1. 1 Chapter 8B Solution Concentrations

2. 2 CHAPTER OUTLINE  Concentration Units Concentration Units  Mass Percent Mass Percent  Using Percent Concentration Using Percent Concentration  Molarity Molarity  Using Molarity Using Molarity  Dilution Dilution  Osmolarity Osmolarity  Tonicity of Solutions Tonicity of Solutions

3. 3 CONCENTRATION UNITS  The amount of solute dissolved in a certain amount of solution is called concentration.  Three types of concentration units will be studied in this class: Mass Percent: Molarity Concentration = amount of solute amount of solution (m/m) and (m/v)

4. 4 MASS PERCENT  Mass percent (% m/m) is defined as the mass of solute divided by the mass of solution. mass of solute + mass of solvent

5. 5 MASS/VOLUME PERCENT  Mass/Volume percent (% m/v) is defined as the mass of solute divided by the volume of solution.

6. 6 Example 1: What is the mass % (m/m) of a NaOH solution that is made by dissolving 30.0 g of NaOH in 120.0 g of water? Mass of solution =30.0 g + 120.0 g= 150.0 g

7. 7 Example 2: What is the mass % (m/v) of a solution prepared by dissolving 5.0 g of KI to give a final volume of 250 mL?

8. 8 USING PERCENT CONCENTRATION  In the preparation of solutions, one often needs to calculate the amount of solute or solution.  To achieve this, percent composition can be used as a conversion factor.  Some examples of percent compositions, their meanings, and possible conversion factors are shown in the table below:

9. 9 Example 1: A topical antibiotic solution is 1.0% (m/v) Clindamycin. How many grams of Clindamycin are in 65 mL of this solution? 65 mL solution x = 0.65 g 1.0 g Clindamycin 100 mL solution

10. 10 Example 2: How many grams of KCl are in 225 g of an 8.00% (m/m) solution? 225 g solution x = 18.0 g KCl 8.00 g KCl 100 g solution

11. 11 Example 3: How many grams of solute are needed to prepare 150 mL of a 40.0% (m/v) solution of LiNO 3 ? 150 mL solution x = 60. g LiNO 3 40.0 g LiNO 3 100 mL solution

12. 12 MOLARITY  The most common unit of concentration used in the laboratory is molarity (M).  Molarity is defined as: Molarity = moles of solute Liter of solution

13. 13 Example 1: What is the molarity of a solution containing 1.4 mol of acetic acid in 250 mL of solution? Vol. of solution = Molarity = = 5.6 M

14. 14 Example 2: What is the molarity of a solution prepared by dissolving 60.0 g of NaOH in 0.250 L of solution? Mol of solute = Molarity =

15. 15 Example 3: What is the molarity of a solution that contains 75 g of KNO 3 in 350 mL of solution? Mol of solute = Vol of solvent = = 0.74 mol

16. 16 USING MOLARITY  Molarity relationship can be used to calculate: Amount of solute: Moles solute = Molarity x volume Volume of solution:

17. 17 Example 1: How many moles of nitric acid are in 325 mL of 16 M HNO 3 solution? Vol. of solution = mol of solute == 5.2 mol 16 1

18. 18 Example 2: How many grams of KCl would you need to prepare 0.250 L of 2.00 M KCl solution? mass of solute = mol of solute == 0.500 mol 2.00 1 = 37.3 g

19. 19 Example 3: How many grams of NaHCO 3 are in 325 mL of 4.50 M solution of NaHCO 3 ? mass of solute = mol of solute == 1.46 mol 4.50 1 = 123 g Vol. of solution =

20. 20 Example 4: What volume (L) of 1.5 M HCl solution contains 6.0 moles of HCl? Vol. of solution == 4.0 L 1 1.5

21. 21 Example 5: What volume (mL) of 2.0 M NaOH solution contains 20.0 g of NaOH? mol of solute = Vol. In L == 0.25 L 1 2.0 = 0.500 mol Vol. In mL == 250 mL

