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Ch. 4: Solution Stoichiometry; Types of Chemical Reactions

Ch. 4: Solution Stoichiometry; Types of Chemical Reactions.

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Ch. 4: Solution Stoichiometry; Types of Chemical Reactions

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  1. Ch. 4: Solution Stoichiometry; Types of Chemical Reactions • Water is a polar molecule: the bond between the oxygen and the hydrogen in water is covalent; but unlike standard covalent molecules, oxygen has a greater attraction for electrons, thus the oxygen becomes slightly negative and the hydrogen becomes slightly positive; because of this unequal charge, water is a polar molecule; • Solubility: because water is a polar molecule, it dissolves ionic substances most readily and covalent substances to a lesser degree because of their non-polar characteristics. • Remember: “Like Dissolves Like.”

  2. Strong vs Weak? • Strong electrolytes (Acids or Bases) are those that dissociate 100% • Weak Electrolytes (Acids or Bases) are those that dissociate only partially or not at all. • Remember: Sulfuric Acid was Strong but Acetic Acid was weak and Sodium Hydroxide was strong while Ammonia was weak

  3. Concentration: amounts of chemicals in a solution • Molarity: moles of solute per liter of solution • M = moles of solute/liter of solution • .2M solution means that there are .2moles of solute/liter solution • Calculate the molarity of a solution prepared by dissolving 11.5g of solid NaOH in enough water to make 1.5 L of solution. • 1st: convert grams to moles: • 11.5g NaOH x 1 mol NaOH= 0.288 mol NaOH 40.00 g NaOH

  4. Concentration continued: • 2nd: calculate molarity by dividing the #moles by the volume of the solution. • Molarity: mol solute/L solution : 0.288 mol NaOH/ 1.50 L solution = 0.192 M NaOH Remember: the volume must be in Liters

  5. Dilution: when water is added to a solution to produce the molarity desired • Key Point: Moles of Solute after dilution= Moles of solute before dilution • Example: Suppose we need to prepare 500ml of 1.00 M acetic acid from a 17.4 stock solution. What volume of stock solution is required? • 1st: Convert 500ml to moles • 500ml x 1 L solution x 1.00 mol HC2H3O2 =.500 mol 1000 ml 1 L Solution HC2H3O2

  6. Continued… • 2nd: Since we need to know the volume that contains .500 moles we calculate: • V x M = moles V x 17.4 mol HC2H3O2 =.500 mol 1 L solution HC2H3O2 • Solve for V = 0.0287 L or 28.7 ml solution • Thus we can take 28.7 ml of 17.4 M HC2H3O2 and dilute it to a total volume of 500 ml to form a 1M solution of acetic acid

  7. Types of Chemical Reactions • Precipitation reactions: what solid will be formed when two solutions are mixed? • Must dissociate the ions to find what can be formed • Ex: Solutions of Potassium Chromate is added to Barium Nitrate; The ions present in solution are K+, CrO4-2, Ba2+, NO3-

  8. Continued: Net Ionic equations • K++ CrO4-2+ Ba2++NO3-  products • What products can be formed? Remember that cations bond with anions • So we get products of potassium nitrate and barium chromate • K++ NO3- + Ba2++ CrO4-2 • But which one forms that precipitate?

  9. Net ionics Not all ionic compounds will dissolve in water (or other polar solvents). Review the solubility rules for ionic compounds dissolving in water. • Any ionic compound whose cation is Group IA or ammonium is soluble, no matter what the anion is. • All nitrates, chlorates, perchlorates, and acetates are soluble in water, no matter what their cation is. • All chlorides, bromides and iodides are soluble in water except silver, lead (II) and mercury (I) • All sulfates are soluble in water except silver, lead (II) and mercury (I), calcium, barium and strontium • All phosphates, phosphites, sulfites, chromates, dichromates, fluorides, permanganates, oxalates, oxides, nitrides, sulfides, cyanides, carbonates and hydroxides are INSOLUBLE in water unless their cation is Group IA Group IIA are only slightly soluble hydroxides

