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
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 H2O molecules surrounds
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Like dissolves like
Two substances form solution
• an attraction between
• particles of solute and solvent.
Water and polar solute
• polar solvent dissolves polar solutes
• water and sugar
• water ionic solutes (NaCl)
• non-polar solvent hexane (C6H14)
• dissolves nonpolar solutes
• (oil or grease)
Substances in water are:
• strong electrolytes produce ions, conduct an electric current
• NaCl(s) + H2O
Na+(aq) + Cl− (aq)
• weak electrolytes produce few ions
• HF(g) + H2O(l)
H3O+(aq) + F- (aq)
• non-electrolytes do not produce ions
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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
Dissolved
solute
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Effect of Temperature on Solubility
Solubility
• Depends on temperature.
• Of most solids increases as temperature increases.
• Of gases decreases as temperature increases.
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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
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Example calculating percentages!
calculation of mass percent
• grams of solute (g KCl) and
• grams of solution (g KCl solution).
g of KCl
=
g of solvent (water)
=
g of KCl solution
=
8.00 g
42.00 g
50.00 g
8.00 g KCl (solute)
x 100 = 16.0%
50.00 g KCl solution
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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
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Learning Check
A solution is prepared by mixing 15.0 g Na2CO3
and 235 g of H2O. Calculate the mass percent
(%m/m) of the solution.
A) 15.0% (m/m) Na2CO3
B) 6.38% (m/m) Na2CO3
C) 6.00% (m/m) Na2CO3
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Solution
C) 6.00% (m/m) Na2CO3
STEP 1 mass solute =
15.0 g Na2CO3
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 Na2CO3 x 100 = 6.00% Na2CO3
250. g solution
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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
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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
1L
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Learning Check
What is the molarity of 325 mL of a solution
containing 46.8 g of NaHCO3?
A) 0.557 M
B) 1.44 M
C) 1.71 M
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Solution
C) 1.71 M
46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole
NaHCO3
84.0 g NaHCO3
0.557 mole NaHCO3 = 1.71 M NaHCO3
0.325 L
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Learning Check
What is the molarity of 225 mL of a KNO3 solution
containing 34.8 g KNO3?
A) 0.344 M
B) 1.53 M
C) 15.5 M
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Solution
B) 1.53 M
34.8 g KNO3 x 1 mole KNO3
101.1 g KNO3
= 0.344 mole KNO3
M = mole =
0.344 mole KNO3 = 1.53 M
L
0.225 L
In one setup:
34.8 g KNO3 x 1 mole KNO3 x 1
= 1.53 M
101.1 g KNO3 0.225 L
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Molarity Conversion Factors
The units of molarity are used as conversion factors
in calculations with solutions.
TABLE 7.8
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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
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Learning Check
How many grams of AlCl3 are needed to prepare
125 mL of a 0.150 M solution?
A)
20.0 g AlCl3
B)
16.7g AlCl3
C)
2.50 g AlCl3
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Solution
C)
2.50 g AlCl3
0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3
1L
1 mole
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Learning Check
How many milliliters of 2.00 M HNO3 contain 24.0 g
HNO3?
A) 12.0 mL
B) 83.3 mL
C) 190. mL
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Solution
24.0 g HNO3 x 1 mole HNO3 x
1000 mL
=
63.0 g HNO3
2.00 mole HNO3
Molarity factor inverted
= 190. mL HNO3
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Dilution
In a dilution
• water is added.
• volume increases.
• concentration decreases.
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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:
C1V1 = C2V2
initial
diluted
For molarity:
M1V1 = M2V2
initial
diluted
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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:
C1= 14.0% (m/v) V1 = 25.0 mL
C2= 2.00% (m/v) V2 = ?
Solve dilution equation for unknown and enter
values:
C1V1 = C2V2
V2
= V1C1
C2
= (25.0 mL)(14.0%) = 175 mL
2.00%
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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?
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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:
C1= 9.00 %(m/v) V1 = 10.0 mL
C2= ?
V2 = 60.0 mL
Solve dilution equation for unknown and enter
values:
C1V1 = C2V2
C2
= C1 V1 = (10.0 mL)(9.00%) = 1.50% (m/v)
V2
60.0 mL
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Dilution Calculations with Molarity
What is the molarity (M) of a solution prepared
by diluting 0.180L of 0.600 M HNO3 to 0.540 L?
Prepare a table:
M1= 0.600 M
V1 = 0.180 L
M2= ?
V2 = 0.540 L
Solve dilution equation for unknown and enter
values:
M1V1 = M2V2
M2
= M1V1
V2
= (0.600 M)(0.180 L) = 0.200 M
0.540 L
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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
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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:
M1= 1.80 M
V1 = 15.0 mL
M2= 0.300 M
V2 = ?
Solve dilution equation for V2 and enter values:
M1V1 = M2V2
V2
= M1V1
M2
= (1.80 M)(15.0 mL) = 90.0 mL
0.300 M
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How colloids and suspension
different than solutions?
Examples
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.
•Fog
•Whipped cream
•Milk
•Cheese
•Blood plasma
•Pearls
• scatter light (Tyndall effect).
Suspensions
• have very large particles.
• settle out.
• can be filtered.
• must be stirred to stay suspended, blood platelets, muddy water
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
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