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

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Presentation on theme: "solution: a homogeneous mixture solute: substance that gets dissolved solvent: substance that does the dissolving tincture: sol’n in which alcohol is."— Presentation transcript:

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

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

4 hydrochloric acid dissolving in water δ+ HCl δ-

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, H 2 SO 4 III. polar covalentex: NH 3, H 2 O 2 cmpds w/ -OH groupsex: sugars, alcohols ethanol: glucose:C 6 H 12 O 6 C 2 H 5 OH

7 compounds that don’t dissolve in water nonpolar covalent cmpds ex: hydrocarbons like C 4 H 10 symmetrical molecules like CCl 4 Conductivity of Solutions Electrolyte: cmpd that dissociates into ions; conducts electricity must be ionic cmpd or an acid ex: CuSO 4, HNO 3

8 nonelectrolyte:doesn’t dissociate; won’t conduct covalent cmpds that are not acids ex: sugar, C 12 H 22 O 11 alcohols, C 2 H 5 OH Sample problems: determine formula, classify as soluble or insoluble, electrolyte or non acetic acidHC 2 H 3 O 2 sol.elec. calcium chlorideCaCl 2 sol.elec. Hexane, C 6 H 14 insol. nonelec

9 silver nitrate methanol hydrochloric acid butane potassium iodide AgNO 3 sol, elec CH 3 OHsol, nonelec HCl sol, elec KI sol, elec C 4 H 10 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 tempdecreasing temp (water) increases solubility of gas Effect of pressureIncreasing 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 KNO 3 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 percent: mass of solute total mass sol’n 100 ppm:parts per million ppb parts per billion Molarity (M): moles solute Liter of sol’n

16 Calculating Molarity 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? 30.0 g g mol1 40 0.75 mol 0.50 L 1.5 M

17 How many moles of HCl are contained in 5.0 liters of a 6.0 M HCl solution? 6.0 M X mol 5.0 L X = 30 mol How many grams of potassium sulfate are needed to dissolve in water to make 250 mL of a 2.0-molar solution? K 2 SO 4 2.0 M X mol 0.25 L X = 0.5 mol 0.5 mol 1 mol 174.3 g 87 g

18 Calculate the molarity of a solution that has 12.5 g of glucose, C 6 H 12 O 6, dissolved in 500 mL of solution. 12.5 g g mol1 180 0.06944 mol 0.5 L 0.14 M 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 0.1125 mol mol 1 g 167 18.8 g

19 Diluting a Solution concentrateddiluted 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)(800.0 mL)M2M2 800 mL M 2 = 0.75 M

20 (12.0 M)V1V1 (1.0 L)(3.0 M) 12.0 M What volume of 12.0 M HCl solution is needed to make 1.0 liter of 3.0 M solution of HCl ? 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. V 1 = 0.25 L (17 M)(25.0 mL)(1000 mL)M2M2 1000 mL M 2 = 0.43 M

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

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

23 electrolytes dissociate into ions; produces a higher concentration of dissolved solute particles NaClNa + + Cl - 1 mol 2 moles ions Nonelectrolytes don’t dissociate C 12 H 22 O 11 (s)C 12 H 22 O 11 (aq) 1 mol Electrolytes have the greater effect on C.P. ‘s

24 examples: Which of these 1-molar solutions has the lowest freezing point? AlCl 3 KI C 12 H 22 O 11 Na 2 CO 3 Al 3+ + Cl - 3 K + + I - C 12 H 22 O 11 Na + + CO 3 2- 2 4 ions 2 ions 1 molecule 3 ions

25 Which of these 1-molal solutions has the highest boiling point? CuSO 4 LiBrMg(NO 3 ) 2 C 6 H 12 O 6 2231 change in F.P. molality freezing pt. constant m = mol/kg solv. ionization factor k f = 1.86°C/m k b = 0.512°C/m

26 Calculate the freezing & boiling point of a 1.5-molal solution of sodium hydroxide. F.P. ΔT = k f m i ΔT = (1.86 °C/m) (1.5 m) (2) NaOH Na + + OH - 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. AlCl 3 Al + + 3 Cl - ΔT = k f m i ΔT = k b m i ΔT = (1.86 °C/m) (2.0 m) (4) ΔT = (0.512 °C/m) (2.0 m) (4) ΔT = 14.88 °C F.P. = -15 °C ΔT = 4.096 °C B.P. = 104 °C


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