Transcript Solutions - Moreau Catholic High School

Solutions
Chpt 16
Mixtures
• In the real world, and in particular in the
laboratory, we generally encounter matter
which consists of more than one component
• i.e., a mixture.
Solutions
• By definition a solution is a homogeneous
mixture, that is, it exists in ONE PHASE
• (there are no apparent boundaries or interfaces
in a solution). Simple binary solutions consist of
two components:
• solvent: the component present in larger
amounts (the "dissolver")
• solute: the component present in smaller
amounts (the "dissolvee”)
Solutions
• More complex solutions contain one solvent
and several solutes.
Solutions
• There are as many types of solutions as there
are possible combinations of the three
ordinary states of matter.
• By far the most common solution encountered
in the lab is the aqueous solution in which
water is the solvent and some solid, other
liquid or gas is the solute.
Solutions
• Unlike pure substances, the properties of
solutions depend to some extent on the
relative proportions of their components, and
are not constant.
• So it is important to specify the composition
of solutions.
Solutions
• There are three semi-quantitative categories
of solution concentration:
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* unsaturated: able to dissolve more solute
* saturated: no more solute will dissolve
* super-saturated: contains more solute
than will dissolve under ordinary conditions
Unsaturated
Unsaturated
• The vast majority of solutions we use in the lab are
unsaturated.
• We recognize them by their visual clarity (you can
always see through a solution).
• The fact that we use different concentrations of the
same solutions (3 M vs. 6 M HCl) also indicates that it is
possible to dissolve more solute into the solvent water.
• When we say a compound is "soluble" in water we are
saying that it will form an unsaturated solution with
water in reasonable proportions.
Saturated
Saturated
• When we say a compound is "insoluble" in
water we are saying that it will not dissolve in
water in reasonable proportions.
• If such compounds form in water a precipitate
is the result.
• When solutions of silver nitrate and sodium
chromate are mixed, insoluble silver chromate
forms.
Super saturated
• A super-saturated solution is not "stable".
Since it contains more dissolved solute per
solvent volume than a saturated solution it
readily reverts to the saturated condition
given the least excuse--mechanical shock,
changes in temperature, etc.
Super saturated
• The most common super-saturated solution
you encounter in your everyday life is soda
water.
• Sealed at high pressure in the factory, carbon
dioxide gas is forced into the water at
concentrations that exceed its solubility at
ordinary room pressure.
Super saturated
• But under the conditions in the sealed bottle
the solution is merely saturated. When the
container is opened there is a rapid reversion
to atmospheric pressure and under those
conditions there is suddenly more carbon
dioxide in the solution than it can hold. At that
moment the solution is super-saturated for
the ambient conditions and gas is released as
the solution reverts to a saturated state at
atmospheric pressure.
Concentration
• These descriptions of solutions give us some
indication of what we can expect, but they do
not help to specify quantitative behavior of
solutions.
• To do that we need what is normally called
"concentration".
Concentration
• The concentration of a solution is a measure
of the amount of solute that is dissolved in a
given quantity of water
• A dilute solution contains a small amount of
solute
• A concentrated solution contains a large
amount of solute
Concentration
• There are many ways to express the
concentrations of solutions, each having its
own particular applications.
• We will look at only three which have the
widest application and will allow us to predict
some of the simpler behaviors of solutions.
1. Molarity (M)
• Molarity is the most common way of
expressing solution concentration
• Molarity (M) = moles of solute (mol)
Liters of solution (L)
The unit is M because mol/L = M!!!
Molarity
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Also know as molar concentration
Eg 6M HCl is read as
“6 molar hydrochloric acid solution”
And means that x moles of HCl has been
dissolved in y Liters of water
Calculating M
• What is the molarity of a solution made from
10.0g of sulfuric acid that is dissolved in a
500mL of water.
What is the molarity of a solution made from 10.0g of
sulfuric acid that is dissolved in a 500mL of water.
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Step 1 – write the equation
M = moles/liters
Step 2 – turn mass to moles
Moles = mass/molar mass = 10.0g/98.1g =
0.101936799 moles
• Step 3 – turn mL into L
• 500 mL | 1 L
= 0.5 L
|1000 mL
What is the molarity of a solution made from 10.0g of
sulfuric acid that is dissolved in a 500mL of water.
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Step 4 – plug in the numbers
M = moles/L
M = 0.101936799 moles/0.5L
M = 0.203873598
0.204 M – a reasonable number of sig figs (3)
Dissolving
• So just what does it mean to "dissolve"?
• In the most general sense dissolving means
that two or more phases merge into one and
all of the various components become evenly
distributed in the mix
• -in other words, the mixture is homogeneous.
• In many cases this change is clearly visible
Dissolving
• Exactly how dissolving happens will vary
depending on the nature of the components.
• Since the most common type of solution we
use in the lab is a mixture of a solid in water, a
closer look at that situation may shed some
light on the general process.
Dilutions
• Diluting a solution reduces the number of moles of
solute per unit volume, but the total number of
moles of solute in solution does not change.
16.2
Making Dilutions
• Making a Dilute Solution
Dilutions
• The total number of moles of solute
remains unchanged upon dilution, so you
can write this equation.
Moles of solute = M1 x V1 = M2 x V2
• M1 and V1 are the molarity and
volume of the initial solution, and M2
and V2 are the molarity and volume
of the diluted solution.
16.2
Making Dilutions
• To prepare 100 ml of 0.40M MgSO4 from a stock
solution of 2.0M MgSO4, a student first measures 20 mL
of the stock solution with a 20-mL pipet.
