Reagents and Solutions
BT 201
Biotechnology Techniques I
Topics to Discuss
• Where do solution recipes come from?
• Concentration of solute: calculations
• Preparing solutions
– Making diluted solutions from
concentrated stock
– Buffers
– Bringing solutions to proper pH
– Solutions with more than one solute
Sources of “Recipes”
•
•
•
•
Scientific Literature
Lab manuals
Handbooks
Manufacturers and suppliers
Terms to Know
• SOLUTES -- substances that are dissolved
• SOLVENTS -- substance in which solutes
are dissolved
– Usually water, if not specifically stated otherwise
• AMOUNT -- how much
• CONCENTRATION -- amount per volume
Concentration
• The fraction where:
– Numerator equals the amount of solute
– Denominator usually equals the total
volume of entire solution
• Solvent + Solute(s)
Amount vs. Concentration
• Each ribbon
represents 2mg of
NaCl
• What is the amount
of NaCl in the tube?
• What is the
concentration of
NaCl in the tube? (in
mg/ml)
5 ml
Expressing Concentration
• WEIGHT PER VOLUME
• MOLARITY
• PERCENTS
a. Weight per Volume % (w/v %)
b. Volume per Volume % (v/v %)
c. Weight per Weight % (w/w %)
Expressing Concentration
• PARTS
Amounts of solutes are expressed as
parts per quantity
a. Parts per Million (ppm)
b. Parts per Billion (ppb)
c. Parts per Thousand (ppt)
d. Percents are same category (pph=%)
Expressing Concentration
• These are not as common in biology
• MOLALITY
– Moles of solute/kilogram of solvent
• NORMALITY
– Gram equivalents/liters of solution
• For compounds with 1 Hydrogen atom like
NaOH and HCl, molarity = normality
– This is not true for all solutes
Weight Per Volume
• A fraction with:
Weight of solute (numerator)
Total volume (denominator)
• So, 10 mg/ml Lysozyme: 10 mg lysozyme in
each ml of solution
• EX. How much lysozyme is needed to make
100 ml of solution at a concentration of
20mg/ml?
Molarity
• Number of moles of a solute that are
dissolved per liter of total solution.
• A 1 M solution contains 1 mole of solute per
liter total volume.
• A mole is Avogadro’s number of atoms:
6.022X1023
• For compounds, the combined atomic
weights of individual atoms are used to
express the mole
What is a mole?
Mole/Weight Relationship Examples Using Helium
Moles Helium
# Helium Atoms
Grams Helium
1/4
1.505 x 1023
1g
1/2
3.01 x 1023
2g
1
6.02 x 1023
4g
2
1.204 x 1024
8g
10
6.02 x 1024
40 g
Making a Solution
• Find the FW of the solute, usually
from label.
• Determine the molarity desired.
• Determine the volume desired.
• Determine how much solute is
necessary by using the formula
• FW X molarity x volume = g solute needed
Making a Solution
• Weigh out the amount of solute.
• Dissolve the solute in less than the
desired final volume of solvent.
• Place the solution in a volumetric flask
or graduated cylinder. Add solvent
until exactly the required volume is
reached, “Bring To Volume”, “BTV”.
Calculating Solute Needed
• How much solute is needed to make 200 ml
of 0.5M NaCl?
• FW X molarity x volume = g solute needed
• (58.44 g/mol) (0.5 mol/L) (0.2 L) = 5.84g
Percent Solutions
• X % is a fraction
numerator is X
denominator is 100
• Three variations on this theme
• Weight/volume
• Weight/weight
• Volume/volume
Weight/Volume
• Grams of solute per 100 ml total
solution
EX. 20 g of NaCl in 100 ml of total solution = 20%
(w/v) solution.
• Common in biology, less common in
chemistry
• EX. How would you prepare 250 ml of
a 5% (w/v) solution?
Calculating a % (w/v) Solution
• How would you prepare 250 ml of a 5%
(w/v) solution?
• Using proportions:
5%= 5g
100 ml
5 g
?
100 ml = 250 ml
Solute needed=12.5 g, BTV 250 ml
Calculating a % (w/v) Solution
• How would you prepare 250 ml of a
5% (w/v) solution?
