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The Nature of Solution

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  • 1.  Solution - homogeneous mixture Solute - substance being dissolved Solvent - present in greater amount (usually liquid, like H20)
  • 2. Solute - KMnO4 Solvent - H2O
  • 3.  Solvation – the process of dissolving solute particles are surrounded by solvent particles solute particles are separated and pulled into solution
  • 4. - - + sugar - + + salt acetic acid NonElectrolyte Weak Electrolyte Strong Electrolyte solute exists as molecules only solute exists as ions and molecules solute exists as ions only View animation online. DISSOCIATION IONIZATION
  • 5.  Dissociation › separation of an ionic solid into aqueous ions NaCl(s)  Na+(aq) + Cl–(aq)
  • 6.  Ionization › breaking apart of some polar molecules into aqueous ions HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
  • 7. “Like Dissolves Like” NONPOLAR POLAR NONPOLAR POLAR
  • 8.  Soap/Detergent › polar “head” with long nonpolar “tail” › dissolves nonpolar grease in polar water
  • 9. UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves concentration SUPERSATURATED SOLUTION becomes unstable, crystals form
  • 10.  Solubility › maximum grams of solute that will dissolve in 100 g of solvent at a given temperature › varies with temp › based on a saturated solution
  • 11.  Solubility Curve › shows the dependence of solubility on temperature
  • 12.  Solids are more soluble at... › high temperatures.  Gases are more soluble at...  low temperatures &  high pressures (Henry’s Law).  EX: nitrogen narcosis, the “bends,” soda
  • 13.  The amount of solute in a solution.  How can we describe or communicate concentration? › Qualitatively through words (“black” coffee vs. “a light amber shade”) › Quantitatively through numbers (2 sugars & 1 cream)
  • 14. With Percent (%)  % by mass › Medicated creams › Proactiv contains 2.5% benzoyl peroxide by mass  % by volume With Moles*  Molarity (M)  Molality (m)  Mole Fraction (X) ** Most often used by chemists ** › Usually used when both solute & solvent are liquids › Gasoline can contain up to 10% ethanol by volume Parts › ppm, ppb - water contaminants (don’t talk about in here)
  • 15. moles of solute molarity (M)  L of solution % m/v  mass of solute (grams) volume of solution (mL) % v/v  volum e of solute total volum e of solution % m/m  mass of solute (grams) mass of solution (grams) Remember to multiply the % formulas by 100 OR move decimal two spots left Moles of solute Mole Fraction  Moles of solution
  • 16. moles of solute molality (m)  kg of solvent 0.25 mol 0.25m  1 kg mass of solvent only 1 kg water = 1 L water M 1V1  M 2V2
  • 17.  Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water mol m kg = 3.2m MgCl2
  • 18.  How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water 1.54 mol NaCl 58.44 g NaCl 1 kg water 1 mol NaCl 1.5 mol 1.5m  1 kg = 45.0 g NaCl
  • 19.  At a restaurant I saw a man put 7 packets of sugar into his cup of coffee. What is the percent by mass of sugar in this poor man’s coffee? › 1 packet of sugar = 4 grams › 1 cup of coffee = 180 grams 28 g sugar (180 + 28) g = 13.5% sugar
  • 20.  Find the mole fraction of sugar in the coffee of that now comatose man who has just come down off his 7 sugar packet high. › 0.08 moles sugar › 10 moles coffee Xsugar = 0.08 moles sugar (10 + 0.08) total moles = 0.0079 Xcoffee = 1 – Xsugar = 1 – 0.0079 = 0.9921
  • 21. What is the mole fraction of each component in a solution in which 3.57 g of sodium chloride, NaCl, is dissolved in 25.0 g of water? First, convert from mass of NaCl to moles of NaCl. 3.57 g NaCl x 1 mole NaCl = 0.0610857139 mole NaCl 58.44247 g NaCl Next, convert from mass of water to moles of water. 25.0 g H2O x 1 mole H2O = 1.387710877 mole H2O 18.01528 g H2O Substitute these two quantities into the defining equation for mole fraction. XNaCl = 0.0610857139 mole NaCl = 0.0421630713 ~ 0.042 (0.0610857139 + 1.387710877) mol solution Xwater = 1.387710877 mole H20 = 0.9578369287 ~ 0.96 (0.0610857139 + 1.387710877) mol solution ** Note the sum of the mole fractions for a solution will equal 1. **
  • 22. Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.  M 1V1  M 2V2
  • 23.  What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3
  • 24.  1.54m NaCl in 0.500 kg of water   mass 45.0 g of NaCl add 0.500 kg of water  500 mL of 1.54M NaCl   mass 45.0 g of NaCl add water until total volume is 500 mL 500 mL water 45.0 g NaCl 500 mL mark 500 mL volumetric flask
  • 25.  250 mL of 6.0M HNO3 by dilution 95 mL of 15.8M HNO3 › measure 95 mL of 15.8M HNO3  combine with water until total volume is 250 mL  Safety: “Do as you oughtta, add the acid to the watta!” 250 mL mark water for safety
  • 26.  Turn in one paper per team.  Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution.  For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.1M concentrate.
