Concentration Based on
Degree of Saturation
Saturation Degree
• Unsaturated Solution
– less than the maximum amount of solute for a given
temperature is dissolved in the solvent.
– There is more available space for solute to dissolve in the
solvent
– No solid remains in flask.
• Saturated solution
– Is one where the concentration is at a maximum - no more
solute is able to dissolve (you begin to see some crystals) at
that temperature.
– A saturated solution represents an equilibrium.
• Supersaturated
– Solvent holds more solute than is normally possible at that
temperature.
– You can see a big amount of solute at the bottom of the flask
Percent Saturation
• Proteins are often purified by differential precipitation with salts ,such as
ammonium sulfate. The salt conc. used to “salt out“ proteins is always
expressed in the terms of percent saturation
• It is the concentration of salt in a solution as a percent of the maximum
concentration possible at a given temperature.
V (ml) = 100 (S2-S1)
1 – S2
– V is the volume of the saturated salt needed.
– S1 is the initial low saturation ( used as a decimal).
– S2 is the final high saturation ( used as a decimal).
– This is to the volume to be added to 100 ml at saturation S1.
Example
• How many ml of a saturated ammonium sulfate solution must be
added to 40 ml of a 20% saturated solution to make the final
solution 70% saturated? Given values:
S1= 20% = 0.2 ,S2= 70% =0.7
(according to the formula, this is to
V (ml) = 100 (S2-S1) = 100 (0.70 – 0.20) = 166.6 ml the volume to be added to 100 ml at
1-S1 1 – 0.70 saturation S1)
100 ml → 166.6 ml
40 m → ?
The volume needed = 40 × 166.6 = 66.6 ml
100
Units Conversion
Expression Symbol Definition
Based on volume:
Molarity M = No. of moles of solute
volume of solution (L)
Normality N = no. of equivalents
volume of solution (L)
= nxM (n= number of OH or H)
Osmolarity O = nx M ( n= number of dissociable ions)
Weight/Vol wt/V% = Wt in gram of solute
% 100ml of solution
Milligram % mg% = Wt in mg of solute
100ml of solution
Vol/Vol% V/V% = volume in ml of a solute
100ml of solution
Based on weight:
Weight/We w/w% = Wt in gram of solute
ight% 100g of solution
Molality m = No. of moles of solute
1000g of solvent
Mole MF MF2 = n2 / (n1 + n2 + n3)
fraction
Based on saturation:
percent saturation V (ml) = 100 (S2-S1)
1 – S2
Preparations of Solutions
Preparation of stock solutions for acids
• The concentrations of many acids are given in the
terms of w/w%
• In order to prepare an acid stock solution we need
to know its density (ρ) or specific gravity, and
calculate the needed volume by :
Wt(g) =V (ml) x ρ x w/w% (as decimal)
Solutions could be prepared either from
1- Solid material 2-Liquid
Preparation of Solutions from Solid Material
After calculating the weight required to prepare any given
solution, you do the following:
Make up the
Weigh the solute Dissolve the solute. solution to a Mix
known volume
Preparation of Solutions from Liquid
• Solutions are often prepared by diluting a more
concentrated stock solution.
A known volume of the stock solution Make up the solution to
Mix
is transferred to a new container a known volume
Dilutions
• Dilution- the procedure for preparing a less
concentrated solution from a more
concentrated one.
• Serial Dilution- the process of diluting a
solution by removing part of it, placing this in
a new flask and adding water to a known
volume in the new flask.
Dilutions
•When a solution is diluted, solvent is added to lower its concentration.
•The amount of solute remains constant before and after the dilution:
moles BEFORE = moles AFTER
Dilutions
To calculate the concentration of diluted solutions:
C1 V1 = C2V2
C1 = concentration of stock
V1 = Volume of stock
C2 = concentration of diluted
V2= Volume of diluted
Example (1)
A bottle of 0.5M standard sucrose stock solution is in the lab. How can
you use the stock solution to prepare 250 mL of a 0.348M sucrose
solution? Given values:
C1= 0.5 M
C1X V1= C2X V2 V1=?
C2= 0.348M
0.5 X V1= 0.348 X 250 V2= 250 ml
0.348 X 250 / 0.5 = 174 ml
i.e: 174 ml of the stock solution will be diluted with water to reach the volume of
250 ml
Example (2)
• Describe how you would prepare 800mL of a 2.0M H2SO4 solution,
starting with a 6.0M stock solution of H2SO4 .
Given values:
C1= 6 M
C1V1 = C2V2 V1=?
6.0 x V1 = 2.0M x 800 C2= 2 M
6.0 x V1 = 1600 V2= 800 ml
V1 = 1600/ 6.0
V1 = 266.6 ml
i.e:266.6 ml of the 6.0M H2SO4 solution should be diluted with water to give a final
volume of 800mL.
Serial Dilution
• A serial dilution is any dilution where the concentration
decreases by the same quantity in each successive step.
• Dilution starts first with stock solution and each diluted
solution produced is used to prepare the next.
• To calculate the concentration: C1 V1 = C2V2
Linear Dilution
• Same stock solution is used to produce samples of
different concentrations.
• To calculate the concentration: C1 V1 = C2V2
Dilution Factor
• Dilution factor refers to the ratio of the volume of the initial
(concentrated) solution to the volume of the final (dilute) solution
• To make a dilute solution without calculating concentrations use a
dilution factor.
• Divide the final volume by the initial volume.
Df=Vf / Vi
– Vi = initial volume (aliquot volume)
– Vf = final volume (aliquot volume + diluent volume)
– DF of 100 = ratio 1:100
Example (1):
• What is the dilution factor if you add 0.1 ml aliquot of a specimen to
9.9 ml of diluent?
– The final volume is equal to the aliquot volume + the diluent volume:
Vf = 0.1 mL + 9.9 mL = 10 mL
– The dilution factor is equal to the final volume divided by the aliquot volume:
Df =10 mL/0.1 mL = 1:100 dilution.
Example(2):
What is the Df when 0.2 ml is added to 3.8 ml diluent?
– Dilution factor = final volume/aliquot volume
– Final volume = 0.2 +3.8 = 4.0 ml
– Aliquot volume = 0.2 ml
– 4.0/0.2 = 1:20 dilution.
Example (3):
• From the previous example if you had 4 tubes what would
be the final dilution of tube 4?
– Since each dilution is 1:20 and we want to know the dilution of the
FORTH tube so in this case it would be 1:20 multiplied FOUR times.
– Df = 1:20 x 1:20 x 1:20 x1:20
– Df = 1:160,000
Examples
Importance of Dilution
• Example: A blood glucose of 800 mg/dl was obtained.
According to the manufacturer the highest glucose result which
can be obtained on this particular instrument is 500 mg/dl.
– The sample must be diluted.
– The serum was diluted 1:10 and retested.
– The result is 80 mg/dL.
– THIS IS NOT THE REPORTALBE RESULT!
– You must multiply by the dilution factor of 10.
– 10 x 80 = 800 mg/dl.
