Water Notes Part 3
5.
CONCENTRATION UNITS AND DILUTION EQTN M 1V1= M2V2 (Ch. 16)
Concentration
Unit
Molarity
Normality
TABLE OF CONCENTRATION UNITS USED IN CHEMISTRY
Abbreviation
Equation Used
Percent Solution
by volume
Percent Sol’n
by mass
Molality
Mole Fraction
Dilution
Equation
BP Elevation for
ionic cmpnds
FP Depression
for ionic cmpnds
BP Elevation
FP Depression
M
N
% sol’n
by volume
% sol’n
by mass
big “m”
XA where A is
chem. formula
M1 M of stock
M2 M of dilute
Kb=0.51°C/m
Tb = Kb m
Kf= -1.86°C/m
Tf = Kf m
For molecular
solutes
M = moles of solute / liters of solution
N = molar equivalent of solute / liters of solution
For acids: N = (# of H+ ions) x M
For bases: N = (# of OH- ions) x M
% by volume = volume of solute / volume of solution x 100
% by mass = mass of solute (g) / solution volume (mL) x 100
OR % by mass = mass of solute (g) / solution mass (g) x 100
m = moles of solute / kg of solvent
Xsolute = moles of solute / sum of moles of solute and solvent
mole fraction is a decimal value, not a percentage
M1V1= M2V2
BPnew = 100°C + (.51)(m)(# of ions)
Must be hotter to boil, tends to stay as liquid water more bonding
FPnew = 0°C - (1.86)(m)(# of ions)
Must be colder to freeze, tends to stay as liquid water
BPnew = 100°C + (.51)(m)
FPnew = 0°C - (1.86)(m)
Concentration is a measurement of the amount of a solute that is dissolved in a given amount of a solvent and can be expressed
at least six different units as shown above. Two qualitative descriptions of concentration are dilute solutions (small amount of
solute dissolved compared to the solvent) and concentrated solutions (a lot of solute dissolved).
6.
COLLIGATIVE PROPERTIES (Ch. 16)
Colligative properties are properties that depend on the number of solute particles dissolved in a given mass of solvent.
They DO NOT depend on the chemical natures of the solute or solvent. It is important here to remember the
differences in the dissolving process of ionic and molecular compounds. 1.0 moles of aluminum chloride (AlCl 3) will
produce 4.0 moles of solute particles (aluminum chloride will dissociate into an aluminum ion and three chloride ions)
and 1.0 moles of glucose (C6H12O6) will produce only one mole of solute particles. Therefore one mole of aluminum
chloride dissolved will have 4 times the colligative effect that one mole of glucose will if they are dissolved in the
same amount of water.
Three important colligative properties of aqueous solutions
are :
1.
2.
3.
Vapor pressure lowering - a nonvolatile solute (one that will not vaporize) will lower the vapor pressure
of water. This is due primarily to the process of solvation. As water molecules bond to the solute particles
it would require more energy to break that bond and allow the water molecule to vaporize to create vapor
pressure.
Freezing point depression - solute particles will lower the freezing point of water. This is due primarily to
the solute particles interfering with the crystal formation of water as it freezes. A common example of this
is putting salt on the roads to melt snow or ice or to prevent ice from forming. Another common example is
the antifreeze in a car's engine (also raises the boiling point to prevent overheating in summer).
Boiling point elevation - nonvolatile solute particles will raise the temperature of boiling water. The forces
here are similar to that of vapor pressure lowering. As water molecules bond to solute particles more
energy is needed to get them to vaporize and it then takes more heat energy to get the vapor pressure of
water to equal the atmospheric pressure to cause the water to boil. A common use of this is when food is
cooked salt is added to the water. This not only adds flavor but raises the temperature of the boiling water
causing the food to cook faster.
CONCENTRATION AND COLLIGATIVE PROPERTIES: DETAILED DESCRIPTIONS AND EXAMPLES
Boiling point elevation and freezing point depression are directly related to the molal concentration by a constant.
these are the molal freezing point constant (Kf = -1.86 ºC/m) and the molal boiling point constant (Kb = .512 ºC/m).
What this means is that for every molal concentration of a nonvolatile solute the freezing point drops by 1.86 ºC and
the boiling point raises by .512ºC. To calculate the amount of change in the freezing point or boiling point elevation
we simply multiply the constant times the molal concentration (Tf = Kfm or Tb = Kbm where Tf and Tb are
the changes in the freezing and boiling points). To find the new boiling point temp. you must add the Tb such as
BPnew = 100°C + (.51)(m) and to find the new freezing point temp. you must substract the Tf such as FPnew = 0°C (1.86)(m) Here is an example using a molecular solute (one that does not dissociate into ions upon dissolving).
Here is an example using an ionic solute - one that will dissociate into ions upon dissolving so it add more solute
particles to the solution. The new equation would be FP new = 0°C - (1.86)(m)(# of ions) where NaNO3 would form 2
ions in solution and something like AlCl3 would form 4 ions in solution (1 Al+3 cation and 3 Cl-1 anions).
