Solution Concentrations CHAPTER OUTLINE

Transcription

1 Chapter 8B Solution Concentrations CHAPTER OUTLINE Concentration Units Mass Percent Using Percent Concentration Molarity Using Molarity Dilution Osmolarity Tonicity of Solutions 2

2 CONCENTRATION UNITS The amount of solute dissolved in a certain amount of solution is called concentration. Concentration = amount of solute amount of solution Three types of concentration units will be studied in this class: Mass Percent: (m/m) and (m/v) Molarity 3 MASS PERCENT Mass percent (% m/m) is defined as the mass of solute divided by the mass of solution. Mass % (m/m) = mass of solute x00 mass of solution mass of solute + mass of solvent 4 2

3 MASS/VOLUME PERCENT Mass/Volume percent (% m/v) is defined as the mass of solute divided by the volume of solution. Mass % (m/v) = mass of solute x00 volume of solution 5 Example : What is the mass % (m/m) of a NaOH solution that is made by dissolving 30.0 g of NaOH in 20.0 g of water? Mass of solution = 30.0 g g = 50.0 g 30.0 g Mass % (m/m)= x00 = 20.0 % 50.0 g 6 3

4 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? 5.0 g Mass % (m/v) = x00 = 2.0 % 250 ml 7 USING PERCENT CONCENTRATION Some In the examples preparation of percent of solutions, compositions, one often their needs to meanings, calculate the and amount possible of conversion solute or solution. factors are shown To achieve in the this, table percent below: composition can be used as a conversion factor. 8 4

5 Example : A topical antibiotic solution is.0% (m/v) Clindamycin. How many grams of Clindamycin are in 65 ml of this solution?.0 g Clindamycin 65 ml solution x 00 ml solution = 0.65 g 9 Example 2: How many grams of KCl are in 225 g of an 8.00% (m/m) solution? 225 g solution x = 8.0 g KCl 8.00 g KCl 00 g solution 0 5

6 Example 3: How many grams of solute are needed to prepare 50 ml of a 40.0% (m/v) solution of LiNO 3? 40.0 g LiNO 50 ml solution x 3 00 ml solution = 60. g LiNO 3 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 2 6

7 Example : What is the molarity of a solution containing.4 mol of acetic acid in 250 ml of solution? Vol. of solution = L 250 ml x = 0.25 L 000 ml Molarity =.4 mol acetic acid 0.25 L = 5.6 M 3 Example 2: What is the molarity of a solution prepared by dissolving 60.0 g of NaOH in L of solution? Mol of solute = mol 60.0 g x =.50 mol 40.0 g Molarity =.50 mol = 6.00 M L 4 7

8 Example 3: What is the molarity of a solution that contains 75 g of KNO 3 in 350 ml of solution? Mol of solute = mol 75 g x 0. g = 0.74 mol Vol of solvent = L 350 ml x = 0.35 L 000 ml 0.74 mol Molarity = = 2. M L 5 USING MOLARITY Molarity relationship can be used to calculate: Amount of solute: moles solute Molarity = volume of solution Moles solute = Molarity x volume Volume of solution: Volume of solution = moles solute Molarity 6 8

9 Example : How many moles of nitric acid are in 325 ml of 6 M HNO 3 solution? Vol. of solution = L 325 ml x = L 000 ml mol of solute = L x 6 mol = 5.2 mol L 7 Example 2: How many grams of KCl would you need to prepare L of 2.00 M KCl solution? mol of solute = L x 2.00 mol L = mol mass of solute = 74.6 g mol x mol = 37.3 g 8 9

10 Example 3: How many grams of NaHCO 3 are in 325 ml of 4.50 M solution of NaHCO 3? Vol. of solution = L 325 ml x = L 000 ml mol of solute = L x 4.50 mol =.46 mol L mass of solute = 84.0 g.46 mol x mol = 23 g 9 Example 4: What volume (L) of.5 M HCl solution contains 6.0 moles of HCl? Vol. of solution = 6.0 mol x.5 L mol = 4.0 L 20 0

11 Example 5: What volume (ml) of 2.0 M NaOH solution contains 20.0 g of NaOH? mol of solute = mol 20.0 g x 40.0 g = mol Vol. In L = mol x 2.0 L mol = 0.25 L Vol. In ml = 000 ml L x L = 250 ml 2 Example 6: How many ml of a M glucose (C 6 H 2 O 6 ) IV solution is needed to deliver 0.0 g of glucose to the patient? mol of solute = mol 0.0 g x 80. g = mol Vol. In L = mol x L mol = 0.85 L Vol. In ml = 000 ml 0.85 L x L = 85 ml 22

12 DILUTION Amount of Volume Solutions When and more concentration are solute water often is prepared added inversely to from a solution, proportional more concentrated remains ones Volume by adding water. This process is called constant increases dilution. Concentration decreases Frozen juice Water Diluted juice 23 DILUTION The amount of solute depends on the concentration and the volume of the solution. Therefore, M x V = M 2 x V 2 Concentrated solution Dilute solution 24 2

13 Example : What is the molarity of the final solution when 75 ml of 6.0 M KCl solution is diluted to 50 ml? Concentration Volume decreases increases M M = 6.0 M x V = M 2 x V 2 M V V = 75 ml (6.0 M)(75 ml) M= 2 = V 2 50 ml M 2 =??? V 2 = 50 ml M 2 = 3.0 M 25 Example 2: What volume (ml) of 0.20 M HCl solution can be prepared by diluting 50.0 ml of.0 M HCl? Volume Concentration increases decreases M M =.0 M x V = M 2 x V 2 M V V = 50.0 ml (.0 M)(50.0 ml) V= 2 = M M M 2 = 0.20 M V 2 =??? V 2 = 250 ml 26 3

14 OSMOLARITY Many Recall important that when properties ionic substances of solutions (strong depend on electrolytes) the number dissolve of particles in water formed they form in solution. several particles for each formula unit. For example: NaCl (s) Na + (aq) + Cl - (aq) formula unit 2 particles 27 OSMOLARITY CaCl 2 (s) Ca 2+ (aq) + 2 Cl - (aq) formula unit 3 particles 28 4

15 OSMOLARITY When covalent substances (non- or weak electrolytes) dissolve in water they form only one particle for each formula unit. For example: C 2 H 22 O (s) C 2 H 22 O (aq) formula unit particle 29 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 30 5

16 Examples: 0.0 M NaCl = 0.0 M CaCl 2 = 2 x 0.0 M particle = 0.20 in osmol solution 3 x 0.0 M = 0.30 osmol 0.0 M C 2 H 22 O 2 = particles x 0.0 M = 0.0 osmol 3 particles in solution Same molarities in solution but different osmolarities 3 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. 32 6

17 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. 33 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). 34 7

18 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). 35 Examples: 0.0 M NaCl = 0.20 osmol hypotonic 0.0 M CaCl 2 = 0.30 osmol isotonic 0.0 M C 2 H 22 O = 0.0 osmol hypotonic 36 8

19 THE END 37 9