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Molarity (Concentration of Solutions) = M

Molarity (Concentration of Solutions) = M. Moles of Solute Moles Liters of Solution L. M = =. solute = material dissolved into the solvent In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc.

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Molarity (Concentration of Solutions) = M

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  1. Molarity (Concentration of Solutions)= M Moles of Solute Moles Liters of Solution L M = = solute = material dissolved into the solvent In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc. are the solutes. In sea water , Water is the solvent, and salt, magnesium chloride, etc. are the solutes. In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

  2. Molar concentration “Molarity” or “M” mol of solute liter of solution NOT: mol solute liter of solvent = M 0.2500 L mark

  3. Making a solution of known molarity Desire 250.0 mL of 0.3128 M CuSO4 • Calculate mass of solute needed • moles = M × V • moles →mass • 159.61 g CuSO4/mole 2. Accurately weigh solute

  4. Do we add exactly 250.00 mL water? • Use volumetric glassware to dissolve solute • Add water until level matches mark on flask 0.07821 moles in 0.2500 L solution

  5. Fig. 3.11

  6. Preparing a Solution - I • Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 l ! • What is the Molarity of the salt and each of the ions? • Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)

  7. Preparing a Solution - II • Mol wt of Na3PO4 = 163.94 g / mol • 3.95 g / 163.94 g/mol = 0.0241 mol Na3PO4 • dissolve and dilute to 300.0 ml • M = 0.0241 mol Na3PO4 / 0.300 l = 0.0803 M Na3PO4 • for PO4-3 ions = 0.0803 M • for Na+ ions = 3 x 0.0803 M = 0.241 M

  8. Concept Check Stoichiometry & Ion Dissociation For a 0.27 M aq. solution of sodium carbonate: • Write the dissociation reaction & identify solute(s). • Find the molarity of Na+(aq) & CO32-(aq). • If you had 185 mL of this solution, how many moles of Na+(aq) & CO32-(aq) would be present? • If you added excess MgBr2(aq), would you expect a rxn.? If so, how many moles of solid would form?

  9. Converting a Concentrated Solution to a Dilute Solution

  10. mol solute liter of solution x = liter of sample mol solute MixVi = mol solute=MfxVf i = initial f = final Add more solvent (“Dilution”) Mass of solute does not change! Constant!

  11. The Dilution Dogma: NEVER FORGET IT! M1V1=M2V2

  12. Dilution of Solutions • Take 25.00 ml of the 0.0400 M KMnO4 • Dilute the 25.00 ml to 1.000 l - What is the resulting Molarity of the diluted solution? • # moles = Vol x M • 0.0250 l x 0.0400 M = 0.00100 Moles • 0.00100 Mol / 1.00 l = 0.00100 M

  13. Concept Check Dilution Calculation How much 0.20 M HCl is needed to make 50 mL of 10 mM HCl solution? Mi×Vi = Mf×Vf (0.20 M HCl)×(L) = (0.010 M HCl)×(0.050 L)  (L) = (0.010 M HCl)×(0.050 L) = 2.5×10-3L 0.20 M HCl

  14. Concept Check Dilution & Solution Examples A) We have a 3.0 M aqueous solution of H2SO4. How do you make 100. mL of 1.4 M H2SO4(aq)? B) Determine how you would make 250. mL of 0.56 M CaCl2(aq)? What is [Ca2+]? [Cl-]? (Reagent is solid is CaCl2·2H20; 147.01 g/mol) C) Predict what would happen in you mixed solutions A and B together. H2SO4(aq) + CaCl2(aq) →?

  15. How could you make 5.0 L of 0.025 M sucrose from a solution which is 0.100 M sucrose? Mix 1.250 L of 0.100 M sucrose with 3.75 L water.

