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Solutions

Solutions. Labs Dry Lab 4 Oxidation-Reduction Equations #9 A Volumetric Analysis #15 Bleach Analysis Chemical equations Chapters 8-11. Will the substances mix?. NaNO 3 and H 2 O Miscible-ionic + polar C 6 H 14 and H 2 O Immiscible-nonpolar + polar I 2 and C 6 H 14

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Solutions

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  1. Solutions Labs Dry Lab 4 Oxidation-Reduction Equations #9 A Volumetric Analysis #15 Bleach Analysis Chemical equations Chapters 8-11

  2. Will the substances mix? • NaNO3 and H2O • Miscible-ionic + polar • C6H14 and H2O • Immiscible-nonpolar + polar • I2 and C6H14 • Miscible-nonpolar + nonpolar • I2 and H2O • immiscible-nonpolar + polar

  3. bent molecule 104.5o Aqueous solutions-solvent is water • Polar molecule-unequal charge distribution gives water ability to dissolve many compounds

  4. Hydration • Positive ends of water molecules are attracted to negatively charged anions • Negative ends of water molecules are attracted to positively charged cations • Strong forces in positive and negative ions of solid are replaced by strong water-ion interactions

  5. Solutionhomogeneous mixture of two or more substances • Solute- • If it and solvent in same phase, one in lesser amount • If it and solvent in different phases, one that changes phase • One dissolved • Solvent- • If it and solute in same phase, one in greater amount • If it and solute in different phases, one retaining phase • One doing dissolving

  6. Electrolytes and Nonelectrolytes • Many aqueous solutions of ionic compounds conduct electricity whereas water itself essentially does not conduct-electrolyte solutions • Dissociation-ionic substances break apart completely into ions when dissolved in water • Ionization-some molecular compounds (nonionic) break apart into ions when dissolved in water

  7. Solvation of ionic compounds • Polarity of water • Ions surrounded by water molecules prevent recombining of ions (solvated) • Ions/shells free to move around, so dispersed uniformly throughout solution

  8. Electrolyte Conductivity Degree Examples of Dissociation Strong high total strong acids (HCl, HNO3) many ionic salts (NaCl, Sr(NO3)3-ions) strong bases (NaOH, Ba(OH)2, other IA/IIA hydroxides) Weak low to partial weak organic acids moderate (tap water, HCO2H- formic acid, C2H3O2) weak bases (NH3) Non none close to zero sugar, sugar solution, AgCl, Fe2O3 (neutral)

  9. Complete following dissociation equations: (all in water) • CaCl2(s)Ca2+(aq) + 2Cl-(aq) • Fe(NO3)3(s)  • KBr(s)  • (NH4)2Cr2O7(s)  Strong, Weak, or Nonelectrolytes • HClO4 • C6H12 • LiOH • NH3 • CaCl2 • HC2H3O2

  10. MolarityM = moles of solute Liter of solution Moles = millimoles = micromoles liter milliliter microliter BUT Moles does not equal millimoles or moles liter liter microliter

  11. What is the molarity of a solution in which 3.57 g of sodium chloride, NaCl, is dissolved in enough water to make 25.0 ml of solution? • Convert from mass of NaCl to moles of NaCl. • 3.57 g NaCl x 1 mol NaCl = 0.061 mol NaCl 58.44 g NaCl  • Convert 25.0 mL to 0.0250 L and substitute these two quantities into the defining equation for molarity. • Molarity = 0.0611 mol NaCl = 2.44 M NaCl l0.0250 L solutionWe read this as 2.44 molar NaCl.   • It is important to understand that molarity means moles of solute per liter of solution, and not per liter of solvent.

  12. How many mL of solution are necessary if we are to have a 2.48 M NaOH solution that contains 31.52 g of the dissolved solid? 31.52 g NaOH 1 mol NaOH = 0.788 mol NaOH 40.00 g NaOH 2.48 M NaOH = 0.788 mol NaOH = .318 L = 318.mL NaOH x L

  13. Determine the molarity of Cl- ion in a solution prepared by dissolving 9.82 g of CuCl2 in enough water to make 600 mL of solution. • 9.82 g CuCl2 1 mol CuCl2 = 0.073 molCuCl2 134.45 g CuCl2 • M = 0.073 mol CuCl2 = 0.1217 M CuCl2 .600 L • Ion = 2 mol Cl- solute 1 mol CuCl2 • 0.1217 M CuCl2 x 2 mol Cl- = 0.243 M L solution 1 mol CuCl2

  14. Calculate the mass of NaCl needed to prepare 175 mL of a 0.500 M NaCl solution. 0.500 M = x mol .175 L x mol = 0.0875 mol NaCl 0.0875 mol NaCl 58.44 g NaCl = 5.11 g NaCl 1 mol NaCl

