Solutions 1

Solutions
Concentration

Solution concentration = how much solute dissolved in solvent

Coffee crystal = solute

Water = solvent
Liquid Coffee = solution

so a solute is dissolved in solvent to make a solution

Sodium chloride in makes salt water

Sucrose (common table sugar) in water makes sugar water
Lemon juice and sucrose (sugar) in water make lemonade

Ways to Express Concentration

1. Percent by mass (%) = (mass solute/ mass of solution) x 100

2. Mole Fraction X
XA = nA/ ntotal = moles of some solute A/ total moles in solution
And if mixture of multiple components A, B, C, … then
1= XA + XB + XC + …

3. Molality (mol/kg) = m
moles of solute/ kg of solvent = m

4. Molarity (mol/L) = M
moles of solute/ liter of solution = M

5. Normality (equiv/ L) = N
equivalents of solute/ liter of solution = N
(Normality used in acid base or redox reactions)
 Solutions 2

Use of symbols for molar mass

MW molecular weight or MM molar mass or FW formula weight are same

Example (ex) water MW = MM = FW = 18.0 g/mol

Concentration Examples
1. Percent by weight (mass)
2. Mole fraction
3. Molality
4. Molarity

Given the mixture below:

23.0g Ethanol M.W. = 46.0g/mol solute CH3CH2OH
85.0g Water M.W. = 18.0 g/mol solvent H2 O
and a total of 100 ml of ethanol/water solution or ethanol(aq)
aq means aqueous (in water)

1. Find Percent by mass (%) ethanol

= [mass ethanol solute/ ( mass of solute + mass solvent)] x 100

= (23/ (85 + 23)) x 100 = 21.3% ethanol and 78.7% water by weight

2. Find Mole Fraction of ethanol

= moles of ethanol/ total moles in solution

(23.0g ethanol) (mol/46.0g) = 0.500 mol ethanol

(85.0g water) (mol/18.0g) = 4.72 mol water

Mole fraction = 0.500/(4.72 + .50) = 0.500/5.22 = 0.096 C2H6O ethanol

0.904 H2O water

3. Find Molality (mol/kg) of ethanol solution

= moles of solute ethanol/ kg of solvent

= 0.50 mol ethanol/0.085 kg water = 5.9 (mol/kg) = 5.9 m
( solution is 5.9 molal )
 Solutions 3

4. Find Molarity (mol/L) of ethanol solution

= moles of solute/ liter of solution

= 0.50 mol/ 0.100 L solution = 5.0(mol/L) = 5.0 M (solution is 5.0 molar)

Note: In very dilute aqueous solution the molarity ~ molality

(approximately equal)
because 1 L water = 1 kg water
that is density of water is 1 kg/L so if very dilute solution 1.0 kg ~ 1.0 L

Normality
Normality = equivalents of solute/ L of solution

Acid-base reactions relates to changes in H+ or OH-

Redox reaction relates to loss or gain (transfer) of electrons

Consider these 1.0 M (molar) solutions and

note the normality may be same or more

N is equal to molarity or small multiple of molarity N = n M where n = integer

so for examples below M multiplied by integer based on H+, OH-, or e-
(in other equations n may represent number of moles:
here n is just an integer 1, 2, 3…)

1M HCl  1H+ + Cl- 1N N= 1 M

H2SO4  2H+ + SO42- 2N N=2M
H3PO4  3H+ + PO43- 3N N=3M
CH3COOH  1H+ + CH3COO- 1N N=1M

1M NaOH  Na+ + 1 OH- 1N N=1M

Ca(OH)2  Ca2+ + 2 OH- 2N N=2M

1M 3e- + Al3+  Al 3N N=3M

2e- + Cu2+  Cu 2N N=2M
1e- + Fe3+  Fe2+ 1N N=1M
 Solutions 4

1 Normal solution (1N) contains one equivalent

N is equivalent of solute/ L of solution = ( g solute/(g/equivalent ) )/ L of solution
Example:
NaOH 40 g/mol (molar mass)
40g/ equivalent (equivalent mass) since NaOH  Na+ + 1 OH-

H2SO4 98 g/mol (molar mass)

49 g/ equivalent (equivalent mass) since H2SO4  2H+ + SO42-
2H+
so
(98g/(49g/equiv))/ (1L ) = 2N (equivalent/liter)

(98g/ (98g/mol))/ (1L ) = 1M (mole/liter)

Equivalent weights (masses) are weights (masses) of substances that are

equivalent in chemical reaction
for example: 49g of H2SO4 will neutralize 40g of NaOH so these are equivalent
weights

Conversion between molarity and molality

To convert from molarity to molality directly, must know density.

