SOLUTIONS - · PDF file13.09.2013 · COMPONENTS OF SOLUTION • SOLUTE –...
Transcript of SOLUTIONS - · PDF file13.09.2013 · COMPONENTS OF SOLUTION • SOLUTE –...
SOLUTIONS
Engr. Yvonne Ligaya F. Musico
SOLUTION
A homogeneous mixture of two or more
substances, the relative proportion of
which may vary within certain limits.
Engr. Yvonne Ligaya F. Musico
COMPONENTS OF SOLUTION
• SOLUTE – component which is in small
quantity
• SOLVENT – component that is in the
greatest abundance and typically determines
the physical state of the solution.
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EXAMPLE
State of
Solution State of Solvent State of Solute Example
Gas
Liquid
Liquid
Liquid
Solid
Solid
Solid
Gas
Liquid
Liquid
Liquid
Solid
Solid
Solid
Gas
Gas
Liquid
Solid
Gas
Liquid
Solid
Air
Oxygen in water
Alcohol in water
Salt in water
Hydrogen in palladium
Mercury in silver
Silver in gold
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SOLUBILITY – the amount of solute needed to form a saturated solution in a given quantity of solvent.
SATURATED – solution that can hold no more of the solute at a particular temperature.
UNSATURATED – solution which contains less amount of solute that is required to saturate it at that temperature.
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FACTORS AFFECTING SOLUBILITY
• Solute-solvent interactions
• Pressure effect
• Temperature effect
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Solute-Solvent Interaction
– The stronger the attraction between solute and solvent molecules, the greater the solubility.
– Polar liquids tend to dissolve readily in polar solvents.
– Non-polar liquids tend to be insoluble in polar liquids.
– Substance with similar intermolecular attractive forces tend to be soluble in one another
– “LIKE DISSOLVES LIKE”
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MISCIBLE VS IMMISCIBLE
• MISCIBLE – liquids that mix in all
proportions.
• IMMISCIBLE – liquids that do not
dissolve significantly in one another.
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Pressure Effect
The solubility of a gas in any solvent is increased
as the pressure of the gas over the solvent
increases.
By contrast, the solubility of solids and liquids are
not appreciably affected by pressure.
The solubility of the gas increases in direct
proportion to its partial pressure above the
solution.
Engr. Yvonne Ligaya F. Musico
Pressure Effect
The relationship between pressure and solubility of gas is expressed by HENRY’S LAW:
gg kPC
Where: Cg = concentration of the gas
k = proportionality constant or Henry’s
law constant
Pg = partial pressure of the gas over the
solution.
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Sample Problem
Calculate the concentration of CO2 in a softdrink
that is bottled with a partial pressure of CO2 at
4.0 atm over the liquid at 25oC. The Henry’s
law constant for CO2 in water at this
temperature is 3.1 x 10-2 mol/L-atm.
Engr. Yvonne Ligaya F. Musico
Solution
= (3.1 x 10-2 mol/L-atm)(4.0
atm)
= 0.12 mol/L or 0.12 M
gg kPC
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Practice Exercise
Calculate the concentration of CO2 in the
softdrink after the bottle is opened and
equilibrates at 25oC under CO2 partial pressure
of 3.0 x 10-4 atm.
Ans.: 9.3 x 10-6 M
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Temperature Effect
– The solubility of most solid solutes in water
increase as the temperature of the solution.
– In contrast to gas solutes, the solubility of gases in
water decreases with increasing temperature.
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Concentration
• It refers to the weight or volume of the solute
present in a specified amount of the solvent or
solution.
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WAYS OF EXPRESSING
CONCENTRATIONS
• QUALITATIVELY
• QUANTITATIVELY
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QUALITATIVELY
• DILUTE – a solution with a relatively small
concentration of solute.
• CONCENTRATED – a solution with a large
concentration.
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QUANTITATIVELY
• Percent by weight
• Percent by volume
• Mole fraction and Mole Percent
• Molality (m)
• Molarity (M)
• Normality
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1. Percent by Weight
100% xsolutionofweight
componentofweightweightby
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Problem 1
A solution is made by dissolving 13.5 g of glucose, C6H12O6, in a 0.1 kg water. What is the weight percentage of solute in a solution?