• By equation:
Total volume required is 250 ml.
5% = 0.05
(0.05) (250 ml) = 12.5g of solute required
Making the Solution
• Weigh out 12.5 g of NaCl. Dissolve it
in less than 250 ml of water.
• In a graduated cylinder or volumetric
flask, bring the solution to 250 ml.
Other % Solutions
• v/v:
• w/w:
ml solute
100 ml solution
g solute
100 g solution
• How would you make 500 g of a 5%
solution of NaCl by weight (w/w)?
Making the Solution
• How would you make 500 g of a 5%
solution of NaCl by weight (w/w)?
– Percent is 5% w/w, total weight desired 500g.
– 5% = 5g/100g
– 5g X 500 g = 25 g = NaCl needed
100 g
– 500 g – 25 g = 475 g = amount of solvent needed
– Dissolve 25 g of NaCl in 475 g of water.
Parts
• Parts may have any units but have to be the
same for all components of the mixture
• A solution of 3:2:1
ethylene:chloroform:isoamyl alcohol
Parts could be:
• 3 liters ethylene
• 2 liters chloroform
• 1 liter isoamyl alcohol
Parts Per ?
• Parts per million (ppm): the number of parts
of solute per 1 million parts of total solution
– 5 ppm chlorine = 5 g of chlorine in 1 million g of
solution,
– 5 mg chlorine in 1 million mg of solution,
– 5 pounds of chlorine in 1 million pounds of
solution
• Parts per billion (ppb): the number of parts of
solute per billion parts of solution.
Conversions
For any solute:
1 ppm in water =
1 microgram
ml
•
•
•
Each star represents 1 mg
of NaCl.
What is the concentration
of NaCl in the tube
expressed as ppm (parts
per million)?
What is the total amount of
NaCl in beaker?
500 ml
1 ppm in water = 1 μg
ml
15 mg/500 ml
= 0.03 mg/ml= 30 μg/ml
Concentration =30 ppm
Preparing a Dilute Solution
From Concentrated Stock
• Use this equation to decide how much
stock solution you will need: C1V1=C2V2
– C1 = concentration of stock solution
– C2 = concentration you want your dilute
solution to be
– V1 = how much stock solution you will
need
– V2 = how much of the dilute solution you
want to make
Example Diluted Solution
• How would you prepare 500 ml of a 1
M solution of Tris buffer from a 3 M
stock of Tris buffer?
–
–
–
–
The concentrated solution =3 M, C1.
The volume of stock needed is unknown, V1
The final concentration required = 1M, C2
The final volume required is 500 ml ,V2
C1 V1 = C2V2  3 M (V1) = 1 M (500 ml)
– V1 =166.67ml of stock, BTV 500 ml
“X” Solutions
• Solution concentration is sometimes
written as “X” value
• “X” is how many times more diluted
the solution is than normal
• Working solution is generally diluted
to 1X, unless state otherwise
Making “X” Solutions
• A can of frozen juice is labeled 4X.
How would you dilute it to make 1L of
drinkable juice?
• Using the C1V1=C2V2 equation:
C1V1 = C2V2
4X (V1 ) = 1X (1L)
V1 = 0.25 L
Use 0.25 L of juice, BTV 1L.
