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VII : Aqueous Solution Concentration, Stoichiometry

VII : Aqueous Solution Concentration, Stoichiometry. LECTURE SLIDES Molarity Solution Stoichiometry Titration Calculations Dilution Problems. Kotz & Treichel: Sections 5.8 -5.9 . Concentrations of Compounds in Aqueous Solutions (Chapter 5, Section 5.8, p. 213).

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VII : Aqueous Solution Concentration, Stoichiometry

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  1. VII: Aqueous Solution Concentration, Stoichiometry • LECTURE SLIDES • Molarity • Solution Stoichiometry • Titration Calculations • Dilution Problems Kotz & Treichel: Sections 5.8 -5.9

  2. Concentrations of Compounds in Aqueous Solutions (Chapter 5, Section 5.8, p. 213) Usually the reactions we run are done in aqueous solution, and therefore we need to add to our study of stoichiometry the concentration of compounds in aqueous solution. We will utilize a very useful quantity known as “molarity,” the number of moles of solute per liter of solution.

  3. Concentration(molarity) = # moles solute L solution If we placed 1.00 mol NaCl (58.4 g) in a 1 L volumetric flask, dissolved it in water, swirled to dissolve and diluted the solution to the 1.00 liter mark you would have a solution that contains 1.00 mol NaCl per liter of solution. This may be represented several ways: Concentration(molarity) = 1.00 mol NaCl / L soln = 1.00 M NaCl = [1.00] NaCl Chemists call this a “1.00 molar solution”

  4. Calculating Molar Amounts • TYPICAL PROBLEMS: • What is the molarity of a solution made by dissolving • 25.0 g of BaCl2 in sufficient water to make up a • solution of 500.0 mL? • How many g of BaCl2 would be contained in 20.0 mL of • this solution? • How many mL of this solution would deliver 1.25 g of • BaCl2? • How many mol of Cl- ions are contained in 10.00 ml of • this solution?

  5. What is the molarity of a solution made by dissolving 25.0 g of BaCl2 in sufficient water to make up a solution of 500.0 mL? 25.0 g BaCl2= ? mol/ L BaCl2 (= ? M BaCl2) 500.0 mL soln Molar mass, BaCl2: 1Ba = 1 X 137.33g = 137.33 2Cl = 2 X 35.45g = 70.90 208.23 g/mol

  6. Molar mass, solute Mass of solute Conversion to L Volume of soltn

  7. How many g of BaCl2 would be contained in 20.0 mL of this solution?(.240 M BaCl2) Question: 20.0 mL soln = ? g BaCl2 Relationships: 1000 mL = 1 L 1 L soln = .240 mol BaCl2 1 mol BaCl2 = 208.23 g BaCl2 20.0 mL soln = ? g BaCl2 mL  L  mol  g

  8. Molarity Molar Mass conversion

  9. How many mL of this solution would deliver 1.25 g of • BaCl2? 1.25 g BaCl2 = ? mL soln .240 mol BaCl2 = 1 L soln 208.23 g BaCl2 = 1 mol BaCl2 Molarity Molar Mass

  10. How many mol of Cl- ions are contained in 10.0 ml of • this solution? Note: BaCl2(aq) ---> Ba2+(aq) + 2 Cl-(aq) 10.0 mL soln = ? Mol Cl- .240 mol BaCl2 = 1000 mL soln 1 mol BaCl2 = 2 mol Cl-

  11. GROUP WORK 7.1: Molarity Problems If 35.00 g CuSO4 is dissolved in sufficient water to makeup 750. mL of an aqueous solution, a) what is the molarity of the solution? b) how many mL of the solution will deliver 10.0 g of CuSO4? c) How many moles of sulfate ion (SO42-) will be delivered in 10.0 mL of the solution? 1 Cu = 1 X 63.55 = 63.55 1 S = 1 X 32.07 = 32.07 4 O = 4 X 16.00 = 64.00 159.62 g/mol

