You have 250mL of a 0.56M solution of sodium acetate. How many mL of 0.50M acetic acid should be added to make a buffer of pH 4.40?
I think this is a buffer system so I would use pH=pka + log base/acid
so would I set this up as
4.40 = 4.74 + log base/ 1.8e-5 from acetic acid?
Yes and no. (acid) is not 1.8E-5; that's ka for acetic acid. You put 250*0.56 for base and mL*0.500 for acid; solve for mL acid.
To create a buffer of pH 4.40, you are correct in using the Henderson-Hasselbalch equation:
pH = pKa + log(base/acid)
In this case, acetic acid is the acid, and you want to find the amount of base (sodium acetate) needed. Let's go through the steps step by step.
Step 1: Find the pKa value of acetic acid.
The pKa of acetic acid is approximately 4.74.
Step 2: Plug the values into the Henderson-Hasselbalch equation:
4.40 = 4.74 + log(base/acid)
Step 3: Rearrange the equation:
log(base/acid) = 4.40 - 4.74
Step 4: Calculate the antilog of both sides:
base/acid = 10^(4.40 - 4.74)
Step 5: Simplify the right side of the equation:
base/acid = 10^(-0.34)
Step 6: Evaluate the right side of the equation:
base/acid = 0.4506
Step 7: Since we know the concentration of the acid (0.50M acetic acid) and the volume of the solution (250mL), we can find the volume of the base needed.
Let's denote the volume of base (sodium acetate) as "x" mL.
Using the equation: (0.50M acetic acid) * (250mL) = (0.4506) * (250mL + x)
Simplifying the equation:
125 = 112.65 + 0.4506x
Step 8: Solve for x:
0.4506x = 125 - 112.65
0.4506x = 12.35
x = 12.35 / 0.4506
Step 9: Calculate x:
x ≈ 27.35 mL
Therefore, you should add approximately 27.35 mL of the 0.50M acetic acid to the 250 mL of the 0.56M sodium acetate solution to create a buffer of pH 4.40.
To answer this question, we need to use the Henderson-Hasselbalch equation, which is pH = pKa + log ([A-]/[HA]).
In this case, the sodium acetate (NaCH3COO) will act as the conjugate base (A-), and acetic acid (CH3COOH) will act as the acid (HA).
First, let's calculate the pKa of acetic acid. The pKa is given by the negative logarithm of the acid dissociation constant (Ka) for acetic acid. However, acetic acid is a weak acid, so we can use the Ka expression to calculate pKa:
Ka = [A-][H+]/[HA]
Ka = (concentration of conjugate base * concentration of H+)/concentration of acid
Given that the concentration of acetic acid is 0.50M, and the concentration of conjugate base (sodium acetate) is 0.56M, we can substitute the values into the Ka expression:
0.50 = (0.56 * [H+])/[HA]
0.50M * [HA] = 0.56M * [H+]
[H+] = (0.50/0.56)M = 0.8929M
Now, let's calculate the pKa:
pKa = -log([H+])
pKa = -log(0.8929)
pKa ≈ 0.050
Next, we can rearrange the Henderson-Hasselbalch equation to solve for the ratio of [A-]/[HA]:
pH = pKa + log ([A-]/[HA])
4.40 = 0.050 + log([A-]/[HA])
4.35 ≈ log([A-]/[HA])
We have the pKa and the desired pH, so we can rearrange the equation to solve for the ratio of [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(4.40 - 0.050)
[A-]/[HA] ≈ 2,381
This means that the concentration of sodium acetate should be approximately 2,381 times higher than the concentration of acetic acid in the buffer solution.
Now, let's consider the volumes. You have 250 mL of a 0.56M solution of sodium acetate. Let's assume you add x mL of the 0.50M acetic acid.
The total volume of the buffer solution will be 250 mL + x mL.
To determine the volume of acetic acid needed, we need to consider the concentration of acetic acid in the final solution. The concentration of acetic acid is given by:
concentration of acetic acid = (0.50M * x mL) / (250 mL + x mL)
The concentration of sodium acetate will be 2,381 times higher than the concentration of acetic acid:
concentration of sodium acetate = 2,381 * concentration of acetic acid
Since the total volume of the buffer solution is 250 mL + x mL, we can substitute the formulas into the Henderson-Hasselbalch equation:
4.40 = 0.050 + log(2,381 * concentration of acetic acid / concentration of acetic acid)
4.35 ≈ log(2,381)
Using logarithmic properties, we can solve for x:
x ≈ (10^(4.35) - 1) * 250 mL
x ≈ 1,722 mL
Therefore, you would need to add approximately 1,722 mL of the 0.50M acetic acid to make a buffer of pH 4.40.