Ex: What is molality of 2.00M NaCl(aq) solution with a density of 1.08 g/mL?

Assume you have 1.000L (can assume convenient amount even if not given)
then 2.00 mol of NaCl is 2.00 mol (58.5 g /mol) = 117 g NaCl

If density is 1.08 g/mL then 1.000 L = 1000 mL (1.08 g/mL) = 1080 g total mass
Water portion is 1080 g total – 117 g NaCl = 963 g H2O and so

m = mol solute / kg solvent = 2.00 mol / 0.963 kg = 2.08 m NaCl(aq)

To convert molar to molal assume 1.000 L of solution

To convert molal to molar assume 1.000 kg of solvent
 Solutions 5

Lab Applications

Below are examples of very useful solution calculations for lab and lecture work

1. Make a solution
Example (ex): How much glucose is required to prepare 200 mL of 0.150 M of
glucose?

Glucose is C6H12O6 so
Molar mass (MM) or molar weight (MW) = 6(12) + 12(1) + 6(16) = 180 g/mol

Moles = (concentration) (volume)

Mol = (conc ) (vol)
Mol = ( mol / L) (L)
notice how units give you equation since molarity M = mol/L

Moles of solute needed = (concentration)(volume)

= (0.150 mol/L)(0.200L)
= 3.00 x 10-2 mol

so mass needed is (3.00 x 10-2 mol)(180 g/mol) = 5.40g glucose

Procedure is to measure out amount needed then add to volumetric flask and add
water to dilute to mark

2. Do a Dilution

ex: How many mL of 18.0 M sulfuric acid are required

to prepare 300 mL of 1.0 M H2SO4

M1V1 = M2V2 note: moles constant since (M)(V) = mol

Before After

(18.0 mol/L)(V1) = (1.0 mol/L)(0.300L)

V1 = 0.0167 L
V1 = 16.7 ml

So add 16.7 mL of 18.0M sulfuric acid to enough water to make 300mL solution

Notice that in the above problem

can use L on both sides or mL because concentration units cancel out
(18.0 mol/L)(V1) = (1.0 mol/L)(300 mL)
V1 = [(1.0 mol/L) / (18.0 mol/L)] (300 mL)
V1 = 16.7 ml
 Solutions 6

REMEMBER to THINK UNITS and THINK EQUATIONS

in all problem based work

Henry’s Law
Concentration of gas dissolved in solution is greater
if pressure of gas above liquid is greater

Cgas = kH Pgas mol/L = (mol/L atm) (atm) or g/L = (g/ L atm) (atm)

where Cgas is concentration of gas dissolved in liquid (mol/L)

Pgas is the pressure of gas above liquid (atm)
kH is the Henry’s law constant that connects these two values (mol/L)/atm
(determine kH experimentally or look up available values in tables

A carbonated beverage is sealed under high pressure of CO2

and this causes more carbon dioxide to dissolve in water.

and so when opened with lower pressure in the air around us than in the can,
the CO2 is less soluble and bubbles out of solution.

ex: Consider a carbonated beverage bottled under 5.0 atm of pressure.

What is the concentration of CO2 when bottled under above pressure
and after opened where pressure of atmospheric CO2 is 0.00039* atm ?
Henry’s law constant kH (CO2 in water) is 0.0313 mol/ (L atm )

Calculate Cgas(mol/L) as 0.16 and 1.2 x10-5 for above pressures using
Cgas = kH Pgas

Coke goes flat (loses fizz) after opened because pressure of CO2 above liquid is
much less than when bottled and sealed.

(*note: atmospheric CO2 is about 390 ppm parts per million (ppm) and so if
exactly 1 atm pressure then move decimal 6 places to left 0.000390
 Solutions 7

Colligative Properties

Colligative properties depend only on concentration of solute

rather than the specific type of solute and include:

Vapor pressure lowering

Freezing point depression
Boiling point elevation
Osmotic pressure

Vapor pressure lowering

Raoult’s Law says

adding nonvolatile solute to solvent causes the vapor pressure of the solute to be
lower.
or Raoult’s Law equation is

Psolution = Xsolvent Posolvent

Psolution = vapor pressure of solution

Xsolvent = mole fraction of solvent in solution
Posolvent = vapor pressure of pure solvent

Since mole fraction (0 < X < 1) is more than 0 and less than 1 for a solution
then Psolution < Posolvent

ex: if vapor pressure of pure solvent is 24 torr then if mole fraction is 0.20 what is
vapor pressure?
Psolution = Xsolvent Posolvent
Psolution = (0.20) (24 torr)
Psolution = 4.8 torr note that only solvent is considered to have vapor
pressure the solute does into go into vapor phase

ex: Water at 100oC has what pressure? and if Xsolute = 0.10 then what
is new vapor pressure of water in this new mixture at 100oC?

