Chapter 18
Solutions
Liquids
• Miscible means that two liquids can
dissolve in each other
–water and antifreeze, water and ethanol
• Partially miscible- slightly
–water and ether
• Immiscible means they can’t
–oil and vinegar
• In general the principle concerning
whether substances will mix is "Like
dissolves Like";
that is, polar (and ionic) substances
will dissolve in polar solvents and
non-polar substances will dissolve in
non-polar solvents,
while non-polar and polar substances
will not mix.
+
Solubility in Water :
2The simple rule to remember is ‘like dissolves like
+
Water is a polar solvent and will dissolve polar molecules and
substances that contain charged particles.
Substances that dissolve in H2O are said to be soluble
E.g. Sugar, ethanol which are polar
most ionic compounds
Site of polarity
can form H-bonding
sucrose
Water is polar, the different ends are attracted to the charged ions.
+
2-
+
Cl-
Na+
Substances that don’t dissolve are called insoluble
E.g. Petroleum (crude oil), which are non-polar
So if you want to dissolve grease which is non-polar, you need to
use a non-polar solvent.
Petroleum in a non-polar organic molecule
Solution formation
• Nature of the solute and the solvent
–Whether a substance will dissolve
–How much will dissolve
• Factors determining rate of solution...
–stirred or shaken (agitation)
–particles are made smaller
–temperature is increased
Temperature and Solutions
• Higher temperature makes the
molecules of the solvent move around
faster and contact the solute harder
and more often.
–Speeds up dissolving.
• Usually increases the amount that will
dissolve (exception is gases)
How Much?
• Solubility- The maximum amount of
substance that will dissolve at a specific
temperature (g solute/100 g solvent)
• Saturated solution- Contains the
maximum amount of solute dissolved
• Unsaturated solution- Can still dissolve
more solute
• Supersaturated- solution that is holding
more than it theoretically can; seed
crystal will make it come out;
In a saturated solution an equilibrium is reached between
dissolving and recrystallization.
Concentration is...
• a measure of the amount of solute
dissolved in a given quantity of solvent
• A concentrated solution has a large
amount of solute
• A dilute solution has a small amount of
solute
–thus, only qualitative descriptions
• But, there are ways to express solution
concentration quantitatively
• Concentration
= amount of solute
amount of solvent (or solution)
Molarity
(M) =
amount of solute in moles
amount of solution in liters
What is the molarity of a solution
with 2.0 moles of NaCl in 250 mL
of solution?
• Molarity= moles/liter
•
= 2.0 mol/ .250 L
•
= 8.0 M
Calculating molarity
• Calculate the molarity of Na2SO4 (s) if you
have 4.2 moles in 500. L of solution.
• Molarity = moles/liter
•
= 4.2 mol/ 500. L
•
= 8.4 x 10-3 M
• Could also have mol/l or molar as units
• How many moles of NaCl are
needed to make 6.0 L of a 0.75 M
NaCl solution?
• Moles = (0.75 mol/l)(6.0 l)
•
= 4.5 mol
Calculate the mass of Na2 SO4 (s) in
1.6 L of a 2.5 M solution.
moles
MxV
Then
Find
mass
Mole x
molar mass
• Molarity = moles/volume
• Moles = (molarity)(volume)
•
= ( 2.5 mol/l) ( 1.6 l)
•
= 4.0 mol
mass
• Need molar mass: 142.02 g/mol
Mass = 568.08g
= 570 g
Making solutions
• 10.3 g of NaCl are dissolved in a
small amount of water, then diluted to
250 mL. What is the concentration?
• How many grams of sugar are
needed to make 125 mL of a 0.50 M
C6H12O6 solution?
Alternate Measures of
Concentration
• Molar concentration is the accepted
method of determining the amount of
substance dissolved in a solvent.
• parts per million (ppm),
• parts per billion (ppb)
• parts per trillion (ppt).
• Part per million
• ppm = mass of solute x 106
mass of solution
• Part per billion
• ppb = mass of solute x 10 9
mass of solution
• Part per trillion
ppt =
mass of solute x 10 12
mass of solution
Molality
moles of solute
molality (m) 
kg of solvent
Molarity = moles of solute
kg of solvent
(independent of temperature)
1 kg of water = 1 liter of water
Molality
• - What is the molality when 0.750 mol is
dissolved in 2.50 L of solvent?
– Molality
= 0.750 mol
2.50 kg
=
0.300m
Molality
• Ex- Suppose you had 58.44 grams of
NaCl and you dissolved it in exactly 2.00
kg of pure water (the solvent). What would
be the molality of the solution?
mol
m
kg
• Step One: convert grams to moles.
• Step Two: divide moles by kg of solvent to
get molality.
• 58.44 grams/mol is the molecular weight
of NaCl
•
58.44g x 1 mol = 1.00 mol.
•
58.44 g
Molality = 1.00 mol
2.0 kg
=0.500 mol/kg (or 0.500 m). or 0.500-molal.
• 1) Calculate the molality of 25.0
grams of KBr dissolved in 750.0 mL
pure water.
• 2) 80.0 grams of glucose (C6 H12 O6,
mol. wt = 180. g/mol) is dissolved
in1.00 kg of water. Calculate the
molality.
• 3) Calcuate the molality when 75.0
grams of MgCl2 is dissolved in 500.0
g of solvent.
1. We can describe the concentration of a solution by using the
mass percent.
Mass percent is the mass of solute present in a given mass of
solution.
Mass Percent =
mass of solute
Mass of solution

