1. Let’s study solutions Solutions homogeneous mixtures of two or more substances solvent & one or more solutes Solutes spread evenly throughout cannot separate filtration Separated evaporation. not visible unless colored

2. Most common solvent “water” polar molecule. forms hydrogen bonds hydrogen atom in one molecule - oxygen atom of different water molecule Na + and Cl - ions surface of a NaCl crystal attracted polar water molecules. hydrated solution many H 2 O molecules surrounds 2

3. Like dissolves like Two substances form solution an attraction between particles of solute and solvent. polar solvent dissolves polar solutes water and sugar water ionic solutes (NaCl) non-polar solvent hexane (C 6 H 14 ) dissolves nonpolar solutes (oil or grease) Substances in water are: strong electrolytes produce ions, conduct an electric current NaCl(s) + H 2 O Na + (aq) + Cl − (aq) weak electrolytes produce few ions HF(g) + H 2 O(l) H 3 O + (aq) + F - (aq) non-electrolytes do not produce ions 3 Water and polar solute

4. Solubility maximum amount of solute dissolves in specific amount of solvent expressed grams of solute in 100 grams of solvent water g of solute 100 g water Unsaturated solutions less than maximum amount of solute dissolve more solute Saturated solutions contain maximum amount of solute than dissolves undissolved solute at bottom of container 4 Dissolved solute

5. Effect of Temperature on Solubility Solubility Depends on temperature. Of most solids increases as temperature increases. Of gases decreases as temperature increases. 5

6. How to calculate percent concentration? Solubility amount of solute amount of solution mass or volume of solute in a solution expressed in “100” percent = amt.solute x 100 amt. solute + amt. solvent a. mass percent (m/m) = g of solute x 100 100 g of solution b. volume % (v/v) = mL of solute x 100 100 mL of solution c. mass/volume % (m/v) = g of solute x 100 100 mL of solution 6

7. Example calculating percentages! calculation of mass percent grams of solute (g KCl) and grams of solution (g KCl solution). g of KCl = 8.00 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g 8.00 g KCl (solute) x 100 = 16.0% 50.00 g KCl solution 7

8. Using percent concentration as conversion factors! How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution? STEP 1 Given: 225 g solution; 10.0% (m/m) NaCl Need: g of NaCl STEP 2 g solution g NaCl STEP 3 Write the 10.0% (m/m) as conversion factors. 10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl STEP 4 Set up using the factor that cancels g solution. 225 g solution x 10.0 g NaCl = 22.5 g NaCl 100 g solution 8

9. Learning Check A solution is prepared by mixing 15.0 g Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (%m/m) of the solution. A) 15.0% (m/m) Na 2 CO 3 B) 6.38% (m/m) Na 2 CO 3 C) 6.00% (m/m) Na 2 CO 3 9

10. Solution C) 6.00% (m/m) Na 2 CO 3 STEP 1 mass solute = 15.0 g Na 2 CO 3 mass solution = 15.0 g + 235 g = 250. g STEP 2 Use g solute/ g solution ratio STEP 3 mass %(m/m) =g solute x 100 g solution STEP 4 Set up problem mass %(m/m) = 15.0 g Na 2 CO 3 x 100 = 6.00% Na 2 CO 3 250. g solution 10

11. What is Molarity (M)? Molarity (M) concentration term for solutions gives moles of solute in 1 L solution moles of solute liter of solution M = m v or m = M x v or v = m M 1.00 M NaCl solution prepared weigh 58.5 g NaCl (1.00 mole) and add water to make 1.00 liter of solution 11

12. Steps for calculation of molarity What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1 Given 6.00 g NaOH in 0.500 L solution Need molarity (mole/L) STEP 2 Plan g NaOH mole NaOH molarity STEP 3 Conversion factors 1 mole NaOH = 40.0 g 1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH STEP 4 Calculate molarity. 6.00 g NaOH x 1 mole NaOH = 0.150 mole 40.0 g NaOH 0.150 mole = 0.300 mole = 0.300 M NaOH 0.500 L 1 L 12

13. Learning Check What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO 3 ? A) 0.557 M B) 1.44 M C) 1.71 M 13

14. Solution C) 1.71 M 46.8 g NaHCO 3 x 1 mole NaHCO 3 = 0.557 mole NaHCO 3 84.0 g NaHCO 3 0.557 mole NaHCO 3 = 1.71 M NaHCO 3 0.325 L 14

