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Solutions

Solutions. solution:. a homogeneous mixture. substance that gets dissolved. solute:. substance that does the dissolving. solvent:. aqueous : water is solvent. sol’n in which alcohol is solvent. tincture:. water is an excellent solvent because of its polarity. salt

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Solutions

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  1. Solutions

  2. solution: a homogeneous mixture substance that gets dissolved solute: substance that does the dissolving solvent: aqueous: water is solvent sol’n in which alcohol is solvent tincture:

  3. water is an excellent solvent because of its polarity salt dissolving in water

  4. hydrochloric acid dissolving in water δ+ HClδ- δ+ HClδ-

  5. “like dissolves like” polar mixes w/ polar; nonpolar w/nonpolar miscible: two liquids that mix ex: water and alcohol immiscible: liquids that don’t mix ex: water and oil

  6. types of compounds that dissolve in water I. Ionic (metal w/nonmetal) ex: NaCl, KI II. Acids (H+ w/ anion) ex: HCl, H2SO4 III. polar covalent ex: NH3, H2O2 cmpds w/ -OH groups ex: sugars, alcohols glucose: C6H12O6 ethanol: C2H5OH

  7. compounds that don’t dissolve in water nonpolar covalent cmpds ex: hydrocarbons like C4H10 symmetrical molecules like CCl4 Conductivity of Solutions Electrolyte: cmpd that dissociates into ions; conducts electricity must be ionic cmpd or an acid ex: CuSO4, HNO3

  8. nonelectrolyte: doesn’t dissociate; won’t conduct covalent cmpds that are not acids ex: sugar, C12H22O11 alcohols, C2H5OH Sample problems: determine formula, classify as soluble or insoluble, electrolyte or non acetic acid HC2H3O2 sol. elec. calcium chloride CaCl2 sol. elec. Hexane, C6H14 nonelec insol.

  9. AgNO3 sol, elec silver nitrate methanol CH3OH sol, nonelec sol, elec hydrochloric acid HCl sol, elec KI potassium iodide butane C4H10 insol, nonelec

  10. saturated, unsaturated, supersaturated saturated sol’n: contains the maximum amt of dissolved solute; equilibrium btwn dissolved and undissolved solute unsaturated sol’n: contains less than max. amt of solute supersaturated sol’n contains more than the normal amt of solute crystallizes completely by adding a “seed” crystal of the solute.

  11. solubility: concentration of a saturated sol’n ex: solubility of NaCl at 25°C is 36g/100 mL Effect of temperature on solubility solid solutes: solubility increases as temp increases

  12. Gases and solubility Effect of temp decreasing temp (water) increases solubility of gas Effect of pressure Increasing pressure (over the water) increases solubility cold, high pressure

  13. solubility curves What is the solubility of ammonium chloride at 70°C ? 61 g/100 mL How many grams of potassium nitrate will dissolve in 100 mL water at 50°C? 83 g How many grams KNO3 will saturate 250 mL water at 50°? 207.5 g

  14. 100 mL of saturated potassium chlorate sol’n is cooled from 80°C to 30°C? How many grams of solute will crystallize? at 80°: 42 g at 30°: 12 g 30 g How many grams stay dissolved? 12 g a sol’n contains 20 g sodium chloride dissolved in 100 mL water at 25°C; is it saturated, unsaturated, or supersaturated? unsaturated

  15. Solution Concentration dilute: contains little solute concentrated: contains a lot of solute ways to indicate concentration mass of solute percent: 100 total mass sol’n ppm: parts per million parts per billion ppb moles solute Molarity (M): Liter of sol’n

  16. Calculating Molarity M= mol L What is the molarity of a sol’n made by dissolving 0.25 mole of sucrose in enough water to make 100.0 mL of sol’n ? M= 0.25 mol 0.1 L 2.5 M What is the molarity of a sol’n that contains 30.0 grams of sodium hydroxide dissolved in 0.50 liter of sol’n? 1 mol 0.75 mol 30.0 g 1.5 M g 0.50 L 40