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Solution
But , weight of solution = weight of solute + weight of solvent = 13.5 g + 100 g = 113.5 g
100% 6126
6126 xsolutionofweight
OHCofweightOHCweightby
%9.111005.113
5.13% 6126 x
g
gOHCweightby
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Problem 2
A commercial bleaching solution contain 3.62%
weight of sodium hypochlorite, NaOCl. What
is the weight of NaOCl in a bottle containing
2500 g of bleaching solution.
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Solution
NaOClgg
NaOClweight
xg
NaOClweight
xsolutionofweight
NaOClweightNaOClweightby
5.90100
)2500)(62.3(
1002500
62.3
100%
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2. Percent by Volume
100% xsolutionofvolume
soluteofvolumevolumeby
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Sample Problem
What is % by volume of 12 ml of alcohol in a 100 ml of wine?
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Solution
%12100100
12%
100%
xml
mlvolumeby
xsolutionofvolume
alcoholofvolumevolumeby
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3. Mole Fraction and Mole Percent
100%
100%
xfractionmoleMole
or
xsolutionofmoles
componentofmolesMole
solutionofmoles
componentofmolescomponentoffractionMole
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NOTE:
The sum of the mole fractions of solute and
solvent must be equal to one (1)
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Sample Problem
Calculate the mole fractions and mole %
of ethyl alcohol, C2H5OH, and water,
H2O, in a solution by dissolving 13.8 g
of alcohol in 27 g of water.
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Solution
molessolventofmolesoluteofmolesolutionofMoles
molemoleg
g
OHMW
OHweightOHMole
molemoleg
g
OHHCMW
OHHCweightOHHCMole
8.15.13.0
5.1/18
27
30.0/46
8.13
2
22
52
5252
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Solution
%3.83)100)(833.0(100%
%7.16)100)(167.0(100%
833.0167.011
833.08.1
5.1
167.08.1
3.0
22
5252
522
2
52
52
52
xOHoffractionMoleOHMole
xOHHCoffractionMoleOHHCMole
OHHCoffractionMoleOHoffractionMole
or
moles
molesOHoffractionMole
moles
molesOHHCoffractionMole
solutionofmoles
OHHCofmolesOHHCoffractionMole
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solventoframlogki
soluteofmolesmMolality )(
4. Molality (m)
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Problem
Calculate the molality of a solution
made by dissolving 262 g of ethylene
glycol, C2H6O2, in 8000 g of water.
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Solution
mkg
molem
molemoleg
g
OHCMW
OHCgOHCofmoles
OHoframlogki
OHCofmolesmMolality
5275.08
22.4
22.4/62
262
)(
262
262262
2
262
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5. Molarity (M)
solutionofLiter
soluteofmolesMMolarity )(
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Problem
Calculate the molarity of a solution made by dissolving 4.0 g of calcium bromide, CaBr2, in enough water to give 200 ml of solution.
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Solution
ML
moleM
molemoleg
g
CaBrMW
CaBrofweightCaBrofmoles
solutionofLiter
CaBrofmolesMMolarity
1.02.0
02.0
02.0/200
0.4
)(
2
22
2
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6. Normality (N)
soluteofweightequivalent
soluteofweightsoluteofequivalent
solutionofLiter
soluteofequivalentNNormality
)(
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Equivalent Weight
• weight of substances that are equivalent to one
another in chemical reaction.
• The equivalent of an oxidizing agent or
reducing agent is the weight of the substance
required to gain or lose 1 mole of electron.
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Equivalent Weight
Equivalent Weight = molecular weight (MW)
f (factor)
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Reacting Capacity of Solute
ACID BASE SALT
f
Total number of H+
replaceable
Total number of replaceable
OH-
Total (+/-) charge
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Sample Problem 1
Calculate the equivalents of:
(a) 20 g of H2SO4
(b) 50 g of Mg(OH)2
(c) 30 g of NaOH
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Sample Problem 2
Calculate the normality of a Na2CO3 solution
containing 13.75 grams in 125 ml solution.