  12. STOICHIOMETRY OF REACTIONS IN AQUEOUS SOLUTION: Chapter 5, Section 5.9 Let’s use our favorite reaction to add another dimension to calculating from balanced equations: How many ml of 3.00 M HCl solution would be required to react with 13.67 g of Fe2O3 according to the following balanced equation: Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 3.00 M 13.67 g ? mL

  13. Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq) 159.70 g/mol 3.00 M 13.67 g ? mL 13.67 g Fe2O3 = ? mL soln 1000 mL soln = 3.00 mol HCl 6 mol HCl = 1 mol Fe2O3 159.70 g Fe2O3 = 1 mol Fe2O3 13.67 g Fe2O3 = ? mL soltn Pathway: g Fe2O3 ---> mol Fe2O3---> mol HCl ---> mL soln

  14. Balanced Equation Molarity Molar Mass

  15. Group Work 7.2: Solution Stoichiometry How many g of Fe2O3 would react with 25.0 mL of 3.00 M HCl? Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq) 159.70 g/mol 3.00 M ? g 25.0 mL 25.0 mL soltn = ? g Fe2O3 mL  mol HCl  mol Fe2O3  g Fe2O3

  16. Limiting Reagent, Solutions Suppose you mixed 20.00 mL of .250 M Pb(NO3)2 solution with 30.00 mL of .150 KI solution. How many g of PbI2 precipitate might you theoretically obtain? Pb(NO3)2 (aq) + 2 KI (aq)---> PbI2(s) + 2 KNO3(aq) .250 M .150 M 461.0 g/mol 20.00 mL 30.00 mL ? g 1Pb = 1 X 207.2 = 207.2 2 I = 2 X 126.9 = 253.8 461.0 g/mol

  17. “A” “B” “C” Pb(NO3)2 (aq) + 2 KI (aq)---> PbI2(s) + 2 KNO3(aq) .250 M .150 M 461.0 g/mol 20.00 mL 30.00 mL ? g Pathway: mL  mol “A”  mol “C”  g “C” mL  mol “B”  mol “C”  g “C” Low number wins!

  18. Pb(NO3)2 (aq) + 2 KI (aq)---> PbI2(s) + 2 KNO3(aq) .250 M .150 M 461.0 g/mol 20.00 mL 30.00 mL ? g 20.00 ml soltn A .250 mol A 1 mol C 461.0 g C = 2.3050 g C 1000 mL 1 mol A 1 mol C = 2.30 g C 30.00 mL soltn B .150 mol B 1 mol C 461.0 g C = 1.037 g C 1000 mL 2 mol B 1 mol C = 1.04 g C

  19. Excess reagent: How many mL of solution A, .250 M Pb(NO3)2, would be required to react with 30.00 mL of solution B, .150 M KI? Pb(NO3)2 (aq) + 2 KI (aq)---> PbI2(s) + 2 KNO3(aq) .250 M .150 M ? mL 30.00 mL Pathway: mL “B” soltn  mol “B”  mol “A”  mL “A” soltn 30.00 mL B soltn = ? mL A soltn

  20. Pb(NO3)2 (aq) + 2 KI (aq)---> PbI2(s) + 2 KNO3(aq) .250 M .150 M ? mL 30.00 mL Molarity equation Molarity 30.00 mL soltn .150 mol B 1 mol A 1000 mL soltn 1000 mL soltn 2 mol B .250 mol A = 9.00 mL soltn A

  21. Group Work 7.3: Solution Stoichiometry #2 How many mL of .150 M KI solution would be required to react with 20.00 mL of .250 M Pb(NO3)2? Pb(NO3)2 (aq) + 2 KI (aq)---> PbI2(s) + 2 KNO3(aq) .250 M .150 M 20.00 mL? mL

  22. Summary 1. Molarity, M: useful description of a solution; gives number of moles of solute in 1 L of solution. Useful to convert mL of solution to moles of solute in equation situations. 2. Molarity of solution can also yield moles per liter of the ions produced when the solute ionizes in water, using the formula of the solute as conversion factor 3. Adds another way to calculate mass or moles from volume (compare to density, #grams of substance or solution in 1 mL or 1 cm3 of solution or substance).