22. 22 Example 6: How many mL of a 0.300 M glucose (C 6 H 12 O 6 ) IV solution is needed to deliver 10.0 g of glucose to the patient? mol of solute = Vol. In L == 0.185 L 1 0.300 = 0.0555 mol Vol. In mL == 185 mL

23. 23 DILUTION  Solutions are often prepared from more concentrated ones by adding water. This process is called dilution.  When more water is added to a solution, Frozen juice Water Diluted juice Volume increases Concentration decreases Amount of solute remains constant Volume and concentration are inversely proportional

24. 24 DILUTION  The amount of solute depends on the concentration and the volume of the solution. Therefore, M 1 x V 1 = M 2 x V 2 Concentrated solution Dilute solution

25. 25 Example 1: What is the molarity of the final solution when 75 mL of 6.0 M KCl solution is diluted to 150 mL? M 2 = 3.0 M M 1 = 6.0 M V 1 = 75 mL M 2 = ??? V 2 = 150 mL M 1 x V 1 = M 2 x V 2 Volume increases Concentration decreases

26. 26 Example 2: What volume (mL) of 0.20 M HCl solution can be prepared by diluting 50.0 mL of 1.0 M HCl? V 2 = 250 mL M 1 = 1.0 M V 1 = 50.0 mL M 2 = 0.20 M V 2 = ??? M 1 x V 1 = M 2 x V 2 Concentration decreases Volume increases

27. 27 OSMOLARITY  Many important properties of solutions depend on the number of particles formed in solution.  Recall that when ionic substances (strong electrolytes) dissolve in water they form several particles for each formula unit.  For example: NaCl (s)Na + (aq) + Cl  (aq) 1 formula unit 2 particles

28. 28 OSMOLARITY CaCl 2 (s)Ca 2+ (aq) + 2 Cl  (aq) 1 formula unit 3 particles

29. 29 OSMOLARITY  When covalent substances (non- or weak electrolytes) dissolve in water they form only one particle for each formula unit.  For example: C 12 H 22 O 11 (s)C 12 H 22 O 11 (aq) 1 formula unit 1 particle

30. 30 OSMOLARITY  Osmolarity of a solution is its molarity multiplied by the number of particles formed in solution. Osmolarity = i x Molarity Number of particles in solution

31. 31 0.10 M NaCl = Examples: 2 particles in solution 2 x 0.10 M =0.20 osmol 0.10 M CaCl 2 = 3 particles in solution 3 x 0.10 M =0.30 osmol 0.10 M C 12 H 22 O 11 = 1 particle in solution 0.10 osmol1 x 0.10 M = Same molarities but different osmolarities

32. 32 TONICITY OF SOLUTIONS  Because the cell membranes in biological systems are semipermeable, particles of solute in solutions can travel in and out of the membranes. This process is called osmosis.  The direction of the flow of solutions in or out of the cell membranes is determined by the relative osmolarity of the cell and the solution.  The comparison of osmolarity of a solution with those in body fluids determines the tonicity of a solution.

33. 33 ISOTONIC SOLUTIONS  Solutions with the same osmolarity as the cells (0.30) are called isotonic.  These solutions are called physiological solutions and allow red blood cells to retain their normal volume.

34. 34 HYPOTONIC SOLUTIONS  Solutions with lower osmolarity than the cells are called hypotonic.  In these solutions, water flows into a red blood cell, causing it to swell and burst (hemolysis).

35. 35 HYPERTONIC SOLUTIONS  Solutions with greater osmolarity than the cells are called hypertonic.  In these solutions, water leaves the red blood cells causing it to shrink (crenation).

36. 36 0.10 M NaCl = Examples: 0.20 osmol 0.10 M CaCl 2 =0.30 osmol 0.10 M C 12 H 22 O 11 =0.10 osmol hypotonic isotonic hypotonic

37. 37 THE END