Chem_2_Ch_4 - Chapter 4 Types of Chemical Reactions and...

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Unformatted text preview: Chapter 4 Types of Chemical Reactions and Solution Stoichiometry The dissolution process... The Partial + and charges on the water molecule make it a unique polar solvent. Water h y d ro g e n b o nd ic e liq uid wa te r Polar water molecules interact with the positive and negative ions of a salt, assisting with the dissolution process. 15.2 Na+ and Cl ions hydrated by H2O molecules. 6 Electrolytes Electrolytic solutions are aqueous solutions that contain dissolved ions. Strong Electrolytes Solutions with a high concentra tion of dissolved ions( i.e. strong acids, strong bases and salts strongly dissociated). NaCl(s) Na+(aq) + Cl(aq) H2O Weak Electrolytes Solutions with a low concentra tion of dissolved ions( i.e. weak acids, weak bases and weakly dissociated salts). HA(s) H+(aq) + A(aq) HA(s) H2O Non-Electrolytes Solutions with no dissolved ions( i.e. molecular species). Sugar(s) Sugar(l) H2O Electrical conductivity A strong electrolyte solution conducts electricity strongly. A weak electrolyte solution conducts electricity weakly. A nonelectrolyte solution does not conduct electricity. Electrical conductivity of aqueous solutions. HCL; a strong acid CH3COOH; a weak acid EtOH; molecular Concentration Units Molarity Molarity (M): a unit of concentration Moles solute M = Liters solution Other measures of concentration Normality (eq/L) ppm and ppb Percent (wt/wt, wt/vol, vol/vol) Density (g/mL) (Solute is the solid dissolved in a solvent; solute + solvent = solution) Question What's the molarity of a solution that has 142.05 g of Na2SO4 completely dissolved in water and brought to a "final " volume of 1.00L? Answer ? M Na2SO4 142.05 g Na2SO4 x 1 mol Na2SO4/142.05 g x 1/ 1 L = 1.0000 M Na SO 2 4 Question What is molarity of propanol, C3H7OH, when 60. mL propanol is mixed with enough H2O to make a final volume of 100 mL? The MW of propanol = 60.11 g/mol and the density of propanol = 0.804 g/mL. a) 8.0 b) 3.5 M c) 7.5 M Answer What is molar concentration of propanol when 60. mL propanol is mixed with H2O to a final volume of 100 mL? a) 8.0 M b) 3.5 M c) 7.5 M Question What is the molar concentration of ethanol, C2H5OH, in a 45.%, by mass, aqueous solution? (an aqueous ethanol solution has a density ~ 1.00 g/mL, ethanol MW 46.08 g/mole) a) 7.8 M b) 5.6 M c) 9.8 M Answer 45%, so 45 g ethanol per 100 g solution C2H5OH 46.08 g/mole, d = 1.00 g/mL 45.g/100g x 1.00g/ml x 1mol./46.08 g x 1000mL/1L = 9.8 M a) 7.8 M b) 5.6 M c) 9.8 M Question A solution of table salt (NaCl) in water is heated to the boiling point. What happens to the molar concentration of NaCl as the solution boils? A) Increases B) Decreases C) Stays the same Answer The water boils away, but the NaCl doesn't. The same number of moles of NaCl is left in a smaller volume of water, increasing the molar concentration. A) Increases B) Decreases C) Stays the same Question You have a sugar solution (Solution A) with concentration x. You pour of this solution into a beaker and add an equal volume of water to make Solution B. How are the molar concentrations of sugar in solutions A and B related? A) They are equal B) A has twice the concentration of B C) A has half the concentration of B D) B has the concentration of A Answer M of sugar in A = M of sugar in B Volume of A = Volume of B Molarity of A = 2 x