A chemist wishes to prepare 250mL of a buffer that is pH = 4.50. Beginning with 100mL of 0.12 mol L^(-1) acetic acid and a supply of 0.10 mol L^(-1) NaOH, explain how this could be done. How much 0.20 mol L^(-1) NaOH must be added to this buffer to raise the pH to 5.1? If the same amount of 0.20 mol L^(-1) NaOH were added to 250 mL of deionized water, what would the new pH be?

I couldn't guess.

How many millimoles acetic acid (HAc) do you have? That's 100 mL x 0.12M = 12.
How much NaOH must be added? That's 0.1 M * x mL = 0.1x

.........HAc + OH^- ==> Ac^- + H2O
initial..12.....0........0.......0
add............0.1x...............
change.-0.1x..-0.1x......0.1x....0.1x
equil..12-0.1x..0.......0.1x.....0.1x

pH = pKa + log(base)/(acid)
4.50 = pKa + log(0.1x)/(12-0.1x)
Solve for x = mL 0.1M NaOH to be added to form buffer of pH = 4.50.

I will leave the last two parts for you; the last one is strictly a strong base problem. The other one follows this same kind of procedure using the Henderson-Hasselbalch equation.
Post your work if you get stuck.

uno for the pKa would it simply be 4.50 as well?

No. It's the pKa for acetic acid. Ka = 1.8E-5. pKa = -log Ka.

To prepare a buffer with a pH of 4.50, a chemist can use a mixture of acetic acid (CH3COOH) and sodium acetate (CH3COONa). Here's how it can be done:

Step 1: Calculate the moles of acetic acid needed.
To prepare 250 mL of buffer, we need to first calculate the moles of acetic acid required using the equation:
moles = concentration (mol L^(-1)) x volume (L)
moles = 0.12 mol L^(-1) x 0.1 L (100 mL/1000 mL)
moles = 0.012 mol

Step 2: Calculate the volume of sodium acetate needed.
To create a buffer, we need equal moles of acetic acid and sodium acetate. Since we have already calculated the moles of acetic acid, the volume of sodium acetate can be calculated using the equation:
volume (L) = moles / concentration (mol L^(-1))
volume (L) = 0.012 mol / 0.10 mol L^(-1)
volume (L) = 0.12 L (120 mL)

To prepare the buffer, mix 100 mL of 0.12 mol L^(-1) acetic acid with 120 mL of 0.10 mol L^(-1) sodium acetate, and then dilute it to a final volume of 250 mL with deionized water. Ensure thorough mixing.

To calculate how much 0.20 mol L^(-1) NaOH is needed to raise the pH to 5.1, we first need to determine the mole ratio between acetic acid (CH3COOH) and sodium acetate (CH3COONa) using the Henderson-Hasselbalch equation.

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

Given that pKa for acetic acid is 4.75, we can use the following equation to calculate the ratio:

ratio = 10^(pH - pKa)
ratio = 10^(5.1 - 4.75)
ratio = 1.995

This indicates that the ratio of [A-] (sodium acetate) to [HA] (acetic acid) should be approximately 1.995 (or 2 for practical purposes).

To raise the pH, we need to add sodium hydroxide (NaOH) to convert some of the acetic acid to sodium acetate. However, we need to maintain the ratio of approximately 2. Therefore, for every 1 mole of acetic acid we convert to sodium acetate, we will need 2 moles of sodium hydroxide.

Step 3: Calculate the moles of sodium hydroxide needed.
moles of sodium hydroxide = (moles of acetic acid) x 2
moles of sodium hydroxide = 0.012 mol x 2
moles of sodium hydroxide = 0.024 mol

Step 4: Calculate the volume of 0.20 mol L^(-1) NaOH needed.
volume (L) = moles / concentration (mol L^(-1))
volume (L) = 0.024 mol / 0.20 mol L^(-1)
volume (L) = 0.12 L (120 mL)

Therefore, to raise the pH to 5.1, approximately 120 mL of 0.20 mol L^(-1) NaOH should be added to the buffer.

If the same amount of 0.20 mol L^(-1) NaOH were added to 250 mL of deionized water, the new pH would be calculated using the Henderson-Hasselbalch equation with the new concentration of acetic acid.

First, calculate the moles of acetic acid:
moles = concentration (mol L^(-1)) x volume (L)
moles = 0.12 mol L^(-1) x 0.1 L (100 mL/1000 mL)
moles = 0.012 mol

Then, calculate the new concentration of acetic acid in the solution after adding NaOH:
new concentration = moles / new volume (L)
new concentration = 0.012 mol / 0.37 L (250 mL/1000 mL)
new concentration = 0.0324 mol L^(-1)

Now, calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 4.75 + log(0.0324/0.012)

Using a logarithmic calculator, the new pH is approximately 4.89.

Therefore, adding the same amount of 0.20 mol L^(-1) NaOH to 250 mL of deionized water would result in a new pH of approximately 4.89.