You are asked to prepare 500. mL of a 0.150 M acetate buffer at pH 5.00 using only pure acetic acid, 3.00 M NaOH, and water. Calculate the quantities needed for each of the following steps in the buffer preparation.

1. Add acetic acid to ~400 mL of water in a 500 mL beaker. How many grams of acetic acid do you need?

2. Add 3.00 M NaOH solution while stirring and monitoring the pH with a pH electrode until the pH is 5.00. What volume of NaOH solution do you calculate that you will need?

By taking the desired moles of the solution (0.15M x 0.5L = 0.075 moles), I was able to calculate the needed Acetic Acid (Question #1 = 4.5g)

(Desired Moles) 0.075 x (Formula Weight) 60.05 = 4.5g Acetic Acid.

Where I am stuck is Question 2. After using HH:

(AC/HAC)=10^(5-4.76), (AC/HAC)=1.74

After this, I multiplied my HAC (0.075) by my 1.74 to achieve my AC (0.13).

I know I need to incorporate this with my Molarity of base (3.00M), but am not sure how. Am I even on the right track?

Thank you very much in advance.

Well, well, well, it seems you're grappling with the complexities of buffer preparation. Don't worry, I'll do my best to help you!

For question 1, you've already done a great job calculating the amount of acetic acid needed. 4.5 grams it is!

Now, let's move on to question 2. You want to use 3.00 M NaOH to adjust the pH of the buffer solution to 5.00, but you're wondering how much of it you'll need.

First, let's calculate the number of moles of acetic acid in your solution. You've rightly determined it to be 0.075 moles.

Now, we can use the Henderson-Hasselbalch equation to find the ratio of acetic acid to acetate ion in the solution. You've found it to be 1.74 (AC/HAC = 1.74).

Multiply this ratio (1.74) by the moles of acetic acid (0.075), and you'll find that you'll need around 0.13 moles of acetate ion.

Since NaOH reacts with acetic acid in a 1:1 ratio, you'll need the same number of moles of NaOH to neutralize the acetic acid and form acetate ion. So, you'll also need approximately 0.13 moles of NaOH.

Now, with the concentration of your NaOH solution (3.00 M), you can calculate the volume needed using the equation: Volume = Moles/Concentration.

Volume = 0.13 moles /(3.00 moles/L) ≈ 0.043 L ≈ 43 mL.

So, you'll need around 43 mL of the 3.00 M NaOH solution to reach a pH of 5.00.

Hope that helps! Good luck with your buffer preparation!

You are on the right track! To calculate the volume of 3.00 M NaOH solution needed, you can use the Henderson-Hasselbalch equation and the information you have already calculated.

The Henderson-Hasselbalch equation is given by:

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

In this case, the pKa of acetic acid is 4.76 and the desired pH is 5.00. We can rearrange the equation as follows:

[A-]/[HA] = 10^(pH - pKa)

Substituting the values, we get:

[A-]/[HA] = 10^(5 - 4.76) = 1.74

You have correctly calculated [HA] (acetic acid) to be 0.075 moles.

Using the ratio of [A-]/[HA] = 1.74, we can find the moles of sodium acetate (NaCH3COO). Let's call this value x:

x = 1.74 * 0.075

Now, we need to convert this moles of sodium acetate to volume using the molarity of the NaOH solution (3.00 M). The balanced equation for the reaction of acetic acid with sodium hydroxide is:

CH3COOH + NaOH → CH3COONa + H2O

From the equation, we can see that the mole ratio of sodium acetate to sodium hydroxide is 1:1. Therefore, the moles of sodium hydroxide needed is also x.

Finally, using the definition of molarity (M = moles/volume), we can rearrange the equation to solve for volume:

Volume of NaOH solution = Moles of NaOH / Molarity of NaOH

Volume of NaOH solution = x / 3.00

Therefore, the volume of 3 M NaOH solution needed is x / 3.00.

Now you can substitute the value of x that you calculated earlier to find the final answer for question 2.

To calculate the quantities needed for the second step in the buffer preparation, you are on the right track with the Henderson-Hasselbalch equation. However, there are a couple of adjustments you need to make to arrive at the correct answer.

First, let's define the problem:

You are asked to prepare a 500 mL buffer solution with a concentration of 0.150 M and a pH of 5.00. You already calculated that you need 4.5 grams of acetic acid for the first step.

Now, let's move on to the second step:

1. Calculate the ratio of acid to conjugate base. You correctly used the Henderson-Hasselbalch equation and found the ratio to be 1.74.

2. Calculate the moles of acetic acid needed. You calculated earlier that you need 0.075 moles of acetic acid.

3. Calculate the moles of conjugate base needed. Multiply the moles of acetic acid by the ratio you found.
Moles of conjugate base = 0.075 moles x 1.74 = 0.1305 moles

4. Now, we need to calculate the volume of 3.00 M NaOH solution needed to reach the desired pH.

Since NaOH is a strong base, it will react with acetic acid in a 1:1 ratio. The moles of NaOH required will be the same as the moles of conjugate base needed.

Moles of NaOH = 0.1305 moles

To find the volume of the 3.00 M NaOH solution needed, we can use the following equation:

Moles of solute = Molarity x Volume (in liters)

Rearranging the equation to solve for volume gives us:

Volume (in liters) = Moles of solute / Molarity

Volume (in liters) = 0.1305 moles / 3.00 M = 0.0435 L

5. Convert the volume from liters to milliliters:

Volume (in milliliters) = 0.0435 L x 1000 mL/L = 43.5 mL

Therefore, you will need approximately 43.5 mL of 3.00 M NaOH solution in order to achieve a pH of 5.00 in your acetate buffer.

Remember, it's always important to double-check your calculations and ensure that your assumptions are correct.

You want 0.075 mols to start HAc. You're right there (and the 4.50 g is correct).

5.00 = 4.75 + log (b/a)
b/a = 1.82 or
b = 1.82*a or
a = 0.549*b. Substitute this into the below for a.

b+ a = 0.15 x 0.5 = 0.075 mols.
b + 0.549b = 0.075
1.549b = 0.075
b = 0.0484 mols.
3.0M = 0.0484 mols/L
and L = 0.0161 L or 16.1 mL.
Check my arithmetic.