molarity of B A B A B B) A has twice the concentration of B Question: molarity and dilution A solution is made by dissolving 0.10 mole of Na2SO4 in water to a final volume of 1.00 L. What is the molar concentration of Na+ ions in the solution? Well... let's consider the stoichiometry. Na2SO4(s) 2Na+(aq) + SO42(aq) 2 4 + For every 1 mol. Na SO there are 2 mol. Na . Answer 0.20 M Na + Question Take 100. mL of the 0.20 M Na+ solution and add water to it until a final volume of 500. mL is reached. What is the new molarity of Na+? Answer 0.040 M Na + "Simple" dilution formula If a solution is diluted from initial volume V1 to final volume V2, then moles1 = moles2 M1V1 = M2V2 Question The density of water at 25C is 1.00 g/mL. The MW of water is 18.02 g/mole. Calculate the molarity of water molecules in pure water. Possibly helpful information: 1000 mL = 1 L Avogadro's number = 6.022 x 1023 Answer 1.00 g/mL x 1000 mL/L x 1mol./18.02g = 55.5 M Did we need to use Avogadro's number? No, we didn't! Question Which of the following must be known in order to calculate the molarity of a salt solution? A) the mass of salt added B) the molar mass of the salt C) the volume of water added D) the total volume of the final solution Answer Which of the following must be known in order to calculate the molarity of a salt solution? A) the mass of salt added B) the molar mass of the salt C) the volume of water added D) the total volume of the final solution Precipitation Reactions Precipitation, ppt, reactions result in the formation of a solid product from soluble (i.e. dissolved) reactants. Na+(aq) + Cl(aq) + Ag+(aq) + NO3(aq) Na+(aq) + NO3(aq) + AgCl(s) Solubility of salts - basic rules (Ch. 4 Table 4.1) Almost all nitrate (NO3) salts are soluble Most salts of Na+, K+, and NH4+ are soluble Most Cl and SO42 salts are soluble (not: AgCl, PbCl2, Hg2Cl2, BaSO4, PbSO4, CaSO4) Most OH, S2, CO32, PO43 salts are only slightly soluble (NaOH, KOH very soluble, Ca(OH)2 somewhat soluble) Types of equations for pptn rxns Balanced Molecular Eq. CoCl2(aq) + 2NaOH(aq) Co(OH)2(s) + 2NaCl(aq) Total Ionic Eq. Co (aq) + 2Cl (aq) + 2Na (aq) + 2OH (aq) Co(OH) (s) + 2Na (aq) + 2Cl (aq) 2 + 2+ + Spectator Ions Na and Cl are spectator ions. They appear in the total ionic equation. They do not take part in the chemical reaction; they are not included in the net ionic equation. Net ionic Co2+(aq) + 2OH(aq) Co(OH)2(s) + Precipitation reactions Pb2+(aq), NO3(aq) K+(aq), I(aq) PbI2(s), NO3(aq), K+(aq) Dissolving lead nitrate: Pb(NO3)2(s) Pb2+(aq) + 2NO3(aq) Dissolving potassium iodide: KI(s) K+(aq) + I(aq) Precipitation reaction: Pb2+(aq) + 2I(aq) PbI2(s) Question 1.50 g of lead(II) nitrate is mixed with 125 mL of 0.100 M sodium sulfate solution. What are the molar concentrations of all ions that remain in solution after the reaction is complete? The final reaction volume is 125 mL. The MW of lead(II) nitrate = 331.2 A Step by Step Approach 1. Write the balanced equations. 2. Find the number of moles of all ions. 3. Determine the limiting reagent. 4. Calculate the amount of product formed. 