  15. Solution Concentration • Amounts of chemical present in solutions-expressed as concentration (amount of solute dissolved in given amount of solvent) • Most common unit of concentration used for aqueous solutions-molarity • # moles of solute per liter of solution

  16. Common Terms of Solution Concentration Stock - routinely used solutions prepared in concentrated form Concentrated - relatively large ratio of solute to solvent (5.0 MNaCl) Dilute - relatively small ratio of solute to solvent (0.01 MNaCl)

  17. To calculate the concentration of each type of ion in a solution • Dissociate solid into its cations/anions • This gives how many moles of ions are formed for each mole of solid • Multiply number of ions by given molarity

  18. Calculate the molarity of all the ions in each of the following solutions: • Always write out the dissociation equation, so you have “ion-to-solute” mole ratio. • 0.25 M Ca(OCl2) • Ca(OCl2)(s)  Ca2+(aq) + 2OCl-(aq) • Molarity of Ca2+ = molarity of Ca(OCl2) = 0.250 M • Molarity of 2OCl- = 2 x molarity of Ca(OCl2) = 0.50 M • 2 M CrCl3 • CrCl3 Cr3+(aq) + 2Cl-(aq)

  19. Standard solution-solution whose concentration is accurately known a. Put a weighed amount of a substance (solute) into a volumetric flask and add a small quantity of water. • c. Add more water (with gentle swirling) until the level of the solution just reaches the mark etched on the neck of the flask. • Then mix solution thoroughly by inverting the flask several times. b. Dissolve the solid in the water by gently swirling the flask (with stopper in place).

  20. Example • To prepare a 0.50 molar (0.50 M) solution of KCl, we first measure out 0.50 mole of solid KCl, that is 37.3 g • When water is added up to this mark, the flask will contain 1.0 L of the KCl solution, and will contain 0.50 mole of KCl • The number of moles of KCl in given amounts of the above solution is easy to find • For instance, 0.10 L of 0.50 M KCl will contain • 0.10 L soln x 0.50 mol KCl= 0.050 mol KCl in 1 L soln

  21. Dilution Used to prepare less concentrated solution from more concentrated solution Adding more water to given amount of solution does not change number of moles of solute present in solution Moles of solute before dilution = moles of solute after dilution Since moles of solute equals solution volume (V) times molarity (M) we have molesi = molesf of MiVi = MfVf i/f stand for initial/final solution, respectively Note that Vf is always larger than Vi

  22. Parts per million (mg/L) • An aqueous solution with a total volume of 750 mL contains 14.38 mg of Cu2+. What is the concentration of Cu2+ in parts per million? • 14.38 mg Cu2+ = 19.2 ppm Cu2+ 0.750 L soln

  23. Molarity to PPM • A solution is 3 x 10-7 M in manganese(VII) ion. What is the Mn7+ 3 x 10-7 M concentration in ppm? 3 x 10-7 mol Mn7+ 54.94 g Mn7+ 1000 mg Mn7+ = 0.0164 = 0.02 ppm Mn7+ L soln 1 mol Mn7+ 1 g Mn7+

  24. Homework: Read 4.1-44, pp. 133-148 Q pp. 180-182, #11 c/d/i, 12, 15c, 16, 20, 23a, 28

  25. Solubility Rules • Solubility of solute: amount dissolved in given quantity of solvent at given temperature • Major ideas: • Many salts dissociate into ions in aqueous solution • If solid forms from combination of selected ions in solution • Solid must contain anion part and cation part • Net charge on solid must be zero • Simple solubility rules used to predict products of reactions in aqueous solutions

  26. Any ionic compound can be broken apart into its cations and anion • Compounds containing 3/more different atoms break apart into appropriate polyatomic ion(s) • Charges of all ions must add up to zero • Subscripts on monatomic ions become coefficients for ions • For polyatomic ions, only subscripts afterparentheses become coefficients • Whether or not ionic compound dissolves to appreciable extent in water depends on which cations/anions make up compound

  27. Soluble-good deal of solid visibly dissolves when added to water Slightly soluble-only small amount of solid dissolves (Ionic compound can have low solubility and still be strong electrolyte) Insoluble-no solid dissolves (relative term-does not mean that no solute dissolves)

  28. NH4+/ Group IA None NO3-, ClO-, ClO4-, HCO3-, (slightly soluble) CrO42- All metallic oxides (O-2) NH4+, Group IA metals All metallic hydroxides (OH-) NH4+, Group IA/IIA from calcium down. Important-NaOH/KOH Ba(OH)2, Sr(OH)2, and Ca(OH)2 marginally soluble

  29. Types of Solution Reactions • Precipitation reactions AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) • Acid-base reactions (Neutralization) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) • Oxidation-reduction reactions Fe2O3(s) + Al(s)  Fe(l) + Al2O3(s)

  30. 2 soluble substances combined AX(aq) + BZ(aq)  AZ +BX Determine, using solubility rules, whether either will form solid (precipitate: insoluble solid that settles from solution) Precipitation Reactions