• Moles of solute = M1 x V1 = M2 x V2
• M1 = 2.0M
• M2 = 0.40M
V1 = 20 mL
V2 = 100 mL
• 2.0M x 0.02L = 0.40M x 0.01L
• 0.04 = 0.04
• Remember M = mol/L
16.2
Making Dilutions
• She then transfers the 20 mL to a 100-mL volumetric
flask.
16.2
Making Dilutions
• Finally she carefully adds water to the mark to make
100 mL of solution.
2. Molality (m)
• The second expression of concentration
• Number of moles of solute dissolved in 1 kg
(1000g) of solvent.
• Molality = moles of solute
•
kilogram of solvent
• Units = mol/kg = m
Molality
• Compare molality and molarity
• Molality = moles of solute per kg of solvent
• Molarity = moles of solute per liter of solvent
• If water is the solvent then 1kg (or 1000g)
equals a volume of 1L or 1000 mL.
Sample calculation
• What is the molality of a solution made using
6.90 g of solid NaCl dissolved in 46.21 g of
water.
• Step 1 – turn mass of solute into moles of
solute
• Moles = mass/molar mass = 6.90g/58.5g/mol
= 0.118 mol
What is the molality of a solution made using 6.90 g of
solid NaCl dissolved in 46.21 g of water?
• Step 2 – turn grams of solute into kg of solute
• 1kg = 1000g
• 46.21g/1000g = 0.04621 kg
What is the molality of a solution made using 6.90 g of
solid NaCl dissolved in 46.21 g of water?
• Step 3 – plug numbers into molarity equation
m= moles of solute/kg of solvent
m= 0.118 mol / 0.04621 kg = 2.55m
Remember m = moles/kg
3. Mole fraction
• The 3rd expression of concentration
• The ratio of moles of solute to total number of
moles of solvent + solute.
• Or – how many moles of solute are present of
every 1 mol of total solution.
Mole fraction
• Mole fraction=
mol A
•
mol A + mol B
• Where A = solute and B = solvent
or:
Mole fraction =
mol solvent
mol solvent + mol solute
Lab time 
• P497 of text book.
• Procedure and analyze sections only
• Do not do the “You’re the chemist” section.
Freezing-point depression and
Boiling-point elevation
• Freezing-point depression ∆Tf
• Boiling-point elevation ∆Tb
• both are directly proportional to m (the molal
concentration)
Freezing-point depression and
Boiling-point elevation
• ∆Tf = the difference between the freezing
point of the solution and the freezing point of
the pure solvent.
• ∆Tb = the difference between the boiling point
of the solution and the boiling point of the
pure solvent
Freezing-point depression and
Boiling-point elevation
• We use constants to show the proportionality
• ∆Tf = Kf x m
∆Tb = Kb x m
• Kf = molal freezing-point depression constant
• = change in freezing-point for a 1-molal solution
of a nonvolatile molecular solute
• Kb = molal boiling-point elevation constant
• = change in boiling-point for a 1-molal solution of
a nonvolatile molecular solute
Freezing-point depression and
Boiling-point elevation
• The units for Kb and Kf are ˚C/m
• For ionic compound both freezing-point
depression and boiling-point elevation depend
on the number of ions produced by each
formula unit
Freezing-point depression ∆Tf
• To calculate:
• ∆Tf = Kf x m
• Remember it’s the solvent we are interested
in, and an aqueous solution is one where a
solute has been dissolved in water, so water is
often the solvent
• Kf for water = 1.86 ˚C/m
Calculate the ∆Tf of an aqueous solution of 10.0g of
glucose in 50.0g of water
• Step 1 – calculate mols of glucose (C6H12O6)
• Mols = mass/ molar mass = 10.0g/180 g/mol
• Mols = 0.0556 mols
• Step 2 – calculate m
• m = mols of solute/ kg of solvent
• m = 0.0556 mol / 0.05 kg = 1.11m
Calculate the ∆Tf of an aqueous solution of 10.0g of
glucose in 50.0g of water
• Step 3 ∆Tf = Kf x m
∆Tf = 1.86 ˚C/m x 1.11m
∆Tf = 2.06˚C
Boiling-point elevation ∆Tb
• To calculate:
• ∆Tb = Kb x m
• Remember it’s the solvent we are interested
in, and an aqueous solution is one where a
solute has been dissolved in water, so water is
often the solvent
• Kb for water = 0.512 ˚C/m
Calculate the ∆Tb of a solution containing 1.25 mol of
calcium chloride dissolved in 1400 g of water
• Step 1 – solute already in mols  (but if were
given in grams we’d have to turn it to mols)
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Step 2 – calculate m
m = mols of solute / kg of solvent
m = 1.25 mol CaCl2 / 1.4 kg
m = 0.893 m CaCl2
Hold on!!! – CaCl2 is an ionic compound……
Calculate the ∆Tb of a solution containing 1.25 mol of
calcium chloride dissolved in 1400 g of water
• CaCl2 will dissociate into Ca2+ and 2Cl- ions
when it is dissolved in water so……
• Step 2a (only for ionic compounds) – multiply
molality by the number of ion particles
• (0.893 m) (3) = 2.68m
Calculate the ∆Tb of a solution containing 1.25 mol of
calcium chloride dissolved in 1400 g of water
• Step 3 ∆Tb = Kb x m
• ∆Tb = (0.512 ˚C/m) (2.68 m)
• ∆Tb = 1.37˚C
• This means the new boiling point of the
solution will be 100˚C + 1.37˚C = 137˚C