Discuss how to prepare 250.0 ml of an ammonium-ammonia buffer (pkb = 4.70), ph = 9 and total concentration of 0.05M using 0.10 M ammonium chloride and 0.10M aqueous ammonia
Need answer tomorrow for pre-lab Thanks
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
pKb = 4.70; therefore, pKa = 14.0-4.70 = 9.3
Solve two equations simultaneously.
Plug in 9.0 for pH and solve for (B/A) and I get 0.501 which you should confirm. If B/A = 0.501 then
(Base) = 0.501*(acid)
The second equation is
(base) + (acid) = 0.05
Solve for base and acid. I get something like
(acid) = 0.0333 M
(base) = 0.0167 M
We have two solutions, both 0.1 M with which to make this. Use a dilution process (formula) to determine how much of each should be taken.
0.1M x (x Liters/0.25 L) = 0.0333 M.
Solve for x = 0.08325 L or 83.25 mL acid.
0.1 M x (x liters/0.25) = 0.0167 M
Solve for x = 0.04175 L or 41.75 mL base.
Therefore, the instructions would be to measure 83.25 mL of NH4Cl and 41.75 mL NH3, place in a 250 mL volumetric flask and make to the mark with water. But you shouldn't stop there. You should go through and prove four things.
1. The final solution is 250 mL.
2. The final solution is 0.05 M.
3. The final (acid) is 0.0333 (or whatever value you find above) and the final (base) = 0.0167 (or whatever value you find above.)
4. The final pH is 9.0.
Thanks!
To prepare a 250.0 ml ammonium-ammonia buffer solution with a pH of 9, a pKb of 4.70, and a total concentration of 0.05 M, you can follow these steps:
1. Determine the mole ratio:
The Henderson-Hasselbalch equation for a buffer solution is:
pH = pKa + log ([A-]/[HA])
Since the pKb is given as 4.70, we can calculate the pKa by subtracting it from 14 (pKa + pKb = 14). Therefore, the pKa is 9.30.
2. Calculate the concentration of ammonium ion (NH4+):
Since NH4+ is in excess, we can assume that the concentration of NH4+ will be equal to the total concentration of the buffer. Therefore, the concentration of NH4+ in the buffer is 0.05 M.
3. Calculate the concentration of ammonia (NH3):
Using the mole ratio from step 1, we can calculate the concentration of NH3 using the equation:
[NH3] = ([A-]/[HA]) * [NH4+]
= 10^(pH - pKa) * [NH4+]
= 10^(9 - 9.30) * 0.05 M (using the given pH of 9)
= 0.0794 M
4. Calculate the volume of each component with the given concentration:
To achieve a total volume of 250.0 ml, the sum of the volumes of NH4+ and NH3 must equal 250.0 ml. Since we know the concentrations of the components, we can calculate their respective volumes.
Let VNH4+ be the volume of 0.10 M ammonium chloride and VNAM be the volume of 0.10 M aqueous ammonia.
Therefore, VNH4+ + VNAM = 250.0 ml.
Now, we can use the formula:
Concentration = (moles of solute) / (volume of solution)
to equate the moles of each component.
For NH4+:
0.10 M = (moles of NH4+) / VNH4+
Moles of NH4+ = 0.10 M * VNH4+
For NH3:
0.10 M = (moles of NH3) / VNAM
Moles of NH3 = 0.10 M * VNAM
Since NH4+ and NH3 have a 1:1 mole ratio, the moles of NH4+ are equal to the moles of NH3:
0.10 M * VNH4+ = 0.10 M * VNAM
Therefore, VNH4+ = VNAM
So, VNH4+ + VNAM = 250.0 ml becomes:
VNH4+ + VNH4+ = 250.0 ml
2 * VNH4+ = 250.0 ml
VNH4+ = 250.0 ml / 2
VNH4+ = 125.0 ml
Now that we know the volume of each component, we can prepare the buffer as follows:
- Measure 125.0 ml of 0.10 M ammonium chloride (NH4Cl) using a graduated cylinder.
- Measure 125.0 ml of 0.10 M aqueous ammonia (NH3) using a separate graduated cylinder.
- Pour both solutions into a clean, dry container and mix them thoroughly.
- Add distilled water to the container until the total volume reaches 250.0 ml.
Congratulations! You have successfully prepared a 250.0 ml ammonium-ammonia buffer solution with a pH of 9 and a total concentration of 0.05 M using 0.10 M ammonium chloride and 0.10 M aqueous ammonia.