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Solution
N = equivalent of solute
Liter of solution
equiv. of Na2CO3 = wt./equiv. wts.
= 13.75 g .
(106 g/mol)/(2 equiv/mol)
= 0.259 equiv/mol
N = (0.259)
(0.125)
N = 2.075 N
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RELATIONSHIP BETWEEN NORMALITY AND
MOLARITY
N = n solute x f
liter of solution
N = molarity (M) x f
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Sample Problem 5
Find:
(a) The molarity (M) of 0.12 N H3PO4
(b) The normality (N) of 0.25 M NaOH
(c) The normality (N) of 2.50 M HCl
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Dilution
• It refers to the reduction of concentration of a
solution.
• The most important thing to remember
concerning dilution is that you are only adding
solvent.
• You are not adding solute when you dilute.
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Dilution
Therefore:
moles of solute before dilution = moles of solute after dilution
or
Number of millimoles of solute before = millimoles of solute after dilution
or
Number of grams of solute before dilution = number of grams of solute after dilution
Since the definition for Molarity is:
Molarity = moles of solute / volume of solution in liters
Solving for moles of solute gives:
moles of solute = M x V of soln in liters
or
millimoles of solute = M x V of solution in ml Engr. Yvonne Ligaya F. Musico
Dilution
If
Moles of solute before dilution = moles of solute after dilution
Then
M x V in liters before dilution = M x V in liters after dilution
or
M1V1 = M2V2
where
M1 = Molarity before dilution
V1 = volume of solution before dilution
M2 = Molarity of solution after dilution
V2 = Volume of solution after dilution Engr. Yvonne Ligaya F. Musico
Dilution
Actually one can use Molarity or mass % as the
concentration term so you could have the
following alternative where mass % is used:
mass % x grams solution before dilution = mass % x grams of solution after dilution
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Sample Problem 1
How would you prepare 500 ml of 3 M HCl
using 6 M HCl from the stockroom.
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Solution
HClMofmlV
M
mlMV
mlMVM
VMVM
6250
6
5003
50036
1
1
1
2211
1. Determine the volume of 6M of HCl
to use applying the dilution equation
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Solution
2. Determine the amount of water to be added to 6M
HCl
Since the total volume after dilution is 500 ml and the
volume of 6M to use is 250 ml then:
Volume of water = 500 - 250 = 250 ml of water
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Sample Problem 2
A chemist starts with 50.0 mL of a 0.40 M NaCl
solution and dilutes it to 1000. mL. What is the
concentration of NaCl in the new solution?
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Solution
MM
mlMM
mlMmlM
VMVM
02.0
1000
504.
1000504.0
2
2
2
2211
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Sample Problem 3
How would you prepare 800 grams of a 3%
Hydrogen Peroxide solution using 10% H2O2
solution
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Solution
1. Determine the mass of 10% H2O2 to be used using the dilution equation mass of H2O2 before dilution = mass of H2O2 after dilution
mass % x mass of solution before dilution = mass % x mass of solution after dilution
10 ( mass of solution before) = (3) (800)
mass of solution before = (3) (800)
10
mass of solution before = 240 grams
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Solution
2. Determine the mass of water to be added
Total mass of 3% = mass of 10% + mass of water added
mass of water to be added = 800 - 240
mass of water to be added = 560 g
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Sample Problem 4
A chemist wants to make 500. mL of 0.050 M
HCl by diluting a 6.0 M HCl solution. How
much of that solution should be used?
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Solution
HClMofmlV
M
mlMV
mlMVM
VMVM
0.617.4
0.6
50005.0
500050.00.6
1
1
1
2211
Engr. Yvonne Ligaya F. Musico
Practice Exercise
1. How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl solution?
2. How would you prepare 1000 ml of a 5% glucose solution using a 20% glucose solution. How much 20% glucose and how much water to be mixed?
3. What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M KOH to 250 mL?
4. What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL of 4.2 M KOH?
5. How much 0.20 M glucose solution can be made from 50. mL of 0.50 M glucose solution?
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Thank You for Listening
Engr. Yvonne Ligaya F. Musico