buffer question?

I suppose the volume of acetic acid should be "100.0 mL" instead of "100.0 L".

Molar mass of CH₃COOK = (12.0×2 + 1.0×3 + 16.0×2 + 39.0) g/mol = 98.0 g/mol
No. of moles of CH₃COOK in the original buffer = (8.203 g) / (98.0 g/mol) = 0.0837 mol
No. of moles of CH₃COOH in the original buffer = (1.00 mol/L) × (100.0/1000 L) = 0.1 mol
In the original buffer, [CH₃COO⁻]/[CH₃COOH] = [CH₃COOK]/[CH₃COOH] = 0.0837/0.1 = 0.837

The equation for the dissociation of CH₃COOH:
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …… pKa
Henderson-Hasselbalch equation:
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
pH = pKa + log(0.837)…… [1]

Let V L be the volume of the 0.500 M NaOH added.

The equation for the reaction between NaOH added and CH₃COOH.
CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l) …… pKa
No. of moles of OH⁻ added = No. of moles of NaOH added = (0.500 mol/L) × (V L) = 0.5L mol
No. of moles of CH₃COOH after addition of NaOH = (0.1 - 0.5V) mol
No. of moles of CH₃COO⁻ after addition of NaOH = (0.0837 + 0.5V) mol
[CH₃COO⁻]/[CH₃COOH] after addition of NaOH = (0.0837 + 0.5V)/(0.1 - 0.5V)

pH will rise when NaOH is added.
Hence, pH after addition of NaOH = pH + 1

Apply Henderson-Hasselbalch equation again:
pH + 1 = pKa + log{(0.0837 + 0.5V)/(0.1 - 0.5V)} …… [2]

[2] - [1]:
1 = log{(0.0837 + 0.5V) / (0.1 - 0.5V)} - log(0.837)
log{(0.0837 + 0.5V) / [(0.1 - 0.5V) × 0.837]} = 1
(0.0837 + 0.5V) / [(0.1 - 0.5V) × 0.837] = 10
0.0837 + 0.5V = (0.1 - 0.5V) × 8.37
0.0837 + 0.5V = 0.837 - 4.185V
4.685V = 0.7533
V = 0.161

Volume of NaOH added = 0.161 L = 161 mL

8.203g of potassium acetate with 100.0L of 1.00M acetic acid
Are you sure about 100.0L - or is it 100.0mL I do not want to do a lot of work for nothing - 100.0L and 8.203g of potassium acetate seems a very low concentration compared to the 1.0M acetic acid.