What volume of 0.5M NaOH must be added to 50mL of .0.1M benzoic acid to make 100.0 mL of 0.05 M benzoate buffer that is pH 4.5?

When you post a problem like this in the future you should include the pKa; otherwise, the answer I get won't be the same as the answer you get since texts differ in tabular values for Ka, Ksp, etc.

Let HB = benzoic acid

I will use pKa 4.2 but you should use the value from your text/notes.

You have 50 mL x 0.1M HB = 5 millimols.

First calculate millmoles acid and base needed.
pH = pKa + log (base)/(acid)
4.5 = 4.2 + log b/a
0.3 = log b/a
b/a = 2 or
base = 2*acid. This is equation 1(two unknowns)
The second equation is
acid + base = 100*0.05 = 5 millimols.

Solve those two equation simultaneously for a and b. I get approximately 1.67 for a and 3.33 for b. Plug that into the following:
.........HB + NaOH ==> NaB + H2O
I........5.....0........0.....0
add............x...............
C.......-x....-x........x.....x
E.......5-x....0........x

So you want x to be NaB(benzoate) or base and above base = 3.33
Therefore, you want to add 3.33 millimoles NaOH. Convert to mL by
M = millimols/mL. YOu know mmols and M, solve for mL. I ran through this quickly and came up with approx 7 mL of the 0.5M NaOH but you should confirm that AND remember that's an estimate and not an exact number.
I always like to check these to make sure of no error; therefore, I recommend you make an ICE chart as above, add the value you get (my estimate of 7 mL) and calculate the values for acid and base, plug these into the HH equation and see if you end up with a pH of 4.5. Post your work if you get stuck.

To find the volume of 0.5M NaOH required to make the buffer, we can follow these steps:

Step 1: Determine the number of moles of benzoic acid in the 50mL solution.
Given that the concentration of benzoic acid is 0.1M and the volume is 50mL, we can calculate the number of moles using the formula:

moles = concentration (M) × volume (L).

Converting the volume to liters:
50 mL = 50 ÷ 1000 = 0.05 L.

Now, the number of moles of benzoic acid:
moles of benzoic acid = 0.1 M × 0.05 L = 0.005 moles.

Step 2: Calculate the number of moles of benzoate required to make the buffer.
Since the buffer concentration is 0.05M and the total volume is 100.0mL, we can calculate the number of moles using the same formula as before:

moles = concentration (M) × volume (L).

Converting the volume to liters:
100.0 mL = 100.0 ÷ 1000 = 0.1 L.

Now, the number of moles of benzoate required:
moles of benzoate = 0.05 M × 0.1 L = 0.005 moles.

Step 3: Determine the change in moles of benzoic acid caused by the reaction.
The change in moles equals the difference in moles of benzoate and benzoic acid:

change in moles = moles of benzoate - moles of benzoic acid.

change in moles = 0.005 moles - 0.005 moles = 0 moles.

Step 4: Calculate the moles of OH- ions required to react with the benzoic acid.
The benzoic acid reacts with NaOH in a 1:1 stoichiometry. Therefore, the number of moles of NaOH required will be the same as the change in moles of benzoic acid:

moles of NaOH = change in moles = 0 moles.

Step 5: Find the volume of 0.5M NaOH required to obtain the desired number of moles.
Using the formula:

moles = concentration (M) × volume (L),

we can rearrange it to find the volume:

volume (L) = moles / concentration (M).

Plugging in the values:
volume (L) = 0 moles / 0.5 M = 0 L.

Therefore, no volume of 0.5M NaOH is required to be added to the solution to make a 0.05M benzoate buffer at pH 4.5.