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Chapter 8B
Solution
Concentrations
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CHAPTER OUTLINE���
§ Concentration Units § Mass Percent § Using Percent Concentration § Molarity § Using Molarity § Dilution § Osmolarity § Tonicity of Solutions
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CONCENTRATION���UNITS���
q The amount of solute dissolved in a certain amount of solution is called concentration.
q 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)
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MASS PERCENT���
q Mass percent (% m/m) is defined as the mass of solute divided by the mass of solution.
mass of soluteMass % (m/m) = x100mass of solution
mass of solute + mass of solvent
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MASS/VOLUME���PERCENT���
q Mass/Volume percent (% m/v) is defined as the mass of solute divided by the volume of solution.
mass of soluteMass % (m/v) = x100volume of solution
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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
30.0 gMass % (m/m)= x100 = 20.0 %150.0 g
= 150.0 g
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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 gMass % (m/v) = x100 = 2.0 %250 mL
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USING PERCENT���CONCENTRATION���
q In the preparation of solutions, one often needs to calculate the amount of solute or solution.
q To achieve this, percent composition can be used as a conversion factor.
q Some examples of percent compositions, their meanings, and possible conversion factors are shown in the table below:
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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
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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
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Example 3:���How many grams of solute are needed to prepare 150 mL of a 40.0% (m/v) solution of LiNO3?
150 mL solution x = 60. g LiNO3 40.0 g LiNO3
100 mL solution
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MOLARITY���
q The most common unit of concentration used in the laboratory is molarity (M).
q Molarity is defined as:
Molarity = moles of solute
Liter of solution
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Example 1:���
What is the molarity of a solution containing 1.4 mol of acetic acid in 250 mL of solution?
Vol. of solution = 1 L250 mL x = 0.25 L
1000 mL
1.4 mol acetic acid0.25 LMolarity = = 5.6 M
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Example 2:���
What is the molarity of a solution prepared by dissolving 60.0 g of NaOH in 0.250 L of solution?
1 mol60.0 g x = 1.50 mol40.0 gMol of solute =
1.50 mol = 6.00 M0.250 LMolarity =
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Example 3:���
What is the molarity of a solution that contains 75 g of KNO3 in 350 mL of solution?
1 mol75 g x101.1 gMol of solute =
Vol of solvent = 1 L350 mL x = 0.35 L1000 mL
0.74 molMolarity = = 2.1 M0.350 L
= 0.74 mol
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USING���MOLARITY���
q Molarity relationship can be used to calculate: moles soluteMolarity =
volume of solution
Amount of solute:
Moles solute = Molarity x volume
Volume of solution: moles soluteVolume of solution =
Molarity
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Example 1:���How many moles of nitric acid are in 325 mL of 16 M HNO3 solution?
Vol. of solution = 1 L325 mL x = 0.325 L
1000 mL
mol of solute = = 5.2 mol mol0.325 L x
L16 1
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Example 2:���How many grams of KCl would you need to prepare 0.250 L of 2.00 M KCl solution?
mass of solute = 74.6 g0.500 mol x1 mol
mol of solute = = 0.500 mol mol0.250 L x
L2.00
1
= 37.3 g
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Example 3:���How many grams of NaHCO3 are in 325 mL of 4.50 M solution of NaHCO3?
84.0 g1.46 mol x1 molmass of solute =
mol of solute = = 1.46 mol mol0.325 L x
L4.50
1
= 123 g
Vol. of solution = 1 L325 mL x = 0.325 L1000 mL
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Example 4:���What volume (L) of 1.5 M HCl solution contains 6.0 moles of HCl?
Vol. of solution = = 4.0 L L6.0 mol x
mol1
1.5
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Example 5:���What volume (mL) of 2.0 M NaOH solution contains 20.0 g of NaOH?
1 mol20.0 g x40.0 g
mol of solute =
Vol. In L = = 0.25 L L0.500 mol x
mol1 2.0
= 0.500 mol
Vol. In mL = 1000 mL0.250 L x1 L
= 250 mL
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Example 6:���How many mL of a 0.300 M glucose (C6H12O6) IV solution is needed to deliver 10.0 g of glucose to the patient?
1 mol10.0 g x180.1 gmol of solute =
Vol. In L = = 0.185 L L0.0555 mol x
mol1
0.300
= 0.0555 mol
Vol. In mL = 1000 mL0.185 L x1 L
= 185 mL
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DILUTION���
q Solutions are often prepared from more concentrated ones by adding water. This process is called dilution.
q 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
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DILUTION���
q The amount of solute depends on the concentration and the volume of the solution.
Therefore,
M1 x V1 = M2 x V2
Concentrated solution Dilute
solution
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Example 1:���What is the molarity of the final solution when 75 mL of 6.0 M KCl solution is diluted to 150 mL?
M2 = 3.0 M
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M VM =V
M1 = 6.0 M
V1 = 75 mL
M2 = ???
V2 = 150 mL
M1 x V1 = M2 x V2
(6.0 M)(75 mL)=150 mL
Volume increases
Concentration decreases
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Example 2:���What volume (mL) of 0.20 M HCl solution can be prepared by diluting 50.0 mL of 1.0 M HCl?
V2 = 250 mL
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M VV =M
M1 = 1.0 M
V1 = 50.0 mL
M2 = 0.20 M
V2 = ???
M1 x V1 = M2 x V2
(1.0 M)(50.0 mL)=0.20 M
Concentration decreases
Volume increases
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OSMOLARITY���
q Many important properties of solutions depend on the number of particles formed in solution.
q Recall that when ionic substances (strong electrolytes) dissolve in water they form several particles for each formula unit.
q For example:
NaCl (s) Na+ (aq) + Cl- (aq)
1 formula unit 2 particles
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OSMOLARITY���
CaCl2 (s) Ca2+ (aq) + 2 Cl- (aq)
1 formula unit
3 particles
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OSMOLARITY���
q When covalent substances (non- or weak electrolytes) dissolve in water they form only one particle for each formula unit.
q For example:
C12H22O11 (s) C12H22O11 (aq)
1 formula unit 1 particle
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OSMOLARITY���
q 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
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0.10 M NaCl =
Examples:���
2 particles in solution
2 x 0.10 M = 0.20 osmol
0.10 M CaCl2 =
3 particles in solution
3 x 0.10 M = 0.30 osmol
0.10 M C12H22O11 =
1 particle in solution
0.10 osmol 1 x 0.10 M =
Same molarities but different osmolarities
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TONICITY OF���SOLUTIONS���
q 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.
q 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.
q The comparison of osmolarity of a solution with those in body fluids determines the tonicity of a solution.
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ISOTONIC���SOLUTIONS���
q Solutions with the same osmolarity as the cells (0.30) are called isotonic.
q These solutions are called physiological solutions and allow red blood cells to retain their normal volume.
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HYPOTONIC���SOLUTIONS���
q Solutions with lower osmolarity than the cells are called hypotonic.
q In these solutions, water flows into a red blood cell, causing it to swell and burst (hemolysis).
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HYPERTONIC���SOLUTIONS���
q Solutions with greater osmolarity than the cells are called hypertonic.
q In these solutions, water leaves the red blood cells causing it to shrink (crenation).
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0.10 M NaCl =
Examples:���
0.20 osmol
0.10 M CaCl2 = 0.30 osmol
0.10 M C12H22O11 = 0.10 osmol
hypotonic
isotonic
hypotonic
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THE END