  16. Chemical Equation Calculation - III Mass Atoms (Molecules) Molecular Weight Avogadro’s Number g/mol 6.02 x 1023 Molecules Reactants Products Moles Molarity moles / liter Solutions

  17. Calculating Mass of Solute from a Given Volume of Solution Volume (L) of Solution Molarity M = (mol solute / Liters of solution) = M/L Moles of Solute Molar Mass (M) = ( mass / mole) = g/mol Mass (g) of Solute

  18. CaCO3(s) + 2 HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) 2 g 10 mL 0.75 M Which is limiting? 2 g CaCO3 x 1 mol CaCl2 = 0.01 mol CaCl2 100 g CaCO3 1 mol CaCO3 0.01 L HCl x 0.75 mol HCl x 1 mol CaCl2= L HCl 2 mol HCl 0.004 mol CaCl2 What is the [Cl-] after the reaction? How many g of CaCO3 remain?

  19. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq) Given 10.0 g Al(OH)3, what volume of 1.50 M HCl is required to neutralize the base? Mass (g) of Al(OH)3 M (g/mol) 10.0 g Al(OH)3 78.00 g/mol Al(OH)3 Moles of Al(OH)3 molar ratio Moles of HCl M ( mol/L) Volume (L) of HCl

  20. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq) Given 10.0 g Al(OH)3, what volume of 1.50 M HCl is required to neutralize the base? Mass (g) of Al(OH)3 M (g/mol) 10.0 g Al(OH)3 78.00 g/mol = 0.128 mol Al(OH)3 Moles of Al(OH)3 0.128 mol Al(OH)3 x = molar ratio Moles HCl Moles of HCl M ( mol/L) Volume (L) of HCl

  21. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq) Given 10.0 g Al(OH)3, what volume of 1.50 M HCl is required to neutralize the base? Mass (g) of Al(OH)3 M (g/mol) 10.0 g Al(OH)3 78.00 g/mol = 0.128 mol Al(OH)3 Moles of Al(OH)3 3 moles HCl moles Al(OH)3 0.128 mol Al(OH)3 x = molar ratio 0.385 Moles HCl Moles of HCl Vol HCl M ( mol/L) Volume (L) of HCl

  22. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq) Given 10.0 g Al(OH)3, what volume of 1.50 M HCl is required to neutralize the base? Mass (g) of Al(OH)3 M (g/mol) 10.0 g Al(OH)3 78.00 g/mol = 0.128 mol Al(OH)3 Moles of Al(OH)3 3 moles HCl moles Al(OH)3 0.128 mol Al(OH)3 x = molar ratio 0.385 Moles HCl Moles of HCl 1.00 L HCl 1.50 Moles HCl Vol HCl = x 0.385 Moles HCl Vol HCl = 0.256 L = 256 ml M ( mol/L) Volume (L) of HCl

  23. Solving Limiting Reactant Problems in Solution - Precipitation Problem - I Problem: Lead has been used as a glaze for pottery for years, and can be a problem if not fired properly in an oven, and is leachable from the pottery. Vinegar is used in leaching tests, followed by Lead precipitated as a sulfide. If 257.8 ml of a 0.0468 M solution of Lead nitrate is added to 156.00 ml of a 0.095 M solution of Sodium sulfide, what mass of solid Lead Sulfide will be formed? Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Writing the balanced equation: Pb(NO3)2 (aq) + Na2S (aq) 2 NaNO3 (aq) + PbS (s)

  24. Volume (L) of Pb(NO3)2 solution Volume (L) of Na2S solution Multiply by M (mol/L) Multiply by M (mol/L) Amount (mol) of Pb(NO3)2 Amount (mol) of Na2S Molar Ratio Molar Ratio Amount (mol) of PbS Amount (mol) of PbS Choose the lower number of PbS and multiply by M (g/mol) Mass (g) of PbS

  25. Volume (L) of Pb(NO3)2 solution Volume (L) of Na2S solution Multiply by M (mol/L) Multiply by M (mol/L) Amount (mol) of Pb(NO3)2 Divide by equation coefficient Amount (mol) of Na2S Divide by equation coefficient Smallest Molar Ratio Amount (mol) of PbS Mass (g) of PbS

  26. Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) =

  27. Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2 Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield:

  28. Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2 Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield: 1 mol PbS 1 mol Pb(NO3)2 Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+2

