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Types of Chemical Reactions & Solutions. Chapter 4. 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

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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 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.

    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 = 0.500 mole HC2H3O2 • V = 0.500 mol HC2H3O2 /17.4mol HC2H3O2 /1L solution = 0.0287L 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 orBa(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? • 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- +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.