  17. How many moles of HCl are contained in 5.0 liters of a 6.0 M HCl solution? X mol M= mol L 6.0 M X = 30 mol 5.0 L How many grams of potassium sulfate are needed to dissolve in water to make 250 mL of a 2.0-molar solution? K2SO4 X mol 2.0 M X = 0.5 mol 0.25 L 174.3 g 0.5 mol 87 g 1 mol

  18. Calculate the molarity of a solution that has 12.5 g of glucose, C6H12O6, dissolved in 500 mL of solution. 1 mol 12.5 g 0.06944 mol 0.14 M 180 g 0.5 L How many moles and grams of potassium iodide would be needed to prepare 0.75 liter of a 0.15 M solution? 0.15 M = x mol 0.75 L X = 0.1125 mol 167 g 0.1125 mol 18.8 g 1 mol

  19. Diluting a Solution M1V1 M2V2 concentrated diluted If 100.0 mL of 6.0 M NaOH solution is diluted with water to 800.0 mL, what is the molarity of the diluted solution? (6.0 M) (100.0 mL) M2 (800.0 mL) 800 mL 800 mL M2 = 0.75 M M2 = 0.75 M

  20. What volume of 12.0 M HCl solution is needed to make 1.0 liter of 3.0 M solution of HCl ? (12.0 M) V1 (3.0 M) (1.0 L) 12.0 M 12.0 M V1 = 0.25 L 25.0 mL of concentrated acetic acid (17 molar) are pipetted into a flask. Water is added to the 1.0 liter Mark. Calculate the molarity of the final solution. (17 M) (25.0 mL) M2 (1000 mL) 1000 mL 1000 mL M2 = 0.43 M

  21. ColligativeProperties Properties that depend on the concentration of a solution 1. Lowering of vapor pressure water + a solute pure water a solution has a lower V.P. than the pure solvent the more concentrated the greater the effect

  22. 2. Boiling Point Elevation 100°C solutions have a higher B.P. than the pure solvent 0°C B> 100°C < 0°C more conc., higher the B.P. 3. Freezing Point Depression solutions have a lower F.P. than the pure solvent

  23. Electrolytes vs. Nonelectrolytes electrolytes dissociate into ions; produces a higher concentration of dissolved solute particles NaCl Na+ + Cl- Electrolytes have the greater effect on C.P. ‘s 2 moles ions 1 mol Nonelectrolytes don’t dissociate C12H22O11 (s) C12H22O11 (aq) 1 mol 1 mol

  24. examples: Which of these 1-molar solutions has the lowest freezing point? AlCl3 Al3+ + Cl- 4 ions 3 KI K+ + I- 2 ions C12H22O11 C12H22O11 1 molecule Na2CO3 Na+ + CO32- 2 3 ions

  25. Which of these 1-molal solutions has the highest boiling point? CuSO4 LiBr Mg(NO3)2 C6H12O6 2 2 3 1 Calculating Freezing & Boiling Point m = mol/kg solv. molality ionizationfactor ΔT = kf m i ΔT = kb m i change in F.P. freezing pt. constant kb = 0.512°C/m kf = 1.86°C/m

  26. Calculate the freezing & boiling point of a 1.5-molal solution of sodium hydroxide. F.P. NaOH Na+ + OH- ΔT = kf m i ΔT = (1.86 °C/m) (1.5 m) (2) ΔT = 5.58 °C F.P. = -5.6 °C B.P. ΔT = (0.512 °C/m) (1.5 m) (2) ΔT = 1.536 °C B.P. = 101.5 °C

  27. Calculate the freezing and boiling point of a 2.0 m aqueous solution of aluminum chloride. AlCl3 Al+ + 3 Cl- ΔT = kf m i ΔT = (1.86 °C/m) (2.0 m) (4) ΔT = 14.88 °C F.P. = -15 °C ΔT = kbmi ΔT = (0.512 °C/m) (2.0 m) (4) ΔT = 4.096 °C B.P. = 104 °C

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