  29. Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2 Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield: 1 mol PbS 1 mol Pb(NO3)2 Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+2 0.012065 Mol Pb+2 = 0.012065 Mol PbS 0.012065 Mol PbS x = 2.89 g PbS 239.3 g PbS 1 Mol PbS

  30. “salt” water Two Types of Acid-Base Reactions 1) A-B Neutralization: LiOH(aq) + HCN(aq)→ LiCN(aq) + H2O(l) H2SO4(aq) + Ca(OH)2(aq)→? gas 2) A-B Reactions with Gas Formation: Na2CO3(aq) + 2HBr(aq)→ 2NaBr(aq) + H2O(l) + CO2(g) Li2SO3(aq) + NaOH(aq)→? Which salts?carbonates  CO2 sulfites  SO2 sulfides  H2S

  31. Concept Check Neutralization Calculation • How much 2.0 M HCl is needed to “neutralize” 2.3 liters of 0.15 M NaOH? HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l) acid + base → “salt” + water or….. How much 2.0 M HCl should be added such that the mol of H+(aq)=mol OH–(aq)? N.I.E.: H+(aq) + OH-(aq)→ H2O(l)

  32. Neutralization Calculations Determine the volume of 0.10 M KOH(aq)solution required to neutralize 25.00 mL samples of three different acids, all at 0.20 M. Acids: 1) Nitric acid 2) Carbonic acid 3) Phosphoric acid • 1) __ HNO3 (aq) + __ KOH (aq)→ • 2) + __ KOH (aq)→ • 3) + __ KOH (aq)→

  33. Titration Calculation 4.49 mL of 0.2500 M NaOH is required to titrate a 25.00 mL sample of H2SO4 to the endpoint. What is the molar concentration of H2SO4 in the sample? NaOH(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l) 2NaOH(aq) + 1H2SO4(aq) Na2SO4(aq) + 2H2O(l) Stoichiometry not 1:1 !!!

  34. 0.2500 M NaOH ∆V = 4.49 mL 25.00 mL [H2SO4] = ? Too far! Starting point At equivalence point

  35. L of NaOH(aq) required to get to endpoint mol NaOH consumed in neutralization mol NaOH liter of solution x = Measured with buret Calculated Known mole ratio [H2SO4] (M) mol H2SO4 25.00 mL sample 2NaOH(aq) + H2SO4(aq)→ Na2SO4(aq)+ 2H2O(l)

  36. L of NaOH(aq) required to get to endpoint mol NaOH consumed in neutralization mol NaOH liter of solution x = 1 mol H2SO4 2 mol NaOH 0.000560 mol H2SO4 present in the initial sample x = (from chemical equation) 2NaOH(aq) + H2SO4(aq)→ Na2SO4(aq)+ 2H2O(l) 0.2500 mol/L x 0.00449 L = 0.00112 mol 0.00112 mol NaOHconsumed by neutralization

  37. molH2SO4 present in initial sample mol liter = = volume initial sample 0.000560 (calculated) [H2SO4] (M) in the initial sample 0.02500 L (known) [H2SO4(aq)] = 0.0224 M

  38. Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas: • ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g) • How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS?

  39. Molar mass of ZnS = 97.47 g • = 0.157 L = 157 mL HCl solution

  40. Quantitative Analysis • The determination of the amount of a substance or species present in a material

  41. Gravimetric Analysis • A type of quantitative analysis in which the amount of a species in a material is determined by converting the species to a product that can be isolated completely and weighed

  42. The figure on the right shows the reaction of Ba(NO3)2 with K2CrO4 forming the yellow BaCrO4 precipitate.

  43. The BaCrO4 precipitate is being filtered in the figure on the right. It can then be dried and weighed.

  44. A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver compound gave 1.788 g of silver chloride, what is the mass percent of silver in the compound?

  45. Molar mass of silver chloride (AgCl) = 143.32 g = 1.346 g Ag in the compound = 85.03% Ag

  46. http://www.learnerstv.com/animation/chemistry/ruther14.swf • http://www.dartmouth.edu/~chemlab/info/resources/mashel/MASHEL.html • http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.swf

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