  31. Chemical equations for precipitation reactions can be written in several ways: • Molecular equation: formulas of compounds are written as usual chemical formulas • Pb(NO3)2(aq) + 2HCl(aq) PbCl2(s) + 2HNO3(aq) • Precipitation reaction: more accurately represented by ionic equation which shows compounds as being dissociated in solution • Pb2+(aq) + 2NO3(aq) + 2H+(aq) + 2Cl–(aq)  PbCl2(s) + 2H+(aq) + 2NO3 (aq) • H+/NO3 ions not involved in formation of precipitate • Spectator ions(identical species on both sides of equation can be omitted from equation because ions not involved in reaction) • Net ionic equationshows only species that actually undergo a chemical change • Pb2+(aq) + 2Cl–(aq)  PbCl2(s)

  32. Homework: Read 4.5-4.6, pp. 148-top 156 Q pg. 182, #30, 34 a/c, 36 b/d, 38

  33. What mass of Fe(OH)3 is produced when 35. mL of a 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M KOH solution? 0.250 M 0.180 M x g Fe(NO3)3(aq) + 3KOH(aq)  Fe(OH)3(aq) + 3KNO3(aq) • 0.250 M Fe(NO3)3 = x mol = 0.00875 mol/1 .035 L • 0.180 M KOH = x mol = 0.00990 mol/3 = .00330 mol .055 L limiting reactant • 0.00990 mol KOH 1 mol Fe(OH)3 106.88 g Fe(OH)3= .35 g Fe(OH)3 3 mol KOH 1 mol Fe(OH)3

  34. Gravimetric Analysis-measurement of weight • An ore sample is to be analyzed for sulfur. As part of the procedure, the ore is dissolved and the sulfur is converted to sulfate ion, SO42-. Barium nitrate is added, which causes the sulfate to precipitate out as BaSO4. The original sample had a mass of 3.187 g. The dried BaSO4 has a mass of 2.005 g. What is the percent of sulfur in the original ore? (present in the 2.005 g) 2.005 g BaSO4 1 mol BaSO4 1 mol S 32.06 g S = 0.275 g S 233.36 g BaSO4 1 mol BaSO4 1 mol S % S = 0.275 g S x 100% = 8.64% S in the ore 3.187 g in ore

  35. Homework: Read 4.7, pp. 156-158 Q pg. 182, #40, 42, 44

  36. Acid-Base Theories Attempt to explain what happens to molecules of acidic/basic substances in solution that gives rise to their characteristic properties.

  37. Acids and Bases • Solutions of acids • Sour taste • Change blue litmus to red (abr) • Dissolve certain metals • React w/carbonates to produce carbon dioxide gas • Solutions of bases • Bitter taste • Feel slippery • Change red litmus to blue (brb) • Properties of acid neutralized by addition of base, and vice versa

  38. SvanteArrhenius proposed this acid-base theory • Acid: substance that produces hydrogen ions (H+, protons) in aqueous solution • HNO3(aq)  H+(aq) + NO3–(aq) • Base: substance that produces hydroxide ions (OH–) in aqueous solution • Ca(OH)2(aq)  Ca2+(aq) + 2OH–(aq) • All are electrolytes undergoing dissociation in aqueous solution • Neutralization is combination of H+/OH– to form water • H+(aq) + OH–(aq)  H2O(l)

  39. Strong acid-completely dissociates into its ions-strong electrolytes HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4 Strong base-soluble ionic compounds containing hydroxide ion (OH-)-strong electrolytes Group IA (Li, Na, K, Rb, Cs)/heavy Group IIA (Ca, Sr, Ba) metal hydroxides, NH2+ Weak acid-dissociates (ionizes) only to a slight extent in aqueous solutions-weak electrolytes Acetic acid, HF Weak base-very few ions are formed in aqueous solutions-weak electrolyte NH3

  40. Acid/base both completely ionized in solution, ionic equation: 2H+(aq) + 2Cl–(aq) + Mg2+(aq) + 2OH–(aq)  2H2O(l) + Mg2+(aq) + 2Cl–(aq) Notice that Mg2+ ion and Cl– ion are spectator ions. Net ionic equation is: 2H+(aq) + 2OH–(aq)  2H2O(l) or H+(aq) + OH–(aq)  H2O(l) (coefficients can be cancelled) Neutralization equation for any strong acid and strong base is: H+(aq) + OH–(aq)  H2O(l) If we had carried out this reaction with two moles of HCl for each mole of base, there would be no left over acid or base, and the solution would be described as neutral.

  41. Brønsted-Lowry Theory • More general theory of acid-base reactions than Arrhenius • Acid: substance that donates proton • Base: substance that accepts proton

  42. All Arrhenius acids are also Brønsted acids All Arrhenius bases are also Brønsted bases

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