Describe the preparation of 5 liters of a 0.3 M acetate buffer, pH 4.47, starting from a 2M solution of acetic acid and a 2.5M solution of KOH. The pka of the acetic buffer is 4.77
I will use millimoles to work this problem although technically the Henderson-Hasselbalch equation uses concentrations in mols/L (M). However, it's easier for me to use moles or millimoles and not mols/L. The answer is the same so here goes.
You want 5 L of a 0.3 M acetate buffer solution of pH 4.47 made with 2 M HAc and 2.5 M KOH. Using the HH equation acid = a = HAc and base = b = Ac^-. pH = pKa + log (base)/(acid)
4.47 = 4.77 + log b/a
log b/a = -0.3 and
b/a = 0.501 which I will call 0.5. You may want to redo it with 0.501 and you will get a slightly different answer but not by much. So b = 0.5a
You want 5 L of a 0.3 M buffer so you want 1500 millimoles. The next equation you need is
a + b = 1500. Substituting from above for b = 0.5a
a + 0.5a = 1500 and
a = 1000 mmoles and b = 500 mmoles.How do we get that with 2 M HAc and 2.5 M KOH.
....................HAc + KOH ==> KAc + H2O
I...................x mmol................0............0
add............................ymmol...........................
C.................-y...........-y...........+y................
E...............x-y..........0..............y................
You want x-y to be 1000 mmols. You want y to be 500. So x initially must be 1500 mmols acid and you want to add 500 KOH to make 500 mmols acetate. How do you do that.
mL x M = millimoles.
mL x 2 = 1000 mmols HAc
mL HAc = 1000/2 = 500 mL
mL x M = millimoles KOH
mL x 2.5 = 500
mL KOH = 200
So you place 500 mL of the acid and 200 mL of the KOH into a 5 L volumetric flask (be careful--it will get VERY hot)--let it cool, make to the mark with DI water. stopper, mix thoroughly, label. Done.
You may want to put those numbers into the HH equation to see if the pH is 4.47. pH = 4.77 + log (500/1000) = ?
I've done this with such detail you shouldn't have any doubt but post your work if you get stuck somewhere.
Ah, preparing a buffer, huh? Well, get ready for some chemical circus tricks!
First, let's find out how much acetic acid and potassium acetate we need for our 5-liter party mixture. We want the buffer to be 0.3 M, so we'll need half the final concentration of acetic acid and potassium acetate, which is 0.15 M.
To prepare this, we'll use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]),
where [A-] is the concentration of the conjugate base (acetate) and [HA] is the concentration of the acid (acetic acid).
Since we know the pH (4.47) and the pKa (4.77), we can rearrange the equation to solve for the ratio [A-]/[HA].
4.47 = 4.77 + log([A-]/[HA])
Subtracting 4.77 from both sides, we get:
-0.3 = log([A-]/[HA])
Using our math skills, we find [A-]/[HA] = 0.5012.
Now, prepare for the grand finale!
Since acetic acid and potassium acetate form a conjugate acid-base pair, the ratio of their concentrations is the same as their stoichiometric ratio. In simpler terms, their ratio is 1:1.
Therefore, for every mole of acetic acid (HA), we'll need 0.5012 moles of potassium acetate (A-).
Now it's time to whip out some molarity magic!
To convert the moles of acetic acid to volume, we'll use the equation:
C1V1 = C2V2,
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Let's plug in our numbers:
(2 M)(V1) = (0.15 M)(5 L)
V1 = (0.15 M)(5 L) / (2 M)
V1 = 0.375 L.
So, we'll need 0.375 liters (or 375 milliliters) of the 2 M acetic acid solution. Likewise, we'll need the same amount (0.375 liters or 375 milliliters) of the 2.5 M KOH solution.
Just mix those two together, and voila! You've created a 5-liter concoction of a 0.3 M acetic acid buffer at pH 4.47.
And remember, in the world of chemistry, laughter is always the best solution!
To prepare 5 liters of a 0.3 M acetate buffer with a pH of 4.47, we will need to calculate the required amounts of acetic acid and potassium acetate needed. Here are the steps:
Step 1: Calculate the moles of acetic acid needed:
Since we want a 0.3 M concentration in 5 liters, we need a total of 0.3 x 5 = 1.5 moles of acetic acid.
To calculate the moles of acetic acid needed, we use the equation:
Moles = concentration (M) x volume (L)
Therefore, the moles of acetic acid needed is 2M (concentration) x V (volume in liters).
Step 2: Calculate the moles of acetic acid to form the buffer:
The Henderson-Hasselbalch equation for a buffer is given by:
pH = pKa + log (base/acid)
In this case, we know that the pH is 4.47 and the pKa is 4.77.