  10. What makes an insoluble precipitate? • Based on these rules: the Potassium nitrate is soluble, why? • Rule #1 and rule #2 • The Barium Chromate is insoluble, why? • Rule #5 • So we write the net ionic equation as: • K++ CrO4-2+ Ba2++NO3-  K++ NO3- + BaCrO4 (s) • Crossing out the like ions we get : • CrO4-2+ Ba2+  BaCrO4 (s)

  11. Molecular Equations; Complete Ionic Equations; Net Ionic Equations; • Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate and aqueous potassium nitrate … • Molecular Equation: KCl (aq) + AgNO3(aq) AgCl (s) + KNO3(aq) • Complete Ionic Equation: K+ + Cl- + Ag+ + NO3+ AgCl (s) + K+ + NO3+ • Net Ionic Equation: Cl- + Ag+ AgCl (s)

  12. Calculations with precipitation reactions • When aqueous solutions of sodium sulfate and lead II Nitrate are mixed, a precipitate forms, determine the precipitate. Write the reaction and predict the products. • Na2SO4 + Pb(NO3)2 NaNO3 + PbSO4 • SO42- + Pb2+ PbSO4 • Is it balanced?

  13. Calculate the mass of lead sulfate formed if 1.25 L of 0.05M lead nitrate and 2.00 L of .025M sodium sulfate are mixed • Calculate the moles of reactants present: • 1.25 L Pb2+ x.05 mols Pb2+ = 0.0625mol Pb2+1 L Pb2+ • 2.00 L SO42- x .025 mols SO42- = 0.05 mol SO42- 1 L SO42- • Which is the limiting reactant? • 1:1 ratio making the SO42- limiting reactant • Calculate the mols of product. • Ratio 1:1 so mols of product produced is 0.05 mol • 0.05 mol PbSO4x 303.3 g PbSO4= 15.2 g PbSO4 1 mol PbSO4

  14. Acid Base Reactions: an acid is a proton donor; a base is a proton acceptor • Often called a neutralization reaction • Most strong acid/ strong base reactions have this net ionic equation: • H+ (aq) + OH-(aq) H2O • You must be careful when working with weak acid/base reactions

  15. Weak Acid/ Strong Base Reactions • Recall acetic acid is a weak acid, when mixed with potassium hydroxide, a strong base, the hydroxide ions have enough strength to rip the H+ right off of the non-dissociated molecules, and it can be assumed this will occur completely. The net ionic equation is: • OH-(aq) + HC2H3O2(aq) H2O (l) + C2H3O2- (aq)

  16. Neutralization Calculations: when just enough base has been added to an acid to react exactly and produce a neutral solution • 28.0 ml of .250 M HNO3 & 53.0 ml of .320M KOH are mixed. Calculate the amount of water formed in the resulting reaction. • HNO3 + KOH  KNO3 + H2O • H+ + OH-  H2O (l) • Calculate mols next: 28.0 ml HNO3 x 1 L x .25 mol H+ = 7.00 x 10-3 mol H+ 1000 ml 1 L 53.0 ml KOH x 1L x .32 mol OH- = 1.70 x 10-2 mol OH- 1000 ml 1L • Ratio 1:1 so HNO3 is limiting so there will 7.00 x 10 -3 mol of water formed

  17. What is the concentration of ions in excess after the reaction has finished? • Since KOH is excess • Original amount- amount used = amount in excess • 1.70 x 10-2 – 7.00 x 10-3 = 1.00 x 10-2 mols water • The volume is needed: add the two solutions 28.0 ml + 53.0 ml = 81.0 ml = .081 L • Molarity of excess OH- ions = 1.00x10-2 mol .081 L = 0.123 M OH- ions

  18. Titration Calculations • You titrate a 25.00 ml amount of your NaOH with 0.1067 M HCl. The sample turns clear after the addition of 42.95 ml of the HCl. What is the molarity of a NaOH Solution? • H+ +OH- H2O • 0.167 H+ x .04295 L H+ = 4.583 x 10 -3mol H+ 1L • From the equation; mols of OH- = H+ for neutralization • M NaOH = mol/L = 4.583 x 10 -3mol OH- = .1833 m NaOH .025 L

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