Transcript Solutions Chemistry I

Solutions
Solutions
Chemistry I
Definitions

A solution is a homogeneous mixture

A solute is dissolved in a solvent.
– solute is the substance being dissolved
– solvent is the liquid in which the solute is dissolved
– an aqueous solution has water as solvent
Dissolution of Solid Solute
What are the driving forces which cause solutes to
dissolve to form solutions?
1. Covalent solutes dissolve by Hydrogen-bonding to water
2. Ionic solutes dissolve by dissociation into their ions.
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Solution and Concentration
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3 ways of expressing concentration
–Molarity(M):
moles solute / Liter solution
–Molality (m) - moles solute / Kg solvent
–Mole Fraction(A) - moles solute / total moles solution
Concentration: Molarity Example
If 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of
solution, what is the molarity of KMnO4?
As is almost always the
case, the first step is to
convert the mass of
material to moles.
0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4
158.0 g KMnO4
Now that the number of moles of substance is known, this can be
combined with the volume of solution — which must be in liters
— to give the molarity. Because 250. mL is equivalent to 0.250 L .
Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M
0.250 L solution
Two Other Concentration Units
MOLALITY, m
mol solute
m of solution =
kilograms solvent
% by mass
% by mass =
grams solute
grams solution
Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in
250. g of H2O. Calculate molality and % by
mass of ethylene glycol.
Ethylene Glycol
Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O.
Calculate m & % of ethylene glycol (by mass).
Calculate molality
1.00 mol glycol
conc (molality) =
 4.00 molal
0.250 kg H2O
Calculate weight %
62.1 g
%glycol =
x 100% = 19.9%
62.1 g + 250. g
Try this molality problem

25.0 g of NaCl is dissolved in 5000. mL of water.
Find the molality (m) of the resulting solution.
m = mol solute / kg solvent
25 g NaCl
1 mol NaCl
= 0.427 mol NaCl
58.5 g NaCl
Since the density of water is 1 g/mL, 5000 mL = 5000 g, which
is 5 kg
0.427 mol NaCl
= 0.0854 m salt water
5 kg water
Dilution
When a solution is diluted, solvent is added to lower its
concentration. A stock solution refers to a more concentrated
solution. When water is added to a stock solution, its concentration
lowers.
The amount of solute remains constant before and after the dilution:
moles BEFORE = moles AFTER
M1V1 = M2V2
Click here for dilution movie
Making a Dilute Solution
Making a dilute solution example
Suppose you have 0.500
M sucrose stock solution.
How do you prepare 250
mL of 0.348 M sucrose
solution ?
Concentration
0.500 M
Sucrose
250 mL of 0.348 M sucrose
A bottle of 0.500 M standard
sucrose stock solution is in the
lab.
Give precise instructions to
your assistant on how to use
the stock solution to prepare
250.0 mL of a 0.348 M sucrose
solution.
Answer

M1V1 = M2V2

(0.500M)V1 = (0.348M)(0.250L)

V1 = (0.348M)(0.250L)
0.500M

V1 = 0.174L or 174 mL

Answer: Obtain 174 mL from the 0.500M stock
solution. Place this into a 250. mL flask. Add water to
the stock solution (76.0 mL).
Miscible solutions
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Miscible –
means the two
liquids dissolve
into each other.
Ex. Water &
ethanol
Immiscible – the
two liquids do
not dissolve into
each other. Ex.
Water and oil
Immiscible liquids
Strong electrolytes

A strong
electrolyte is a
solute that will
dissociate
completely into
ions in an
aqueous
solution.
Weak electrolytes

A weak electrolyte
partially dissociates.
Most solute particles
stay intact and only
a few dissociate into
ions. Acetic acid, as
shown on the left, is
an example of a
weak electrolyte.
Nonelectrolytes

A non-electrolyte is
a solute that does
not dissociate into
ions. The solute
particle stays intact.
An example is
sugar. Most
covalent solutes are
non-electrolytes
Like dissolves like. Polar solutes will
dissolve in polar solvents. Nonpolar solutes
will dissolve in nonpolar solvents.
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Ethylene glycol has polar properties so it
will dissolve in water.
Oil, a nonpolar substance, will not
dissolve in water, a polar substance.
3 Stages of Solution Process

Separation of Solute
– must overcome inter-molecular forces (IMF) or ion-ion
attractions in solute
– requires energy, ENDOTHERMIC ( + DH)

Separation of Solvent
– must overcome IMF of solvent particles
– requires energy, ENDOTHERMIC (+ DH)