Molarity (abbreviated M) is concentration expressed as moles of solute per liter of solution (mol/L) and is the most
common unit used for acid and base concentrations. It is really nothing more than a ratio of one thing to another
(moles of solute to liters of solution). Molarity is calculated by dividing the number of moles of solute by the numbers
of liters of solution, not solvent. Here is an example of calculating molarity.
What is the molarity (M) of a solution which has 2.5 moles of solute in enough water to form .75 L of solution?
M = mol/L = 2.5 mol/ .75 L = 3.3 M
Sometimes you may need to calculate the number of moles first as shown here. What is the molarity of a solution
formed by dissolving 5.5 g of NaCl in enough water to form 1.2 L of solution?
You will also need to be able to calculate moles of solute (and then grams of solute) and volume of solution from the
equation for molarity. To solve for moles of solute rearrange the equation M = mol/L to get mol = L x M. Mass can
then be solved for by multiplying by the molar mass. Here is an example.
To solve for volume of solution we can rearrange the equation for molarity to get L = mol/M. Here is an example.
Dilution of Stock Solutions: Often, in the laboratory, certain diluted molar concentrations need to be prepared from
existing concentrated stock solutions. This is called dilution and the formula to calculate molarity from existing
solutions by dilution is : M1 x V1 = M2 x V2. You do not need to have units for V in liters for dilution problems, it
will work equally well with any volume unit, such as mL or cm3. Do be sure volume units are the same if you are
calculating for molarity. Here is a sample problem using dilution.
Normality: Normality (N) is another ratio that relates the amount of solute to the total volume of solution. It is
specifically defined as the number of equivalents per liter of solution:
normality =
number of equivalents / 1 L of solution
There is a very simple relationship between normality and molarity: N = n × M (where n is an integer) For an acid
solution, n is the number of H+ provided by a formula unit of acid. Example: A 3 M H2SO4 solution is the same as a 6
N H2SO4 solution because each mole of H2SO4 that dissociates in water forms two mole equivalents of H + ions so the
normal concentration is twice the molar concentration.
For a basic solution, n is the number of OHprovided by a formula unit of base. Example: A 1 M Ca(OH)2 solution is the same as a 2N Ca(OH)2 solution.
Remember! The normality of a solution is NEVER less than the molarity (the normality is either equal or a multiple of
times greater than the molarity).
Percent concentration is another way to quantitatively express concentration. Percent concentration is the fraction
of the solute to solvent ratio as expressed in percent. To calculate the percent composition you divide the amount of
solute by the amount of solution and then multiply by 100%. The equation is (solute/solution) x 100%. Units are in
percents with solute and solvent units indicated in parentheses. Percent composition can be a ratio of volume of
solute to volume of solution or mass of solute to volume of solution. Here is an example of a volume/volume (v/v)
percent composition problem. In the problem below units are the same for volume so no conversions are necessary
and it is not necessary to indicate the units of volume in parentheses (% comp. would be the same if it were L/L or
mL/mL).
Here is a mass/volume (m/v) percent composition problem :
Here we needed to indicate units to communicate the concentration. If units were not indicated the concentration could not be
known. 7.3% (kg/mL) would be a 1 000 times more concentrated than 7.3 % (g/mL)!
m
Molality ( ) is concentration expressed as moles of solute per kilogram of solvent (not solution). To calculate molality
you divide the number of moles of solute by the number of kilograms of solvent. Here is a sample problem. In this
problem the solute is given in grams so it will need to be converted into moles and the mass of the solvent is given in
grams which will need to be converted into kg.
You will need to also be able to calculate moles of solute, mass of solute and mass of solvent from the above
equation. It will be very similar to the problems given for molarity above so I will not give you examples here.
Although their spellings are similar, molarity and molality cannot be interchanged. Compare the molar and molal
volumes of 1 mol of solute dissolved in CCl4 (density = 1.59 g / mL). By definition, a 1 M solution would contain 1
mol of solute in exactly 1.00 L of CCl4 , and a 1 m solution would contain 1 mol of solute in 629 mL of CCl4 .
Whereas, 1 kg of CCl4 x (1000 g / 1 kg) x (mL / 1.59 g) = 629 mL CCl4
Mole fraction is yet another way of expressing concentration. Mole fraction is the ratio of moles of the parts of a
solution (solute or solvent) to the total number of moles of the solution (sum of solute and solvent). The mole fraction
of the solvent is the number of moles of the solvent divided by the total number of moles of the solution. Here is an
example (Moles are given here, but you will usually have to calculate the number of moles first).
The sum of the mole fractions should be very close to wholeness (1). It may be a little off due to the use of
significant figures. Note also that there are no units for mole fraction as moles cancel out in the equation.
More Calculations involving the rearrangement of the FP depression or BP elevation equations. If an analytical chemist
wishes to determine the molality of a solution or continue on to identify molecular mass of an unknown nonvolatile
molecular (not ionic) solute he can experimentally determine the changes in freezing point or boiling point and then
rearrange the equations listed above. This is a 1 step (for molality) or three step process (for molecular mass) :
Step 1. Calculate the molality of the solution.
Step 3. Calculate the molecular mass of the solute.
Step 2. Calculate the moles of the solution.
The 3 steps are shown & explained in the following problem.