Chap 08B-Solution Concentrations

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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


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 =


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


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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


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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Chap 08B-Solution Concentrations

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