describe the preparation of 2litres of 0.2M acetate buffer of pH 5.5,starting from 2.5M solution of acetic acid and 2.5M solution of KOH{4.77}
I assume the 4.77 has nothing to do with the problem or perhaps that was meant to be pKa for acetic acid. I've always used 4.74 for pKa acetic acid.
You want the solution to be 0.2M so if you start with 0.2M HAc it will react with the KOH to produce acetate. Whatever you lose of the HAc will be formed into Ac^- and the total will be 0.2M which is what you want. I will work in mols and not M. You want 2L of 0.2M which is 0.4 mol.
........HAc + KOH ==> KAc + H2O
I.......0.4...0........0......0
add...........x..................
C........-x..-x........x
E.....0.4-x...0........x
5.5 = 4.77 + log (x/0.4-x)
Solve for x = mols KOH to be added.
Convert to L of 2.5M to get that number of mols.
To start, calculate L of 2.5M acetic acid to give 0.4 mol.
I obtained approx 160 mL acetic acid and approx 135 mL KOH and those are approximate. You should confirm all of this. Also, I like to plug in the numbers at the end and see if the final pH is in fact 5.5 AND that the molarity of the final solution is 0.2M.
To prepare a 2-liter solution of 0.2M acetate buffer with a pH of 5.5, you will need to follow a step-by-step process involving the acetic acid and potassium hydroxide solutions.
Here's how you can approach it:
Step 1: Calculate the moles of acetic acid and KOH required:
The molar ratio of acetic acid to acetate (conjugate base) is 1:1, so you will need the same number of moles of acetic acid and acetate in your buffer solution.
Moles of acetic acid required:
Molarity of acetic acid = 2.5M
Volume of solution = 2 liters
Moles = Molarity x Volume
Moles of acetic acid = 2.5M x 2 L = 5 moles
Moles of KOH required:
Since we need to prepare a buffer solution of pH 5.5, we need to calculate the moles of KOH required to adjust the pH.
pKa of acetic acid = 4.77 (given)
pH of buffer solution = 5.5 (given)
pH = pKa + log(Acetate/Acetic acid)
Rearranging the equation and solving for (Acetate/Acetic acid):
Acetate/Acetic acid = 10^(pH - pKa)
Acetate/Acetic acid = 10^(5.5 - 4.77) = 3.56
Since the moles of acetic acid and acetate should be equal, we need to calculate the total moles required:
Total Moles = Moles of acetic acid = 5 moles
Moles of KOH required = Total Moles - Moles of acetic acid = 5 - 5 = 0 moles
Step 2: Calculate the volumes of the stock solutions required:
Now that we have the moles of acetic acid and KOH required, we need to calculate the volumes of the stock solutions to achieve these quantities.
For acetic acid:
Molarity of acetic acid stock solution = 2.5M
Moles of acetic acid in stock solution = 5 moles
Volume of acetic acid stock solution required = Moles/Molarity = 5 moles / 2.5M = 2 L
For KOH:
Molarity of KOH stock solution = 2.5M
Moles of KOH in stock solution = 0 moles (as we calculated earlier)
Volume of KOH stock solution required = Moles/Molarity = 0 moles / 2.5M = 0 L
Step 3: Preparation of the acetate buffer:
To prepare the acetate buffer, you will mix the calculated volumes of acetic acid and KOH stock solutions:
- Take 2 liters of the acetic acid stock solution.
- Add 0 liters of the KOH stock solution (you don't need to add any).
- Mix them thoroughly to ensure complete mixing.
Step 4: Adjusting the pH:
To achieve the desired pH of 5.5, you can use a pH meter and adjust the pH accordingly using small amounts of either acetic acid or KOH, depending on the initial pH. However, in this case, since the calculated moles of KOH required were zero, no further adjustment is needed.
You have now prepared a 2-liter solution of 0.2M acetate buffer with a pH of 5.5 using the given acetic acid and KOH stock solutions.