  • 27.  Colligative Property › property that depends on the concentration of solute particles, not their identity
  • 28.  Freezing Point Depression (tf) › f.p. of a solution is lower than f.p. of the pure solvent  Boiling Point Elevation (tb) › b.p. of a solution is higher than b.p. of the pure solvent
  • 29. Freezing Point Depression View Flash animation.
  • 30. Boiling Point Elevation Solute particles weaken IMF in the solvent.
  • 31.  Applications › salting icy roads › making ice cream › antifreeze  cars (-64°C to 136°C)  fish & insects
  • 32. t = k · m · n t: change in temperature ( C) k: constant based on the solvent ( C·kg/mol) m: molality (m) n: # of particles
  • 33.  # of Particles › Nonelectrolytes (covalent)  remain intact when dissolved  1 particle › Electrolytes (ionic)  dissociate into ions when dissolved  2 or more particles
  • 34.  At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? GIVEN: WORK: b.p. = ? m = 0.73mol ÷ 0.225kg tb = ? tb = (3.60°C·kg/mol)(3.2m)(1) kb = 3.60°C·kg/mol tb = 12°C m = 3.2m b.p. = 181.8°C + 12°C n=1 b.p. = 194°C tb = kb · m · n
  • 35.  Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. GIVEN: f.p. = ? tf = ? kf = 1.86°C·kg/mol WORK: m = 0.48mol ÷ 0.100kg tf = (1.86°C·kg/mol)(4.8m)(2) m = 4.8m n=2 tf = kf · m · n f.p. = 0.00°C - 18°C tf = 18°C f.p. = -18°C
  • 36.      Once you write the molecular equation (synthesis, decomposition, etc.), you should check for reactants and products that are soluble or insoluble (determine states of matter, which stay the same throughout reaction). Use a solubility table to tell us what compounds are aqueous (dissolved in water), solids, liquids, and/or gases We usually assume the reaction is in water If the compound is soluble (does dissolve in water), the the compound splits into its ions If the compound is insoluble (does NOT dissolve in water), then it remains as a compound
  • 37. 1. 2. 3. Write a balanced chemical (molecular) equation Consult the solubility rules (along with strong acids and strong bases) and assign the correct state of matter symbol annotation Write the Total Ionic Equation (T.I.E) a. 4. Eliminate spectator ions a. b. 5. All compounds that are annotated (aq) break up into individual cations and anions in that order Spectator ions-those ions that do not participate in the chemical reaction but are present in the reaction mixture Spectator ions are in the same form on each side of the equation arrow Write the Net Ionic Equation (N.I.E) a. The convention is to write the cation first followed by the anion on the reactants side
  • 38.  Strong Acids › Hdrochloric acid - HCl › Nitric acid - HNO3 › Sulfuric acid - H2SO4 › Hydrobromic acid HBr › Hydroiodic acid - HI › Chloric acid - HClO3 › Perchloric acid HClO4 A strong acid is an acid that ionizes completely  Strong Bases › Lithium Hydroxide – Li(OH) › Sodium Hydroxide – Na(OH) › Potassium Hydroxide – K(OH) › Calcium Hydroxide – Ca(OH)2 › Rubidium Hydroxide – Rb(OH) › Strontium Hydroxide – Sr(OH)2 › Cesium Hydroxide – Cs(OH)
  • 39. Solubility Rules
  • 40. Potassium chromate mixes with lead (II) nitrate Molecular Equation: K2CrO4 + Pb(NO3)2  Soluble Soluble PbCrO4 + 2 KNO3 Insoluble Total Ionic Equation: 2 K+ + CrO4 -2 + Pb+2 + 2 NO3-  PbCrO4 (s) + 2 K+ + 2 NO3Net Ionic Equation: CrO4 -2 + Pb+2  PbCrO4 (s) Soluble
  • 41. Balanced Chemical Equation: Pb(NO3)2(aq) + 2NaI(aq)  PbI2(s) + 2NaNO3(aq) “Complete Ionic” Equation: Pb2+(aq) + 2NO3-(aq) + 2Na+(aq)+ 2I-(aq)  PbI2(s) + 2Na+(aq) + 2NO3(aq) Cancel the “spectator ions” that appear on both sides of the arrow Pb2+(aq) + 2NO3-(aq) + 2Na+(aq)+ 2I-(aq)  PbI2(s) + 2Na+(aq) + 2NO3(aq) “Net Ionic” Equation: Pb2+(aq) + 2I-(aq)  PbI2(s)
  • 42. States of matter assigned in the molecular equation stay the same throughout the T.I.E. and the N.I.E.  Solids, liquids, and gases DO NOT ionize  Gases keep their subscript (it doesn’t become a coefficient like other compounds)  › For example: H2, Br2, Cl2, etc. stay like this  They don’t become 2H-1, 2Br -1, 2Cl -1

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