Interaction of Solute & Solvent
– attractive bonds form between solute particles and solvent
particles
– “Solvation” or “Hydration” (where water = solvent)
– releases energy, EXOTHERMIC (- DH)
Dissolution at the molecular level?
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Consider the dissolution of NaOH in H2O
Heat of solution
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the heat evolved or absorbed when one mole of a
substance is dissolved in a large volume of a solvent.
∆H = q
mol
Remember: q = (mass)(specific heat)(∆T)
Heat of Solution
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If 0.860 g copper(II) bromide is dissolved in 100.0 mL
of water, the temperature changes from 23.10°C to
23.41°C. What is the heat of solution for copper(II)
bromide? Assume the specific heat of water, 4.184
J/g•°C.
The reaction is:
– CuBr2 (s) → Cu+2 (aq) + 2Br – (aq)

Was this an endothermic or exothermic reaction?
Answer
To determine the amount of heat absorbed (q), use the heat equation.
Remember that the heat energy is absorbed by the entire solution that
includes the solvent as well as the solute. Recall that the density of
water is 1.00 g/mL, so 100.0 mL of water is also 100.0 g water.
 q = (mass)(specific heat)(T)
 q = (100.860 g)(4.184 J/g•°C)(23.41 °C – 23.10 °C)
 q = (100.860)(4.184)(0.31)
 q = 130.8 J
 Since the temperature increases (exothermic), the ∆H value will be
negative.
 mol CuBr2 = 0.860 g x 1 mol
= 0.00385 mol so
223.35 g
 ∆H = –(130.8 J)
= 33974 J/mol
0.00385 mol
33974 J/mol x 1 kJ /1000 J = –33.974 kJ/mol
 ∆ Hsoln = –3.4 x 104 J/mol or –34kJ/mol (2 sig figs)

Heat of solution Ex. #2

What is the heat of solution and the dissolution
reaction if when 1.893 g lithium fluoride dissolves in
250.0 mL water, the temperature changes from
23.69 °C to 19.59 °C?

The reaction is LiF (s) → Li+1(aq) + F-1(aq)
Is this an endothermic or exothermic reaction?

Answer
The moles of lithium fluoride that dissolved are
 1.893 g LiF x 1 mol
= 0.07298 mol LiF
25.94 g
The heat evolved is determined from
 q = (mass solution)(specific heat water)(T)
 q = (251.9 g)(4.184 J/g•°C)(23.69 °C – 19.59 °C)
 q = (251.9)(4.184)(4.10)
 q = 4321 J
 The heat of solution will be positive because the temperature of the
solution decreased (endothermic).
 ∆H = +q/ mol = –(4321 J)/ 0.07298 mol = +59,211 J/mol
 = +5.92 x 104 J/mol OR
 +5.92 x 104 J/mol x 1 kJ
= +59.2 kJ/mol
1000 J
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Rate of solution