100%
=
grams of solute

100%
(grams of solute) + (grams of solvent)
Example 1:
We make a solution by dissolving 1.0g of
sodium chloride in 48g of water.
What is the mass percent of the solute?
Mass Percent = 1 .0 g solute
49 g solution
= 2.0% NaCl
x100%
Example 2:
A solution is prepared by dissolving
1.00g of ethanol, C2H5OH, with 100.0g
of water.
What is the mass percent of ethanol in
this solution?
Mass Percent = 1 .00 g ethanol 100%
101 g solution
= 0.990% C2H5OH
Example 3:
Cow’s milk typically contains 4.5% by mass of the sugar
lactose, C12H22O11. Calculate the mass of lactose present in
175 g of milk.
Mass of solution (milk) = 175g
Mass percent of solute (lactose) = 4.5%
Mass Percent =
4.5% =
grams of solute
grams of solution

100%
grams of solute
175 g

100%
(4.5%)(175 g) = 0.045  175g
100%
Grams of solute = 7.9g lactose
Grams of solute =
Dilution
Adding water to a solution
M 1 V1 = M 2 V2
Dilution
• The number of moles of solute doesn’t
change if you add more solvent!
• The # moles before = the # moles after
• M1 x V 1 = M 2 x V 2
• M1 and V1 are the starting
concentration and volume.
• M2 and V2 are the final concentration
and volume.
• Stock solutions are pre-made to known
Molarity
Ex:1 - 53.4 mL of a 1.50 M solution of
NaCl is on hand, but you need some
0.800 M solution. How many mL of
0.800 M can you make?
M1V1 = M2V2
(1.50 mol/L)(53.4 mL) = (0.800 mol/L) (x)
• (1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x)
(0.800 mol/L)
(0.800 mol/L)
•
100. mL = x
• Ex:2 - 100.0 mL of 2.500 M KBr solution
is on hand. You need 0.5500 M. What is
the final volume of solution which
results?
•
M1 V1 = M2 V2
(2.500 mol/L)(100.0 mL)=(0.5500mol/L)(x)
•
454.55 mL = x
•
454.6 mL = x
• Ex: 3 - 53.4 mL of a 1.50 M (mol/l)
solution of NaCl is on hand, but you need
some 0.500 M solution. How many ml of
0.500 M can you make?
•
M1V2 = M2V2
• (1.50 mol/L) (53.4 ml) = (0.500 mol/L) (x)
•
160. ml = x
•
Cb Vb = Ca Vb
• This is the same as the dilution equation,
concentration before and volume before is
equal to the concentration after and the
volume after.
Practice
a. 2.0 L of a 0.88 M solution are diluted
to 3.8 L. What is the new molarity?
b. You have 150 mL of 6.0 M HCl. What
volume of 1.3 M HCl can you make?
c. Need 450 mL of 0.15 M NaOH. All
you have available is a 2.0 M stock
solution of NaOH. How do you make
the required solution?