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?

See your other post above.

To calculate the volume of 0.5M NaOH required to make the buffer solution, we need to consider the moles of benzoic acid and sodium benzoate present in the solution.

1. Start by calculating the moles of benzoic acid present in the 50mL of 0.1M solution:
Moles of benzoic acid = Concentration (M) x Volume (L)
Moles of benzoic acid = 0.1 mol/L x 0.050 L
Moles of benzoic acid = 0.005 mol

2. Next, calculate the moles of sodium benzoate required to form the 0.05M benzoate buffer at a pH of 4.5. To do this, we'll use the Henderson-Hasselbalch equation:

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

Since the pH and pKa are given, we can rearrange the equation to solve for the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)

[A-]/[HA] = 10^(4.5 - 4.2) (assuming the pKa of benzoic acid is 4.2)
[A-]/[HA] = 2.51189

Now, let's assume x moles of benzoic acid have reacted with x moles of NaOH to form the buffer. This means that x moles of benzoic acid have turned into x moles of sodium benzoate.

3. Equating the moles of benzoic acid to sodium benzoate:
0.005 mol benzoic acid = 0.005 mol sodium benzoate

Since [A-]/[HA] = 2.51189, we can write:
2.51189 = (0.005 + x) / x

Solving for x:
2.51189x = 0.005 + x
2.51189x - x = 0.005
1.51189x = 0.005
x = 0.003308 mol

4. Now that we know the moles of NaOH needed to form the sodium benzoate, we can calculate the volume using the concentration of NaOH:
Volume of NaOH = Moles of NaOH / Concentration (M)
Volume of NaOH = 0.003308 mol / 0.5 mol/L
Volume of NaOH = 0.006616 L

Since we need the volume in mL, we multiply by 1000:
Volume of NaOH = 0.006616 L x 1000 mL/L
Volume of NaOH = 6.616 mL

So, 6.616 mL of 0.5M NaOH must be added to 50 mL of 0.1M benzoic acid to make 100 mL of 0.05M benzoate buffer at pH 4.5.