5. Calculate the amount of reactants used up. 6. Calculate conc. of remaining reactants. Question 1.50 g of lead(II) nitrate is mixed with 125 mL of 0.100 M sodium sulfate solution. What are the concentrations of all ions that remain in solution after the reaction is complete? The final reaction volume is 125 mL. The MW of lead(II) nitrate = 331.2 Answer Lead(II) nitrate: Pb(NO3)2 Pb2+ + 2NO3 Sodium sulfate: Na2SO4 2Na+ + SO42 Reaction: Pb2+(aq) + SO42(aq) PbSO4(s) Pb(NO3)2: MW = 331.22, 1.50 g x mol./331.22 g = 0.00453 moles Pb2+ 0.00453 moles Pb2+ x 2/1 = 0.00906 moles NO3 Na2SO4: 0.125 L of 0.100 M = 0.0125 moles SO42 0.0125 moles SO42 x 2/1= 0.0250 moles Na+ Answer - Cont. So Pb2+ is limiting, 0.00453 moles Pb2+, and we form 2+ 0.00453 moles of PbSO4, using up all the Pb 2+ 2 Answer - Cont. NO3 doesn't react, 0.00906 moles/0.125 L = 0.0725 M 3 NO (Spectator ion) + Na+ doesn't react, 0.0250 moles/0.125 L = 0.200 M Na (Spectator ion) SO 2 was initially 0.0125 moles and 0.00453 moles Question A student makes 100.0 mL of a lead(II) nitrate solution but forgets to cap it. When he returns next week, the volume is only 80.0 mL. He also forgot to write down the concentration of the solution. His lab partner says not to worry. They measure out 2.00 mL of the solution and add an excess of Con. NaCl solution, producing a precipitate. When dried, the solid weighs 3.407 g. What was the concentration of the original lead(II) nitrate solution? A Step by Step Approach lead(II) nitrate is Pb(NO ) 32 What happened to reduce the volume from 100.0 to 80.0 mL? Some water evaporated. But, all the lead(II)nitrate is still there. What was the solid formed when sodium chloride was added? We know NaCl is soluble in water and almost all Answer How much lead (II) was in the precipitate? Atomic wts: Pb 207.20, Cl 35.45 MW of PbCl2 = 278.10 g/mole So 3.407 g of PbCl2 contains 3.407 g x mol/278.1 g x 1mol Pb/1mol PbCl2= 0.0123 mol +2 Pb This is the # of mol. of Pb+2 in the 2.00 mL of the evaporated sample (i.e. 80.0 mL). The molarity of Pb+2 in 80.00 mL is: 0.0123 moles/0.002 L = 6.15M Pb+2 Original volume was 100.0 mL, so original molarity of lead(II) nitrate = 0.080L x 6.15mol. Pb+2/L x 1/0.100 L x 1mol. Pb(NO3)2/1mol. Pb+2 4.92 M Pb(NO ) = Answer - Cont. Question 0.010 moles of sodium sulfate are dissolved in 100 mL of water in a beaker. Then, a solution of barium nitrate is added slowly to the beaker. A precipitate forms. What is the precipitate? Na2SO4(s) 2Na+(aq) + SO42(aq) Ba(NO3)2(s) Ba2+(aq) + 2NO3(aq) Ba2+(aq) + SO42(aq) BaSO4(s) Question - cont. What will a plot of grams of precipitate formed (yaxis) versus volume of barium nitrate solution added (xaxis) look like? BaSO4 (s) BaSO4 (s) BaSO4 (s) Ba(NO3)2 Vol. Ba(NO3)2 Vol. Ba(NO3)2 Vol. (no more ppt forms after all the sulfate is used up) Question - cont. How many moles of barium nitrate must be added before the amount of precipitate formed stops increasing? Initial moles of SO42 present = 0.010 Each mole of barium nitrate contains one mole of Ba2+ BaSO4 is the formula, so moles of Ba2+ from Barium nitrate needed to react with all the sulfate = 0.010 moles Acid Base Reactions A reaction between an acid and a base, which produces water and/or a salt. HCl + NaOH H2O + NaCl HCl + NH3 NH Cl Acids and bases Arrhenius definition: Acid produces H+ when dissolved in water Base produces OH when dissolved in water BrnstedLowry definition: Acid is a proton (H+) donor Base is a proton acceptor 3 + + (H O and H are equivalent) 3 + What is the structure of proton in water? H3O+ surrounded by 4 H2O H9O4+ cluster in water Strong and weak acids Strong acids ionize completely in H2O to form H+; they form strong electrolytic solutions. HCl + H2O H3O+ + Cl or HCl H + Cl + Weak acids remain mostly in unionized form; they form