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Stoichiometry of Precipitation Reactions Procedure

Stoichiometry of Precipitation Reactions Procedure Example: ??? g NaCl needed to precipitate all Ag + from 1.50L of 0.100 M AgNO 3 ? Net Ionic Equation Ag + (aq) + Cl - (aq) ----- --> AgCl(s).

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Stoichiometry of Precipitation Reactions Procedure

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  1. Stoichiometry of Precipitation Reactions • Procedure • Example: ??? g NaCl needed to precipitate all Ag+ from 1.50L of • 0.100 M AgNO3? • Net Ionic Equation Ag+(aq) + Cl-(aq) -------> AgCl(s)

  2. C. Example: ??? g PbSO4 form when 1.25L 0.0500M Pb(NO3)2 and 2.00L 0.0250M Na2SO4 are mixed? • Net ionic equation: Pb2+(aq) + SO42-(aq) -------> PbSO4(s) • Acid-Base Reactions • Acid Base models • Arrhenius model: correct, but too limiting • Acid = H+ producer in water • Base = OH- producer in water • Bronsted-Lowry model: includes Arrhenius model, but more general • Acid = proton (H+) donor • Base = proton acceptor • Example: HC2H3O2 + H2O -------> C2H3O3- + H3O+ acid base

  3. Neutralization Reactions • Water is a nonelectrolyte (not highly ionized in solution) • H+(aq) + OH-(aq) -------> H2O(l) • Reaction of an acid and a base (usually to form water) = Neutralization • H+ is a strong acid and will completely react with any weak base present • H+(aq) + NH3(aq) -------> NH4+(aq) • OH- is a strong base and will completely react with any weak acid present • OH-(aq) + HC2H3O2(aq) -------> H2O(l) + C2H3O2-(aq) • Similar to a precipitation reaction, except the product is a liquid (H2O) • Example: ??? L of 0.100M HCl to neutralize 25.0 ml 0.350M NaOH? • Net Ionic Equation: H+(aq) + OH-(aq) -------> H2O(l)

  4. D. Example: ??? mol H2O is formed when 28.0ml of 0.250M HNO3 and 53.0ml of 0.320M KOH are reacted? What is the concentration of H+/OH-left? • Net Ionic Equation: H+(aq) + OH-(aq) -------> H2O(l) • Calculate how much water is formed • H+ was limiting, so calculate how much OH- is left over and concentration • Acid-Base Titrations • Definition: volumetric analysis of the concentration of an unkown • Titrant =solution of known concentration whose volume is measured • Analyte =solution whose concentration is to be determined • Equivalence Point = amount of titrant just reacts with all analyte • Indicator = changes color at endpoint of the titration

  5. Acid-Base Titration results in neutralization of all of the analyte • a. H+(aq) + OH-(aq) -------> H2O(l) • Phenolphthalein indicator is colorless in acid and pink in base • A buret accurately measures the amount of titrant added The flask contains acid of unknown concentration and phenolphthalein. The buret contains base of a known concentration Base is added dropwise. The faint pink color goes away as you stir. The endpoint hasn’t been reached yet. The endpoint has now been reached as the pink color persists. Measuring the volume of base dispensed aids calculation.

  6. Standardizing the base solution: Example: • 1.3009g KHP is dissolved in distilled water (doesn’t matter how much) and titrated with 41.20ml unknown NaOH solution. • HP-(aq) + OH-(aq) -------> H2O(l) + P2-(aq) • Calculating the concentration of the analyte: Example: • 0.3518g of sample was titrated with 10.59ml of 0.1546M NaOH. Calculate the mass percent of HC7H5O2 (benzoic acid) in the sample • HC7H5O2(aq) + OH-(aq) -------> H2O(l) + C7H5O2-(aq)

  7. F. Reactions that give off gases • Sometimes the product of a reaction is not a solid, but a gas • We can still observe that something happened: bubbles form

  8. III. Oxidation-ReductionReactions • Definition = Reactions in which electrons are transferred (aka Redox Rxns) • Example: 2Na(s) + Cl2(g) -------> 2NaCl(s) • Each Na loses one electron to become Na+ (Oxidation) • Each Cl gains one electron to become Cl- (Reduction) • Oxidation States = tool for keeping track of electrons in Redox reactions • The total charge on all atoms must match the molecule or ion charge • Oxidation states of a few elements help us calculate for all others