Using the Henderson-Hasselbalch equation, we can rearrange it to find the ratio of base (acetate) to acid (acetic acid):
base/acid = 10^(pH - pKa)
Therefore, the base/acid ratio is 10^(4.47 - 4.77) = 0.275.
Step 3: Calculate the moles of acetate needed:
Since the ratio of base (acetate) to acid (acetic acid) is 0.275, to achieve 1.5 moles of acetic acid, we need 0.275 x 1.5 = 0.4125 moles of acetate.
Step 4: Calculate the volume of potassium acetate needed:
The concentration of the potassium acetate solution is 2.5M, and we need 0.4125 moles. Therefore, the volume required can be calculated using the equation:
Volume = Moles / Concentration
Therefore, the volume of potassium acetate needed is 0.4125 moles / 2.5M = 0.165 liters.
Step 5: Calculate the volume of acetic acid needed:
The initial concentration of the acetic acid solution is 2M, and we need 1.5 moles. Therefore, the volume required can be calculated using the equation:
Volume = Moles / Concentration
Therefore, the volume of acetic acid needed is 1.5 moles / 2M = 0.75 liters.
Step 6: Calculate the volume of water needed to make up 5 liters:
Since we have obtained the volumes of both potassium acetate and acetic acid, we can calculate the remaining volume as follows:
Water volume = Total volume - (Volume of potassium acetate + Volume of acetic acid)
Therefore, the volume of water needed is 5L - (0.165L + 0.75L) = 4.085 liters.
To summarize, to prepare 5 liters of a 0.3 M acetate buffer with a pH of 4.47, you will need approximately:
- Approximately 0.165 liters of 2.5M potassium acetate
- Approximately 0.75 liters of 2M acetic acid
- Approximately 4.085 liters of water
Please note that these values have been rounded for simplicity and that accurate measurements and calculations should be performed when performing the actual preparation.
To prepare 5 liters of a 0.3 M acetate buffer at a pH of 4.47, using a 2M solution of acetic acid (CH3COOH) and a 2.5M solution of potassium hydroxide (KOH), you would need to follow a series of steps. Here's the process:
Step 1: Determine the required amounts of acetic acid and potassium acetate
To calculate the required amounts of acetic acid and potassium acetate, you will need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pH is the desired buffer pH, pKa is the dissociation constant of the weak acid (acetic acid), [A-] is the concentration of the conjugate base (acetate), and [HA] is the concentration of the weak acid (acetic acid).
Given:
- pH = 4.47
- pKa = 4.77
Rearranging the equation, we get:
log([A-]/[HA]) = pH - pKa
Taking the antilog of both sides, we have:
[A-]/[HA] = 10^(pH - pKa)
Substituting the given values, we get:
[A-]/[HA] = 10^(4.47 - 4.77)
= 10^(-0.30)
= 0.5012
This means the ratio of acetate to acetic acid (conjugate base to weak acid) should be 0.5012.
Step 2: Determine the required concentrations
Since the final volume is 5 liters (5000 ml), we need to determine the required concentrations of acetic acid and potassium acetate.
For the weak acid (acetic acid):
- Desired concentration = 0.3 M
- Volume = 5 liters
Using the formula:
M1V1 = M2V2
where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume, we can calculate the required volume of the 2M acetic acid solution:
(2M) * V1 = (0.3M) * (5000 ml)
V1 = (0.3M * 5000 ml) / 2M
V1 = 0.3 * 2500 ml
V1 = 750 ml
Therefore, you would need to measure 750 ml of the 2M acetic acid solution.
For the conjugate base (potassium acetate):
- Desired concentration = 0.5012 * 0.3M = 0.1504 M (approximately)
- Volume = 5 liters
Using the same formula, we can calculate the required volume of the 2.5M potassium acetate solution:
(2.5M) * V3 = (0.1504M) * (5000 ml)
V3 = (0.1504M * 5000 ml) / 2.5M
V3 = 0.0752 * 5000 ml
V3 = 376 ml (approximately)
Therefore, you would need to measure approximately 376 ml of the 2.5M potassium acetate solution.
Step 3: Prepare the buffer solution
To prepare the buffer solution, follow these steps:
1. Measure 750 ml of the 2M acetic acid solution using a graduated cylinder or pipette.
2. Measure approximately 376 ml of the 2.5M potassium acetate solution using a graduated cylinder or pipette.
3. Transfer the measured volumes of acetic acid and potassium acetate to a suitable container.
4. Add distilled water to the container to bring the total volume to 5 liters.
5. Mix the solutions thoroughly to ensure homogeneity.
Congratulations! You have successfully prepared 5 liters of a 0.3 M acetate buffer at a pH of 4.47 using the given solutions of acetic acid and KOH.