weak electrolytic solutions. HC H O H+ + C2H3O2 Strong bases dissociate completely to form OH; they form strong electrolytic solutions. NaOH Na+ + OH Weak bases are hydrolyzed by water to a small degree to form a little OH; they form weak electrolytic solutions. NH3 + H2O NH4+ + OH and the OH can react with H+ from acids Strong and weak bases Acid-base reactions H+ reacts with OH to form H2O Example: reaction of aqueous acetic acid, HC2H3O2 (a weak acid) with aqueous potassium hydroxide, KOH (a strong base) HC2H3O2(aq) + KOH(aq) H O(l) + 2 KC2H3O2 (aq) Monoprotic and polyprotic acids Monoprotic acids have only one acidic proton, can produce only one H+ per molecule of acid HC2H3O2 H+ + C2H3O2 HC2H3O2 + OH C2H3O2 + H2O Polyprotic acids have more than one acidic proton, can produce more than one H+ per molecule of acid H SO4 2H+ + SO42 H SO4 + 2OH SO42 + 2H O 2 Titrations Reaction of an acid with enough base to neutralize it (all the H+ reacts with OH to form water). or Reaction of a base with enough acid to neutralize it (all the OH reacts with H+ to form water). Titrations The fundamental relation: At the equivalence point (i.e. end point) of a titration, + moles H = moles OH enough base (or acid) has been added to react with all the acid (or base) originally present Question How many grams of NaOH are needed to neutralize 20.0 mL of 0.150 M HCl solution? NaOH + HCl NaCl + H2O 20.0mL HCl x 1L/1000mL x 0.150 mol/L HCl x 1 mol NaOH/1 mol HCl x 40g/mol NaOH = 0.120 g NaOH + Question 40 mL of 0.50 M H2SO4 (a diprotic acid) is required to neutralize 20 mL of NaOH solution. What is the molarity of the NaOH? + moles H = moles OH when neutralized H2SO4 + 2NaOH Na2SO4 + H2O = 2.0 M NaOH Question A 2.20 g sample of an unknown acid having empirical formula C3H4O3 is dissolved in 1.0 L of water. A titration required 25.0 mL of 0.500 M NaOH to react completely with all the acid present. Assuming the unknown acid is monoprotic, what is the molecular formula of the acid? Answer Moles OH added = moles H+ originally present = 0.0250 L x 0.500 moles/L = 0.0125 moles Amount of acid originally present = 2.20 g MW = 2.20 g/0.0125 moles = 176 g/mole Empirical formula C3H4O3 has MW = 88g/mole So molecular formula must be twice empirical formula, C H O (i.e. n = 176/88.07 = 2) "Redox" reactions involve the transfer of electrons from one element to another Simple example: 2Na(s) + Cl2(g) 2NaCl(s) NaCl contains Na ions and Cl ions + Oxidation-Reduction Reactions Oxidation-reduction reactions Most "redox" reactions are a bit more complicated; we need a systematic way to keep track of which atoms gain / lose the electrons. Oxidation states: a method of assigning electrons to elements in compounds. Treats bonds as ionic (electrons are owned by one atom, not shared between two bonded atoms as in covalent bonds); this is somewhat unrealistic, but it's still useful. Oxidation states (ox. numbers) Rules for assigning electrons to atoms within molecules or ions: 1. Neutral elements/diatomics: ox. state = 0 2. Monatomic ions: ox. state = charge 3. H has an ox. state +1 in covalent compounds with nonmetals 4. O has ox. state 2 in covalent compounds except peroxides (OO), where it is 1. Oxidation states - continued 5. Otherwise, in binary compounds the more electronegative element has negative ox. number equal to its charge in ionic compounds. 