Buffer Solutions
• Laboratory buffers: solutions to help
maintain a biological system at proper
pH
• pKa of a buffer: the pH at which the
buffer experiences little change in pH
with addition of acids or bases = the pH
at which the buffer is most useful
Buffers
• Some buffers change pH as their
temperature and/or concentration
changes
• Tris buffer is temperature sensitive
• Some buffers are sensitive to dilution
• Phosphate buffer is sensitive to
dilution
Adjusting pH of Buffers
• To set the buffer to a pH value
– Try to get close to its pKa
– Most useful for the biological system the
buffer is to be used with
• Often done using NaOH or HCl
– Not the method used for all buffers (see
textbook regarding phosphate buffers)
Bringing Solution to pH
• Adjust pH at the temperature you plan to use the
buffer
• Mix solutes with most of the solvent volume, but
don’t BTV
• Stir solution and test initial pH
• Add a small amount of acid or base (usually HCl or
NaOH, or recipe may specify)
• Stir and re-check pH, repeat until pH is correct
without overshooting
• BTV and re-check pH
Multi-Solute Solutions
RECIPE I
RECIPE II
•
•
•
•
•
•
• 1 M MgCl2
• 0.1 M Tris
• This recipe gives the final
concentration of each
solute and is more
difficult to follow
• Requires calculation of
solute needed
Na2HPO4 6 g
KH2PO4 3 g
0.4% glycerol 10 ml
Dissolve in water
BTV 1 liter
This recipe lists
amounts of solutes
and is easy to follow
Making Multi-Solute Solutions
• Example “recipe”:
–
–
–
–
0.1 M NaCl
1 mM MgSO4
0.2 M Tris, pH 7.5
0.01% gelatin
• Two strategies can be used to make:
with stock solutions or without
Preparation Without Stock
Solutions
• Prepare a solution of 0.2 M Tris, pH
7.5
• Decide how much buffer to make
• 1 liter of 0.2 M Tris requires 24.2 g of
Tris base (MW = 121.1 g/mole)
• Dissolve Tris in about 700 ml of water
and bring pH to 7.5
• Do not bring Tris to volume
Preparation Without Stock
Solutions
• Calculate grams of each of the other
solutes required
– 1 liter of 0.1 M NaCl requires 5.84 g of
NaCl
• Add to Tris buffer
– 1 liter of 1 mM MgSO4 requires 1/1000 of
its MW
• MgSO4 comes in more than one hydrated
form
• Weigh out correct amount.
• Add to Tris buffer
Preparation Without Stock
Solutions
• 0.01% gelatin is 0.1 g in 1 L
– Weigh out and add to Tris buffer.
– Stir to dissolve
• BTV 1 L
• Check and record pH
Preparation Using Stock
Solutions
• Solutes are prepared separately as
concentrated stock solutions.
• Stocks are combined to dilute one
another to the needed final
concentrations
Preparation Using Stock
Solutions
•
Prepare a stock solution of Tris buffer at pH 7.5
a. No set rule as to concentration of stock.
b. Make 1 L of 1 M stock by dissolving121.1 g of Tris base
in 900 ml of water
c. Bring pH to 7.5
d. BTV 1L
•
Prepare a stock solution of magnesium sulfate
a. Make 100 ml of 1M stock by dissolving 0.1 MW of
MgSO4 in water
b. BTV 100 ml
Preparation Using Stock
Solutions
•
Prepare a stock solution of NaCl
a. Make 100 ml of 1M stock by dissolving 5.84g
in water
b. BTV 100 ml
• Prepare a stock solution of gelatin
a. Make a 1% stock solution by dissolving 1g
in water
b. BTV 100ml
Preparation Using Stock
Solutions
• To make the final solution, combine
the right amounts of each stock.
– Use the C1V1 = C2V2 equation four times,
once for each solute, since stock solutions
are being diluted
Preparation Using Stock
Solutions
• To make 1000 ml buffer, combine:
200 ml
1 ml
100 ml
10 ml
Tris stock
MgSO4 stock
NaCl stock
Gelatin stock
• BTV 1L
• Check and record pH.
Solution Making
• Either strategy is correct
• Making stock solutions is less time
consuming than weighing out
ingredients every time
Disclaimer
• This workforce solution was funded by a grant awarded under the
President’s Community-Based Job Training Grants as implemented by
the U.S. Department of Labor’s Employment and Training
Administration. The solution was created by the grantee and does not
necessarily reflect the official position of the U.S. Department of
Labor. The Department of Labor makes no guarantees, warranties, or
assurances of any kind, express or implied, with respect to such
information, including any information on linked sites and including,
but not limited to, accuracy of the information or its completeness,
timeliness, usefulness, adequacy, continued availability, or
ownership. This solution is copyrighted by the institution that created
it. Internal use by an organization and/or personal use by an
individual for non-commercial purposes is permissible. All other uses
require the prior authorization of the copyright owner.