PROBLEMS:
1. Make a 750ml solution at pH 7.4 with the following concentrations:
100 mM CaCl2
20 mM Tris (buffer with m.w. = 121 amu)
pH 7.4
2. I made a solution by adding 74g of NaCl and 1.4g of Hepes (buffer with m.w. = 238
amu) to a FINAL volume of 400ml. What is the concentration of NaCl and Hepes in
this solution?
3. A) Explain how to make a 1.5L solution of 200mM BaCl2 and 35mM CHES (buffer
with m.w. = 207 amu).
B) To what pH value should you adjust your solution? You will need to use the
buffer chart in the PowerPoint to figure this out.
4. I made a 700ml solution consisting of 400mM NaCl. I took 200ml of that solution
and added 800ml of water to it. What is the NaCl concentration of this new solution?
ANSWERS:
1. Make a 750ml solution at pH 7.4 with the following concentrations:
100 mM CaCl2
20 mM Tris (buffer with m.w. = 121 amu)
pH 7.4
CaCl2:
You should realize that 100mM CaCl2 = .1M CaCl2 = .1 mol/L CaCl2
Therefore you want 0.1 mol in every litre. However, you are only making 750 ml = 0.75L
.1 mol/L = x mol/.75 L
x = .075 mol of CaCl2
How many grams do I now need to mass out to have .075 mol of CaCl2?
Use the periodic table to figure out how many protons and neutrons in CaCl2:
Ca = 40 amu
Cl = 35 amu (x 2 because there are two of them in CaCl2) = 70 amu
Therefore, CaCl2 has a total of 110 protons and neutrons or 110 amu
Because 1mol x 1amu = 1g…
1 mol x 110 amu = 110 g
So…1 mol of CaCl2 = 110 g of CaCl2
1 mol CaCl2 / 110g = .075 mol / x g
x = 8.25 g
TRIS:
You should realize that 20mM Tris = .02M Tris = .02 mol/L Tris
Therefore you want 0.02 mol in every litre. However, you are only making 750 ml = 0.75L
.02 mol/L = x mol/.75 L
x = .015 mol of Tris
How many grams do I now need to mass out to have .015 mol of Tris?
You were given the atomic mass = 121 amu
Because 1mol x 1amu = 1g…
1 mol x 121 amu = 121 g
So…1 mol of Tris = 121 g of Tris
1 mol Tris / 121g = .015 mol / x g
x = 1.82 g
MAKING THE SOLUTION:
We now take the 8.25g of CaCl2 and the 1.82g of Tris, and dissolve them in a volume
of water 10% of so less than the final volume of 750ml. Therefore, I would dissolve them in
about 650ml of water. This will bring the volume up to perhaps 670ml as the CaCl2 and
Tris have volume.
pH Adjustment:
Use acid or base combined with a pH meter to adjust the pH to 7.4. After this, bring
the volume to the final 750ml…done.
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2. I made a solution by adding 74g of NaCl and 1.4g of Hepes (buffer with m.w. = 238 amu)
to a FINAL volume of 400ml. What is the concentration of NaCl and Hepes in this solution?
First thing to do is figure out how many moles of NaCl and Hepes I added:
NaCl
HEPES
1 mol NaCl/58g = x mol/74g
x = 1.3 moles
1 mol HEPES/238g = x mol/1.4g
x = 5.9 x 10 -3 or .0059 moles
Now we need to account for the volume or 400ml.
NaCl
1.3 moles are present in 400ml
We need to know how many would be in a liter to determine moles/L or M
1.3 moles/400ml = x moles/1000ml
x = 3.25 moles
or
1.3 moles / .4L = x moles / 1L
Therefore there are 3.25 moles/Litre or 3.25M NaCl
HEPES
0.0059 moles are present in 400ml
We need to know how many would be in a liter to determine moles/L or M
0.0059 moles/400ml = x moles/1000ml
x = 0.015 moles
or
0.0059 moles / .4L = x moles / 1L
Therefore there are 0.015 moles/Litre or 0.015M HEPES = 15mM HEPES
CONCLUSION
The solution: 3.25M NaCl, 15mM HEPES
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3. A) Explain how to make a 1.5L solution of 200mM BaCl2 and 35mM CHES (buffer with
m.w. = 207 amu).
B) To what pH value should you adjust your solution? You will need to use the buffer chart
in the PowerPoint to figure this out.
PART A is the same rationale as Number 1:
BaCl2
200mM BaCl2 = .2M = .2 mol/L
.2 mol/L = x mol / 1.5L
x = .3 mol
1 mol / 207 g = 0.3 mol / x g
x = 62 g BaCl2
CHES:
35mM CHES = .035M = .035 mol/L
.035 mol/L = x mol / 1.5L
x = .0525 mol
1 mol / 207 g = 0.0525 mol / x g
x = 10.9 g CHES
MAKING THE SOLUTION:
We now take the 62g of BaCl2 and the 10.9g of CHES, and dissolve them in a volume
of water 10% of so less than the final volume of 1.5L. Therefore, I would dissolve them in
about 1350ml of water. This will bring the volume up to perhaps 1400ml as the BaCl2 and
CHES have volume.
