s)  4Ag(CN)-2(aq) + 2H2O(l) + 4OH OH- Eliminate water molecules formed: O2(g) + 4H2O(l) 8CN-(aq) + 4Ag(s)  4Ag(CN)-2(aq) + 2H2O(l) +4OH- -2H2O H2O 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.

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