Calculate the volume of 2 M acetic acid needed to prepare 50 ml of 0.1 M acetate buffer.
Are you adding NaOH to acetic acid or are you adding sodium acetate to acetic acid.
I'm adding NaOH
......CH3COOH + OH^-==> CH3COO^- + H2O
If we started with 2.5 mL of 2M CH3COOH that would be 2.5 x 2 = 5 mmoles. Then we add 2.5 mmoles NaOH which gives us at equilibrium 2.5 mmoles acetate and 5.0-2.5 = 2.5 mmoles CH3COOH.
Total is 2.5 mmoles CH3COOH + 2.5 mmoles CH3COONa = 5 mmoles total/50 mL = 0.1M
To calculate the volume of 2 M acetic acid needed to prepare 50 ml of 0.1 M acetate buffer, we first need to understand the concept of buffer solutions and the Henderson-Hasselbalch equation.
A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it. It consists of a weak acid and its conjugate base (or vice versa) in roughly equal amounts. In this case, we have acetic acid (CH3COOH) as the weak acid and acetate (CH3COO-) as the conjugate base.
The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution and is given by:
pH = pKa + log([A-]/[HA]),
where pH is the desired pH of the buffer, pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, we want to prepare a buffer with a pH of 0.1 M, so we need to calculate the volume of 2 M acetic acid required to achieve this concentration of acetate buffer.
Here's how we can calculate it:
1. Determine the pKa value of acetic acid: The pKa of acetic acid is approximately 4.75.
2. Use the Henderson-Hasselbalch equation to calculate the concentration of acetate ([A-]) and acetic acid ([HA]) required to achieve a pH of 0.1 M. The equation can be rearranged as follows:
0.1 = 4.75 + log([A-]/[HA])
Rearranging the equation yields:
log([A-]/[HA]) = 0.1 - 4.75
Taking the antilog of both sides of the equation:
[A-]/[HA] = 10^(0.1 - 4.75)
[A-]/[HA] = 10^(-4.65)
[A-]/[HA] = 3.5481 × 10^(-5)
3. Since the concentration of acetic acid ([HA]) is not given explicitly, we assume it to be equal to x.
Therefore, [A-] = 3.5481 × 10^(-5) * x
4. Since the total volume of the buffer is 50 ml, the sum of the volumes of acetate and acetic acid will be 50 ml. Therefore:
x + 3.5481 × 10^(-5) * x = 50
(1 + 3.5481 × 10^(-5)) * x = 50
x = 50 / (1 + 3.5481 × 10^(-5))
x ≈ 49.9998 ml
5. Finally, calculate the volume of 2 M acetic acid needed. Since the original concentration of acetic acid is 2 M, we can use the following equation:
2 M * V = x * 2 M
V ≈ 49.9998 ml
Therefore, the volume of 2 M acetic acid needed to prepare 50 ml of 0.1 M acetate buffer is approximately 49.9998 ml.