You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 4.90 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer.

Question 1: 1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.
My Solution: .25*.5=.125*60.05=7.5g.
Question 2:2. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 4.90 at a final volume of 500 mL? (Ignore activity coefficients.)
10^(4.9-4.76)=1.38=a/b
so a=(1.38*.125)b
.1725b+b=.125moles
1.1725b=.125moles
b=.107moles.
3M=.107mole so mL=35.7. However this is incorrect.. Any help would be amazing!

HOw is it .1725b +b

To answer Question 2 correctly, you need to consider the Henderson-Hasselbalch equation and the stoichiometry of the reaction between acetic acid and sodium hydroxide.

The Henderson-Hasselbalch equation is given by:

pH = pKa + log([A-]/[HA])

In this case, the pKa of acetic acid is 4.76, and you want a pH of 4.90. Let's calculate the ratio of [A-] (acetate) to [HA] (acetic acid):

4.90 = 4.76 + log([A-]/[HA])

0.14 = log([A-]/[HA])

Taking the antilog, we find:

[A-]/[HA] = 10^0.14

[A-] = [HA] * 10^0.14

Now, let's calculate the number of moles of acetic acid required (Question 1) and the number of moles of sodium hydroxide to be added (Question 2):

1. Moles of acetic acid required:
0.250 M * 0.500 L = 0.125 moles

2. Moles of sodium hydroxide required:
Since the [A-] to [HA] ratio is 10^0.14, the moles of [A-] and [HA] will be the same. Therefore, you need 0.125 moles of sodium hydroxide.

Now, let's calculate the volume of 3.00 M NaOH needed (remember to use the stoichiometry of the reaction):

Moles of NaOH = Moles of acetic acid = 0.125 moles
Volume of NaOH = (Moles of NaOH) / (Concentration of NaOH)
Volume of NaOH = 0.125 moles / 3.00 M
Volume of NaOH = 0.0417 L or 41.7 mL

So, to achieve a buffer with a pH of 4.90 at a final volume of 500 mL, you would need to add 41.7 mL of 3.00 M NaOH to the acetic acid.

To determine the correct volume of 3.00 M NaOH required to achieve a buffer with a pH of 4.90 at a final volume of 500 mL, we need to consider the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

In this case, acetic acid (HA) is the weak acid and acetate (A-) is its conjugate base.

1. First, let's find the moles of acetic acid required to prepare the buffer:
moles of acetic acid (HA) = volume (L) * concentration (M)
moles of acetic acid = 0.500 L * 0.250 M = 0.125 moles

2. Next, we need to determine the moles of acetate (A-) required to achieve the desired buffer pH:
pH = pKa + log([A-]/[HA])
4.90 = 4.76 + log([A-]/0.125)
0.14 = log([A-]/0.125)

Using the logarithmic property, we can rewrite the equation as:
[A-]/0.125 = 10^0.14
[A-] = 0.125 * 10^0.14
[A-] ≈ 0.176 M

3. Now, let's find the moles of acetate (A-) required:
moles of acetate = volume (L) * concentration (M)
moles of acetate ≈ 0.500 L * 0.176 M = 0.088 moles

4. The moles of NaOH required should be equal to the difference in moles of acetate and acetic acid to achieve the desired pH:
moles of NaOH = moles of acetate - moles of acetic acid
moles of NaOH = 0.088 - 0.125
moles of NaOH ≈ -0.037 moles

Note: The negative sign indicates that a base (NaOH) is needed to neutralize a portion of the acetic acid.

5. Finally, let's calculate the volume of 3.00 M NaOH required to supply the moles needed:
volume (L) = moles / concentration (M)
volume (L) = (-0.037 moles) / (3.00 M)
volume (L) ≈ 0.0123 L = 12.3 mL

Therefore, the correct volume of 3.00 M NaOH to add to the acetic acid to achieve a buffer with a pH of 4.90 at a final volume of 500 mL is approximately 12.3 mL.