  23. TITRATION CALCULATIONS Titration: Procedure in which measured increments of one reactant are added to a known amount of a second reactant until some indicator signals that the reaction is complete. This point in the reaction is called “the equivalence point.” Indicators include many acid/base dyes, potentiometers, color change in one reagent...

  24. Titrations are run with buret and Erlenmeyer flasks as seen on the CD ROM or demo just observed... • Two types of problems are typically encountered: • Standardizing an acid or a base solution • (determining the exact molar concentration of • an acid or base solution) • Determining amount of acidic or basic material • in a sample ( or other substances detectable by • some indicator)

  25. Ex 5.13 type:Standardizing an acid solution: Suppose a pure, dry sample of Na2CO3 weighing 0.379 g is dissolved in water and titrated to the equivalence point with 35.65 mL of HCl solution. What is the molarity of the HCl solution? Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g) 105.99 g/mol 35.65 mL soln 0.379 g M= ? = #mol HCl / L soltn 2 Na = 2 X 22.99 = 45.98 1 C = 1 X 12.01 = 12.01 3 O = 3 X 16.00 = 48.00 105.99 g/mol

  26. Calculations for Standardization 1) Calculate moles of solute from balanced equation, using information about known reagent 2) Calculate M, using volume of solution required for titration of known to equivalence point

  27. Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g) 105.99 g/mol 35.65 mL soln 0.379 g M=? = Mol HCl / L soltn Step One: Calculate Moles of HCl from equation .379 g Na2CO3 = ? moles HCl

  28. Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g) 105.99 g/mol 35.65 mL soln 0.379 g .00715 mol M=? Calculated in Step One Step Two: Calculate molarity: M, HCl soln = ? = # mol HCl = L soln .00715 mol HCl 1000 mL = .201 mol HCl = .201 M HCl 35.65 mL soltn 1 L L soltn

  29. Group Work 7.4: Suppose a pure, dry sample of Na2CO3 weighing 0.437 g is dissolved in water and titrated to the equivalence point with 39.85 mL of HCl solution. What is the molarity of the HCl solution? Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g) 105.99 g/mol 39.85 mL soln 0.437 g M= ? = #mol HCl / L soltn #1: Find moles of HCl from g Na2CO3 #2: Find Molarity (moles HCl / volume soltn)

  30. Standardization of a Base: Let’s use the 0.201 M HCl which we standardized in the first problem to determine the exact molar concentration of a sodium hydroxide solution (“standardize the base”!): Problem: If titrating 25.00 mL of NaOH to a phenolphthalein endpoint required 31.25 mL of .201 M HCl, what is the molar concentration of the base?

  31. If titrating 25.00 mL of NaOH to a phenolphthalein endpoint required 31.25 mL of .201 M HCl, what is the molar concentration of the base? NaOH(aq) + HCl(aq) ---> H2O(l) + NaCl(aq) 25.00 mL 31.25 mL M=? .201 M Steps: 1) calculate moles NaOH reacted 2) calculate M, moles NaOH/ volume soln

  32. NaOH(aq) + HCl(aq) ---> H2O(l) + NaCl(aq) 25.00 mL 31.25 mL M=? .201 M Step 1) 31.25 mL HCl soln = ? mol NaOH Step 2)Calculate molarity:

  33. Group Work 7.5 A 30.0 mL sample of vinegar requires 39.35 mL of .843 M NaOH solution for titration to the equivalence point. What is the molar concentration of the acetic acid in vinegar? CH3CO2H(aq) + NaOH(aq) ----> H2O + NaCH3CO2(aq) 30.0 mL 39.35mL .843 M M=?mol CH3CO2H /L Step 1) solve for moles, CH3CO2H Step 2) solve for M CH3CO2H

  34. Let’s do another base solution, standardizing it with Potassium acid phthalate, KHC8H4O4, a popular solid for this purpose. (Structure, next slide!) It reacts with strong bases according to the following net ionic equation: KHC8H4O4 (aq) + NaOH (aq) ------> H2O(l) + KNaC8H4O4 (aq)