Xsolvent = 1 - Xsolute Psolution = Xsolvent Posolvent

0.90 0.90 atm (or 684 torr)
 Solutions 8

Freezing point depression and Boiling point elevation

On phase diagram for water below can observe

the normal freezing point (liquid  solid)
and normal boiling point (liquid  gas) that occurs at 1.00 atm pressure

http://www.naturalsci.gardner-webb.edu/Faculty/vtotten/PChem/h2ophase.gif

In a solution (mixture):
the freezing point is lower than pure solvent
the boiling point is higher than pure solvent

ΔTf freezing point depression

ΔTb boiling point elevation
 Solutions 9

FP Depression and BP Elevation Equations used are:

ΔTf = Kf Cm Cm = concentration in molality (m = mol solute/kg solvent)

ΔTb= Kb Cm

water solvent constants: Kf = -1.86 oC/m (freezing point depression

constant)
Kb = 0.51 oC/m (boiling point elevation constant)

Does not matter what solute you use but use Kf and Kb values for solvent
Those would be given on exam and may need to be looked up in textbook for
homework

ex: in car radiators we add ethylene glycol to raise boiling point for summer
driving and lower freezing point in winter driving.

Ethylene glycol (antifreeze)

(http://www.inchem.org/documents/pims/chemical/pim227.htm)

So what is the freezing point of 621 g of ethylene glycol in 2000g of water?

MW = 62.1 g/mol
621g = 10.0 mol
2000g of water in radiator (2.00 kg)

Cm = 10.0 mol/ 2.00 kg = 5.00 m

ΔT = (-1.86 oC/m)(5.00m) = -9.30 oC

Note if Kf given as positive ( as it is in some tables of data)

Then have to change to negative since fp is always decreased

and since water freezes at 0.00 oC then this would be lower by

Tsolution = Tsolvent + ∆T = 0.00 + (- 9.30) = - 9.30 oC

And solution would freeze at - 9.30oC

 Solutions 10

If instead of molecular solid an ionic solid is dissolved then the effect may be
larger since we must count all the things dissolved in solution
And we use equations:

ΔTf = i Kf Cm

ΔTb = i Kb Cm where i is number of actual of effective different ions dissolved

Consider a 5.00 m solution of NaCl(aq) then 5.00 m in Na+ and 5.00 m in Cl–
so a 5.00 m solution of NaCl(aq) would have i = 2 (1 Na+ and 1 Cl– )

ΔTf = i Kf Cm

ΔT = (2) (-1.86 oC/m) (5.00m) = -18.60 oC

And for 5.00 CaCl2 (aq) since Ca2+ and Cl- and Cl- then i=3

ΔTf = i Kf Cm

ΔT = (3) (-1.86 oC/m) (5.00m) = -27.90 oC

In some problems value of i maybe given and not be an integer because there is
some clustering of ions and i is less then you would expect to calculate. If this is
the case just use the given value of i given, otherwise calculate based on number
of ions.
 Solutions 11

Osmotic Pressure

(http://www.chem.arizona.edu/~salzmanr/480a/480ants/colprop/colprop.html)

Semipermeable membrane means solute molecules or ions cannot go through

but solvent molecules can go through.

Solvent molecules such as water will go through membrane to dilute solution

unless a pressure equal to the osmotic pressure is applied to stop the flow.

Pressure needed to stop flow is: Osmotic pressure = π

 Solutions 12

Osmotic pressure flow direction

Think of solvent (normally water) going from

where there is more water (solvent side)
to where there is relatively less water (solution side)

Osmosis is process by which solvent molecules move through membrane from

more solvent (lower concentration)
into side with less solvent (higher concentration )

With the injection of large amounts of fluid such as I.V. fluids

must use isotonic solutions

Hypertonic greater concentration than cell fluid

Isotonic  same concentration as cell fluid
Hypotonic  lower concentration than cell fluid

http://www.sirinet.net/~jgjohnso/homeostasis.html

ex: Contact lens solution is made to be isotonic with corneal cells in eye

Remember solvent moves to make two sides less different

by diluting more concentrated side.
ex: two sugar solutions of diff conc observe (1.0M water | water 5.0M )
 Solutions 13