Rate of solution is how quickly a solute dissolves in
a solvent. Factors determining the rate of solution
are:
– Surface Area: When a solute dissolves, only the surface of
the solute comes in contact with the solvent. Therefore, the
more the surface area of the solute, the faster it dissolves.
– Stirring: When you're dealing with solid and liquid solutes,
stirring brings fresh parts of the solvent into contact with the
solute and particles are forced to connect.
– Temperature: Increasing the temperature also generally
increases the amount of solute the solvent can hold (solid
and liquid solutes but not gas solutes).
Factors Affecting Solubility
1. Nature of Solute / Solvent. - Like dissolves like (IMF)
2. Temperature i) Solids/Liquids- Solubility increases with Temperature
Increase K.E. increases motion and collision between solute / solvent.
ii) gas - Solubility decreases with Temperature
Increase K.E. result in gas escaping to atmosphere.
3. Pressure Factor i) Solids/Liquids - Very little effect
Solids and Liquids are already close together, extra pressure will not
increase solubility.
ii) gas - Solubility increases with Pressure.
Increase pressure squeezes gas solute into solvent.
The effect of pressure on gas solubility.
Solubility Curves
Solubilities of
several ionic solid
as a function of
temperature. MOST
salts have greater
solubility in hot
water.
A few salts have
negative heat of
solution,
(exothermic
process) and they
become less
soluble with
increasing
temperature.
Solubility curves
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1.
2.
3.
Each point on any curve represents a saturated
solution. A saturated solution cannot dissolve any
more solute at that temperature.
An unsaturated solution will dissolve more solute at a
certain temperature until it becomes saturated.
The solubility of KNO3 is 30 g/100 g of water at 20 Co.
How many more grams of solute must be added to
keep the solution saturated at 50 Co.
How many grams of Ce2(SO4)3 will precipitate out of
solution when it is cooled from 70 Co to 10 Co?
(assume 100 g of solvent)
What mass of KClO3 is required to saturate 50 g of
water at 70 Co?
Definitions
SUPERSATURATED SOLUTIONS
contain more solute than is possible
to be dissolved
Supersaturated solutions are unstable.
The supersaturation is only
temporary, and usually
accomplished in one of two ways:
1. Warm the solvent so that it will
dissolve more, then cool the
solution
2. Evaporate some of the solvent
carefully so that the solute does not
solidify and come out of solution.
Click here to
watch movie on
supersaturated
solutions
Temperature & the Solubility of Gases
The solubility of gases DECREASES at higher temperatures since the
Intermolecular forces can not be formed between the gas and the solvent.
Pressure and Soft Drinks
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Soft drinks contain “carbonated
water” – water with dissolved
carbon dioxide gas.
The drinks are bottled with a CO2
pressure greater than 1 atm.
When the bottle is opened, the
pressure of CO2 decreases and
the solubility of CO2 also
decreases.
Therefore, bubbles of CO2
escape from solution.
Colligative Properties
Dissolving solute in pure liquid will change all physical
properties of liquid, Density, Vapor Pressure, Boiling
Point, & Freezing Point
Colligative Properties are properties of a liquid that
change when a solute is added.
The magnitude of the change depends on the number
of solute particles in the solution, NOT on the identity
of the solute particles.
Normal Boiling Process
Extension of vapor pressure concept:
Normal Boiling Point: BP of Substance @ 1atm
When solute is added, BP > Normal BP
Boiling point is elevated when solute inhibits solvent from
escaping.
The magnitude of the boiling-point elevation is proportional to
the number of solute particles dissolved in the solvent.
Elevation of Boiling point occurs in
the beaker on the right.
Boiling Point Elevation
DTb = (Tb -Tb°) = i ·m · kb
Where, DTb = BP. Elevation
Tb = BP of solvent in solution
Tb° = BP of pure solvent
m = molality ,
kb = BP Constant
i = Van’t Hoff factor
Some Boiling Point Elevation and Freezing Point Depression Constants
Solvent
Normal bp (°C)
pure solvent
Kb
(°C/m)
Normal fp (°C)
pure solvent
Kf
(°C/m)
Water
Benzene
Camphor
Chloroform
100.00
80.10
207
61.70
+0.5121
+2.53
+5.611
+3.63
0.0
5.50
179.75
- 63.5
1.86
4.90
39.7
4.70
(CH3Cl)
van’t Hoff factor (i)
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The van’t Hoff factor is a whole number integer that
has the value of 1, 2, 3, etc. It is never zero or a
negative number.
This number represents the number of particles
based on the dissociation of a solute in water.
The van’t Hoff factor for any covalent solute is always
1. Covalent compounds are always nonmetals only.
The van’t Hoff factor for any ionic solute is equal to
the number of ions in that compound. Ionic
compounds always involve 1 metal.
van’t Hoff factor examples
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Determine the number of particles that are
represented in each of the following substances:
1
1) C6H12O6
2
2) NaCl
3) Al2(SO4)3
5
4) Ca(C2H3O2)2 3
5) CCl4
1
6) CaCl2 3
7) Al(NO3)3 4
8) C7H6O2
1
Note: the van’t Hoff factor is 2
x2
Ex. 2 Bioling pt. elevation

What is the boiling point elevation when 11.4 g of
ammonia (NH3) is dissolved in 200g of water? Kb for
water is 0.52 Co/m.
Solution:

First find moles of solute: 11.4 g
= 0.669 mol
17.04 g/mol
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Find molality of solution:
– Determine the kg of solvent: 200 g x 1kg
= 0.2000 kg
1000 g
– Molality: 0.669 mol/0.2000 kg = 3.35 mol/kg
– ∆T = Kf x m x i
– ∆T = 0.52 m/Co x 3.35 m x 1
– ∆T = 3.37 Co so the boiling pt of this solution will be 103.37 Co
Freezing Point Depression
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Normal Freezing Point: FP of Substance @ 1atm
When solute is added, FP < Normal FP
FP is depressed when solute inhibits solvent from crystallizing.
When solution freezes the solid form is almost
always pure.
Solute particles does not fit into the crystal lattice
of the solvent because of the differences in size.
The solute essentially remains in solution and
blocks other solvent from fitting into the crystal
lattice during the freezing process.
Change in Freezing Point
Pure water
Ethylene glycol/water
solution
The freezing point of a solution is LOWER than
that of the pure solvent
x1
Freezing pt. depression problem.

How many grams of pyrozole (C3H4N2) must be
added to 451 g of benzene to lower the freezing point
by 5.00 Co? Kf for benzene is 5.12 Co/m.
Answer

First analyze what you are given:
– Benzene is the solvent. So,
– 451 g x 1 kg = 0.451 kg
1000 g
–
–
–
–
∆T = 5.00 Co
Kf = 5.12 Co/m
i = 1 since solute is covalent
Substitute your givens into the equation:
∆T = Kf x i x m
5.00 Co = 5.12 Co/m x 1x m
m = 0.977mol/kg
Solve for mass using the equation for molality:
m = mol/kg so 0.977 mol/kg x 0.451 kg = 0.954 mol
Answer
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From the previous slide mol = 0.954 mol
Remember mol = grams/molar mass. To solve for
mass in grams multiply mol and molar mass.
0.954 mol x 68.09 g/mol = 65.0 grams