  35. Group Work 7.6 If a .896 g sample of potassium acid phthalate is dissolved in water and titrated to the equivalence point with 16.95 mL of NaOH solution, what is the molarity of the NaOH soln? 1K = 1 X 39.10 = 39.10 8C = 8 X 12.01 = 96.08 5H = 5 X 1.008 = 5.04 4O = 4 X 16.00 = 64.00 204.22 g/mol KHC8H4O4(aq) + NaOH(aq) ---> H2O (l) + K NaC8H4O4 (aq) 204.22 g/mol .896 g16.95 mL M=? mol NaOH /L

  36. “Redox” titration You wish to determine the weight percent of copper in a copper containing alloy. After dissolving a sample of an alloy in acid, an excess of KI is added, and the Cu2+ and I- ions undergo the reaction: 2Cu2+ (aq) + 5 I- (aq) -----> 2 CuI(s) + I3- (aq) The I3- which is produced in this reaction is titrated with sodium thiosulfate according to the equation: I3- (aq) +2 S2O32-(aq) -----> S4O62-(aq) + 3 I- (aq) If 26.32 mL of 0.101 M Na2S2O3 is required for titration to the equivalence point, what is the wt % of Cu in .251 g alloy?

  37. 2Cu2+ (aq) + 5 I- (aq) -----> 2 CuI(s) + I3- (aq) 63.55 g/mol produced 0.251 g alloy by reaction g Cu =? % Cu, alloy=? I3- (aq) + 2 S2O32-(aq) -----> S4O62-(aq) + 3 I- (aq) ? mol 26.32 mL .101 M Na2S2O3

  38. I3- (aq) + 2 S2O32-(aq) -----> S4O62-(aq) + 3 I- (aq) ? mol 26.32 mL .101 M Na2S2O3

  39. 2Cu2+ (aq) + 5 I- (aq) -----> 2 CuI(s) + I3- (aq) 63.55 g/mol g=? .00133 mol % Cu, alloy=? 0.251 g alloy % Cu, alloy=? = 1.69 g Cu X 100 = 67.3% 0.251 g alloy

  40. DILUTION PROBLEMS Suppose you would like to make up a more “dilute” solution (less moles/L) from a more “concentrated” one (more moles /L). This is how you might proceed: a) figure out how many total moles of solute you want in the desired volume of the dilute solution b) figure out what volume of the more concentrated solution will deliver this number of moles c) measure out the concentrated solution and add water to make up the desired volume of the dilute solution. (DEMO!)

  41. Describe how you might makeup 750. mL of 1.00 M H2SO4 from a concentrated sulfuric acid solution which is 12.0 M H2SO4 . a) Calculate desired number of moles in dilute soln: b) Calculate volume of concentrated solution which will contain this number of moles:

  42. c) Measure out 62.5 mL of the concentrated soln and carefully add it to sufficient water to make up 750. mL of solution. NEVER ADD WATER TO CONCENTRATED ACIDS; ALWAYS ADD CON ACIDS TO WATER, SLOWLY...

  43. Formula for Dilution Problems: Since we can calculate moles as follows: mL X # moles = moles of solute 1000 mL and since: moles, con soln = moles, dil soln We can say: mL, con X M, con = mL, dil X M, dil and do the last problem “the easy way”:

  44. Describe how you might makeup 750. mL of 1.00 M H2SO4 from a concentrated sulfuric acid solution which is 12.0 M H2SO4 . mL, con X M, con= mL, dil X M, dil ? mL X 12.0 M H2SO4 = 750. mL X 1.00 M H2SO4 mL con = 750 mL dil X 1.00 M H2SO4 12.0 M H2SO4 mL con = 62.5 mL Punch line the same: measure out 62.5 mL con acid and dilute to 750. mL.

  45. Group Work 7.7: Dilution Problem How many mL of 2.35 M AgNO3 solution are required to makeup 2.00 L of .100 M AgNO3 solution? Describe how you would makeup this new solution. mL con X M con = mL dil X M dil ? 2.35 M 2.00L .100 M

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