6. Sum of ox. numbers must equal total charge on compound (i.e. 0) or polyatomic ion (i.e + or charge). Electronegativity increases from the lower left to the upper right of the periodic table (i.e. B < F). Oxidation states - examples NaNO3: contains Na+ and NO3 ions Na: +1 by Rule 2 O: 2 by Rule 4 N: +5 by Rule 6 (total charge balance) BrF3: covalent, F more electronegative F: 1 by Rule 5 Br: +3 by Rule 6 (total charge balance) CH3OH: covalent compound H: +1 by Rule 3 O: 2 by Rule 4 C: 2 by Rule 6 Example: combustion of ethanol Oxidation - reduction reactions 2 +1 2 +1 0 +4 2 +1 2 ox. states C2H5OH + 3O2 2CO2 + 3H2O 2 carbon atoms go from 2 to +4 C is oxidized; C2H5OH is reducing agent 6 oxygen atoms go from 0 to 2 O is reduced; O2 is oxidizing agent Total change in oxidation state must be zero for balanced reaction! Redox rxns Step by Step Method one: by inspection 1. Assign oxidation states to all elements 2. Decide which elements are oxidized and reduced, and the changes in oxidation state 3. Choose coefficients for species containing oxidized and reduced elements so that total change in oxidation state is zero 4. Balance all remaining species Question Mg(s) + HCl(aq) Mg2+(aq) + Cl(aq) + H2(g) Mg goes from 0 to +2 H goes from +1 to 0 Cl remains 1 Balance oxidation states: Mg(s) + 2HCl(aq) Mg2+(aq) + Cl(aq) + H2(g) Balance everything else: Mg(s) + 2HCl(aq) Mg2+(aq) + 2Cl(aq) + H2(g) Redox rxns Step by Step Method two: halfreactions (acidic solution) 1. Assign all oxidation states, decide which elements are oxidized and reduced, and write separate eqs for oxidation and reduction rxns 2. For each eq, balance (a) all elements except H and O, (b) 1st O using H2O, (c) 2nd H using H+, (d) 3rd charge using electrons 3. Multiply one or both eqs by smallest whole number(s) to equalize number of electrons 4. Add halfrxns and cancel any species Question Acidic Solution Br(aq) + MnO4(aq) Br2(l) + Mn2+(aq) Br is oxidized from 1 in Br to 0 in Br2 Mn is reduced from +7 in MnO4 to +2 in Mn2+ Oxidation halfrxn: 2Br Br2 + 2e Reduction halfrxn: MnO4 + 5e Mn2+ and balance it for acidic solution: MnO + 5e Mn2+ + 4H O Answer Multiply by integers to balance electrons, and add 5 x (2Br Br2 + 2e) 2 x (MnO4 + 8H+ + 5e Mn2+ + 4H2O) ______________________________________________________ 10Br + 2MnO4 + 16H+ 5Br2 + 2Mn2+ + 8H2O The electrons cancel on both sides of the eqn. Check: atoms balance, charges balance (+4 on reactant and product sides) Balancing redox reactions Halfreaction method in basic solution: Treat it like an acidic solution to get a final balanced equation. Then add enough OH ions to both sides of the equation to cancel out any H+ that appears The H+ and OH combine to form H2O. Cancel any water molecules that appear on both sides of the equation Question - Basic Solution CrO42(aq) + AsH3(g) Cr(OH)3(s) + As(s) Cr is reduced from +6 in CrO42 to +3 in Cr(OH)3 As is oxidized from 3 in AsH3 to 0 in As Oxidation halfrxn: AsH3 As + 3e and balance it as if in acidic solution: Answer Reduction halfrxn: CrO42 + 3e Cr(OH)3 and balance it for acidic solution: CrO42 + 3e Cr(OH)3 + H O 2 Answer Cont. Now, add enough OH to both sides of equation to convert all H+ to H2O: AsH3 + CrO42 + 2H O As + Cr(OH)3 + H2O + 2 2OH And finally cancel out the extra water: 2 ...
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This note was uploaded on 04/27/2009 for the course CHEM chem1A taught by Professor Hooker during the Spring '09 term at UCSB.

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