B) You figure out part B by looking at the pKa of CHES.
Recall the definition of pKa: The pH value at which the acid is exactly half dissociated. For
example, looking at the simple buffer like carbonic acid:
H2CO3  H+ + HCO3- pKa = 6.4
Let’s say that you dissolved 1 mole of H2CO3 in water. The acid would dissociate as shown
above and reach equilibrium.
If you set the pH to 6.4, because this is the pKa, you would have an equal amount of H2CO3
and HCO3-. Since you added 1 mole of H2CO3, and it dissociates in a 1:1 ratio with HCO3-, you
would expect to have in your solution 0.5 moles of H2CO3 and 0.5 moles of HCO3-.
Therefore, H2CO3 works best at pH 6.4 because
1. There is plenty of acid (H2CO3) to release H+ is you remove H+ from solution
using a base like NaOH. Recall what is happening…you add NaOH, which will
dissociate to Na+ and OH-. The OH- will combine with H+ to form water. You are
therefore removing H+ and there will be fewer collisions then between H+ and
HCO3- slowing down the reaction to the left. This will cause equilibrium to shift
to the right as a greater amount of H2CO3 dissociates thereby replacing most of
the H+ the base took.
2. There is plenty of base (HCO3-) to absorb H+ if you add an acid like HCl to the
solution. If you add H+, you are increasing the number of collisions between H+
and HCO3- simply because there are more H+ to collide. Therefore, you are
increasing the reaction rate to the left and most of the added H+ will combine
with HCO3- to form H2CO3. Therefore the HCO3- acts like a sponge for H+.
pH Adjustment:
To what pH should we adjust this solution? Well the pKa of CHES is around 9.5.
Therefore the pH would need to be set within + or – 1 pH unit of the pKa or between 8.5
and 10.5. Ideally 9.5 would work best in terms of buffering ability, but perhaps your
application needs a pH a little bit away from this value.
Remember that buffers are not magic and only resist pH change...they do not prevent it.
How much is resists depends on the concentration of the buffer (1M tris buffers a lot better
than 10mM tris) as well as the pH of the solution (a pH value closer to the pKa buffers the
best).
4. I made a 700ml solution consisting of 400mM NaCl. I took 200ml of that solution and
added 800ml of water to it. What is the NaCl concentration of this new solution?
This one is a dilution problem.
Concentration don’t change when you simply take some of the solution:
The solution is 400mM. No matter what volume I take…200ml (like in the
problem)…100ml…10 microliters…you are not changing the concentration of the solution.
Therefore, you took 200ml of solution that has a concentration of 400mM NaCl. That 700ml
volume doesn’t matter…
You then add 800 ml to it, which would bring the solution up to 1000ml or 1L.
Answering the Question:
You can do this a few ways, but what you must realize is that you added water increasing
the volume. However, you did not add any more NaCl. Therefore the concentration better go
down…
Initial volume = 200ml
Final volume = 1000ml
Initial concentration = 400mM
Final concentration = x mM
Don’t memorize any formulas!!....think!
You took the volume from 200ml to 1000ml. By what fold did you increase the volume?
1000ml final/200ml intial or the final is 5X greater than the initial.
If you increase the volume 5X, but did not change the amount of NaCl…what should happen
to the concentration of NaCl?
Working out the problem logically:
The solution is 400mM or .4M or .4 mol/L or .4 mol / 1000ml
You took 200ml
.4 mol/1000ml = x mol / 200ml
x = .08 mol
Therefore you have 0.08 mol of NaCl in that 200ml volume you took. Now you added water
and brought the volume to 1000ml.
Therefore you have 0.08 mol in 1000ml:
0.08mol/1000ml = 0.08mol/L = 0.08M = 80 mM
That is your concentration of NaCl in the new solution….80mM.
Looking at the relationship between volume and concentration:
Let’s look at it again then:
Initial volume = 200ml
Final volume = 1000ml
Initial concentration = 400mM
Final concentration = x mM
You took the volume from 200ml to 1000ml. By what fold did you increase the volume?
1000ml final/200ml intial or the final is 5X greater than the initial.
If you increase the volume 5X, but did not change the amount of NaCl…what should happen
to the concentration of NaCl?
Well you now know the final concentration…its 80mM…or 5X less than your initial!!
Therefore, if you increase the volume 5X then you decrease the concentration 5X!
Conclusion: Whatever fold you increase the volume, that is the same fold that you decrease
the concentration. If volume increase 3X then concentration decreases 3X…understand,
don’t memorize…