  9. B. Example: • CO2 • O = -2 so we have a total of -4 coming from the 2 oxygens • C must be +4 to balance the negative charge • SF6 • F = -1 so we have a total of -6 coming from the 6 fluorines • S must be +6 to balance the negative charge • NO3- • O = -2 so we have -6 coming from the 3 oxygens • N must be +5 in order to give us an overall 1- charge • C. Recognizing what is happening in Redox Reactions Radius decreases Radius increases

  10. CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g) • (-4)(+1) (0) (+4)(-2) (+1)(-2) • CH4 -----> CO2 + 8e- Carbon = oxidized, CH4 = reducing agent • 1(-4) 1(+4) • 2O2 + 8e- -----> CO2 + 2H2O Oxygen = reduced = oxidizing agent • 2(0) 2(-2) 2(-2) • Example: 2Al(s) + 3I2(s) -----> 2AlI3(s) • (0) (0) (+3)(-1) • a. Al is oxidized = reducing agent • b. I is reduced; I2 is the oxidizing agent • 3. Example: Oxidized? Reduced? Oxidizing agent? Reducing agent? • 2PbS(s) + 3O2(g) -----> 2PbO(s) + 2SO2(g) • 2PbO(s) + CO(g) -----> Pb(s) + CO2(g)

  11. Balancing Redox Equations: Half-Reaction Method in Acidic Solution • MnO4-(aq) + Fe2+(aq) -----> Fe3+(aq) + Mn2+(aq) (acidic solution) • Identify and write equations for the two half-reactions • MnO4- -----> Mn2+ (this is the reduction half-reaction) • (+7)(-2) (+2) • Fe2+ -----> Fe3+ (this is the oxidation half-reaction) • (+2) (+3) • Balance each half-reaction • Add water if you need oxygen • Add H+ if you need hydrogen (since we are in acidic solution) • Balance the charge by adding electrons • MnO4- -----> Mn2+ + 4H2O • 8H+ + MnO4- -----> Mn2+ + 4H2O • (+7) (+2) • 5e- + 8H+ + MnO4- -----> Mn2+ + 4H2O Balanced! • Fe2+ -----> Fe3+ + 1e- Balanced!

  12. 3. Equalize the number of electrons in each half-reaction and add reactions • 5(Fe2+ -----> Fe3+ + 1e-) = 5Fe2+ -----> 5Fe3+ + 5e- • 5e- + 8H+ + MnO4- -----> Mn2+ + 4H2O • 5Fe2+ + 8H+ + MnO4- ------> 5Fe3+ + Mn2+ + 4H2O • Species (including e-) on each side cancel out (algebra) • Check that the charges and elements all balance: DONE! • Example: H+ + Cr2O72- + C2H5OH -----> Cr3+ + CO2 + H2O • Balancing Redox Equations: Half-Reaction Method in Basic Solution • Follow the Acidic Solution Method until you have the final balance eqn • H+ can’t exist in basic solution, so add enough OH- to both sides to turn all of the H+ in H2O • Example: Ag + CN- + O2 -----> Ag(CN)2- (basic solution) • Ag + CN- -----> Ag(CN)2- (oxidation half-reaction) • Becomes: Ag + 2CN- -----> Ag(CN)2- + 1e- Balanced • O2 -----> (reduction half-reaction) • Becomes: 4e- + 4H+ + O2 -----> 2H2O Balanced

  13. Equalize the electrons in each half-reaction and add reactions • 4(Ag + 2CN- -----> Ag(CN)2- + 1e-) • Becomes: 4Ag + 8CN- -----> 4Ag(CN)2- + 4e- • Add to: 4e- + 4H+ + O2 -----> 2H2O • Gives: 4Ag + 8CN- + 4H+ + O2 -----> 4Ag(CN)2- + 2H2O DONE! • Add OH- ions to both sides to remove H+ ions • 4Ag + 8CN- + 4H+ + O2 + 4OH- -----> 4Ag(CN)2- + 2H2O + 4OH- • 4Ag + 8CN- + 4H2O + O2 -----> 4Ag(CN)2- + 2H2O + 4OH- • Cancel water molecules appearing on both sides of the equation • 4Ag + 8CN- + 2H2O + O2 -----> 4Ag(CN)2- + 4OH- • Check that everything balances REALLY DONE!

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