15. Learning Check What is the molarity of 225 mL of a KNO 3 solution containing 34.8 g KNO 3 ? A)0.344 M B)1.53 M C)15.5 M 15

16. Solution B)1.53 M 34.8 g KNO 3 x 1 mole KNO 3 = 0.344 mole KNO 3 101.1 g KNO 3 M = mole = 0.344 mole KNO 3 = 1.53 M L 0.225 L In one setup: 34.8 g KNO 3 x 1 mole KNO 3 x 1 = 1.53 M 101.1 g KNO 3 0.225 L 16

17. Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions. 17 TABLE 7.8

18. Calculations Using Molarity How many grams of KCl are needed to prepare 125mL of a 0.720 M KCl solution? STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl STEP 2 Plan L KCl moles KCl g KCl 18

19. Learning Check How many grams of AlCl 3 are needed to prepare 125 mL of a 0.150 M solution? A) 20.0 g AlCl 3 B) 16.7g AlCl 3 C) 2.50 g AlCl 3 19

20. Solution C) 2.50 g AlCl 3 0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl 3 1 L 1 mole 20

21. Learning Check How many milliliters of 2.00 M HNO 3 contain 24.0 g HNO 3 ? A) 12.0 mL B) 83.3 mL C) 190. mL 21

22. Solution 24.0 g HNO 3 x 1 mole HNO 3 x 1000 mL = 63.0 g HNO 3 2.00 mole HNO 3 Molarity factor inverted = 190. mL HNO 3 22

23. D ilution In a dilution water is added. volume increases. concentration decreases. 23

24. Comparing Initial and Diluted Solutions In the initial and diluted solution, the moles of solute are the same. the concentrations and volumes are related by the following equations: For percent concentration: C 1 V 1 = C 2 V 2 initial diluted For molarity: M 1 V 1 = M 2 V 2 initial diluted 24

25. Dilution Calculations with Percent What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution? Prepare a table: C 1 = 14.0% (m/v)V 1 = 25.0 mL C 2 = 2.00% (m/v)V 2 = ? Solve dilution equation for unknown and enter values: C 1 V 1 = C 2 V 2 V 2 = V 1 C 1 = (25.0 mL)(14.0%) = 175 mL C 2 2.00% 25

26. Learning Check What is the percent (% m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL? 26

27. Solution What is the percent (%m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL? Prepare a table: C 1 = 9.00 %(m/v)V 1 = 10.0 mL C 2 = ?V 2 = 60.0 mL Solve dilution equation for unknown and enter values: C 1 V 1 = C 2 V 2 C 2 = C 1 V 1 = (10.0 mL)(9.00%) = 1.50% (m/v) V 2 60.0 mL 27

28. Dilution Calculations with Molarity What is the molarity (M) of a solution prepared by diluting 0.180L of 0.600 M HNO 3 to 0.540 L? Prepare a table: M 1 = 0.600 MV 1 = 0.180 L M 2 = ?V 2 = 0.540 L Solve dilution equation for unknown and enter values: M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 = (0.600 M)(0.180 L) = 0.200 M V 2 0.540 L 28

29. Learning Check What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? A) 27.0 mL B) 60.0 mL C) 90.0 mL 29

30. Solution What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? Prepare a table: M 1 = 1.80 MV 1 = 15.0 mL M 2 = 0.300MV 2 = ? Solve dilution equation for V 2 and enter values: M 1 V 1 = M 2 V 2 V 2 = M 1 V 1 = (1.80 M)(15.0 mL) = 90.0 mL M 2 0.300 M 30

31. How colloids and suspension different than solutions? Solutions contain small particles (ions or molecules). are transparent. do not separate. cannot be filtered. do not scatter light. Colloids have medium size particles. cannot be filtered. can be separated by semipermeable membranes. scatter light (Tyndall effect). Suspensions have very large particles. settle out. can be filtered. must be stirred to stay suspended, blood platelets, muddy water Examples Fog Whipped cream Milk Cheese Blood plasma Pearls

32. What is osmosis? water (solvent) flows from lower solute concentration into higher solute concentration. level of solution with higher concentration rises concentrations of two solutions become equal with time Osmotic pressure produced by solute particles dissolved in solution. equal to pressure that prevent flow of additional water into more concentrated solution. greater as number of dissolved particles in solution increases. isotonic solution same osmotic pressure hypotonic solution has a lower osmotic pressure hypertonic solution has a higher osmotic pressure 32