Osmotic pressure calculation

Osmotic pressure π V = n R T or π = M R T
n = moles of solute,
V = volume of solution (L)
R = gas constant ( 0.08206 L atm / mol K )
T = temperature in (K, Kelvin)
π = osmotic pressure (atm)

M = n/V (mol/L) or molarity

ex: How much glucose needed in 1.00 L solution to make the solution isotonic
with blood given that Blood (red blood cells) π = 7.7 atm and Temp=37oC or
T = (37 + 273 ) = 310K
(Glucose MM=MW= 180g/mol where MM=molar mass and MW=molar weight)

πV=nRT
(7.7 atm)(1.00 L) = (n) (.08206 L atm/ mol K) (310 K)
0.303 mol = n

or (.303 mol)(180 g/mol) = 54 g 54 g of glucose

Application of Reverse osmosis is to Purify salt water (desalination)

Force water through membrane by applying pressure above osmotic pressure
Salt will not go through membrane but only water

ex: San Diego CA http://www.sdcwa.org/issue-desal

reverse osmosis info http://en.wikipedia.org/wiki/Reverse_osmosis

Persian Gulf War in 1990s– Saudi desalination plants shut down so oil dumped
into water by Iraq would not destroy filters in reverse osmosis facilities. Reverse
osmosis is used to get pure water from ocean water and requires special filters.
 Solutions 14

Colligative Properties of Electrolyte Solution (ions in solution)

Presence of ions applies in all cases of colligative properties

ΔTf in solutions of
-1.86 1m glucose C6H12O6  1m C6H12O6 expected amount for 1m
-3.72 1m NaCl  1m Na+ 1m Cl- so 2x as much lowering
2+ -
-5.58 1m CaCl2 1m Ca 2m Cl so 3x as much lowering

In some problems value of i maybe given and not be an integer because there is
some clustering of ions and i is less then you would expect to calculate. If this is
the case just use the given value of i given, otherwise calculate based on
expected number of ions.

Ex: approximate as NaCl i = 2 CaCl2 i = 3

Colligative properties depend on amount of solute molecules or ions added to

solvent.

Count everything in solution true for osmotic pressure

so expect 1.0 M solution of NaCl would have 2x osmotic pressure
of 1.0 M of C6H12O6

expect effective conc = 2 mol/L with both Na+ and Cl- ions
1 mol/L C6H12O6 with molecules
 Solutions 15

Colloids
Particles (collections of molecules) suspended in another medium
1 phase ( s, l, g ) suspended in another ( s, l, g )

example phases general name for

smoke solids in gas aerosol
milk butterfat liquid in water liquid emulsion
marshmallow NO2 in solid foam
fog water in air aerosol

Can show it is colloid and not solution by Tyndall effect (the scattering of light )
ex: water droplets in air form fog or cloud
because this colloid mixture scatters light

In water a substance can be hydrophobic (water fearing) or

hydrophilic (water loving)

Polar and Nonpolar: Like dissolves Like

Substances can be
hydrophobic (water fearing) or hydrophilic (water loving)

ex: oil is nonpolar (hydrophobic)

water is polar ( hydrophilic)

polar molecule has more negative and more positive side ( H2O HCl )
nonpolar molecule the charges are uniformly distributed ( CH4 C10H22 )

larger molecules may have polar and nonpolar regions

soap has a nonpolar portion and polar portion

“tail” nonpolar “head” polar

 Solutions 16

Like dissolves like means:

polar solutes dissolve in polar solvents
nonpolar solutes dissolve in nonpolar solvents

1) Example
Given that:
purple iodine I2 is nonpolar
blue food coloring dye is polar
yellow cooking oil is nonpolar
colorless water is polar

We observe the following:

I2 in water NOT dissolve water remains colorless
I2 in oil dissolves forms red solution
blue dye in water dissolves forms blue solution
blue dye in oil NOT dissolve oil remains yellow

2) Example
Small layer of oil in jar of water do not mix
Layer of yellow oil stays on top of colorless layer of water
Shake and they still separate into two layers

Add enough soap or detergent and shake and the oil and water will mix
together. Soap and detergent molecules have polar portion and nonpolar portion.
More info on soaps and detergents at:
http://jan.ucc.nau.edu/~doetqp-p/courses/env440/env440_2/lectures/lec19/lec19.html
 Solutions 17

An Association colloid (micelle) is formed with soap or detergent

Soap molecules form a sphere (micelle) around oil with polar portions extended
out to dissolve in water. Soap (nonpolar/polar parts) used to remove oil or grease.
The hydrophobic part attracts to nonpolar grease or oil and the hydrophilic part is
attracted to polar water

For more information see: http://en.wikipedia.org/wiki/Surfactant