Chemistry - adding a strong acid to buffer, how will this change pH?

Let y M be the molarity of CH₃COOH in the original buffer.
Then, the molarity of CH₃COO⁻ in the original buffer = (0.100 - y) M

Consider the dissociation of CH₃COOH:
CH₃COOH() ⇌ CH₃COO⁻(aq) + H⁺(aq) …… pKa = 4.740

Henderson-hasselbalch equation :
pH = pKa + log([CH₃COO⁻]/[ CH₃COOH])
5.000 = 4.740 + log{(0.100 - y)/y}
log{(0.100 - y)/y} = 0.260
(0.100 - y)/y = 10⁰˙²⁶⁰
0.100 - y = 10⁰˙²⁶⁰y
(1 + 10⁰˙²⁶⁰)y = 0.100
y = 0.100 / (1 + 10⁰˙²⁶⁰) = 0.0355
[CH₃COOH] = 0.0355 M
[CH₃COO⁻] = (0.100 - y) M = 0.0645 M

When HCl added :
CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq)
Final volume of the solution = (2.00 × 10² + 6.20) mL = (200 + 6.20) mL = 206.2 mL
[CH₃COO⁻] = {0.0645 × (200/206.2) - 0.300 × (6.20/206.2)} M = 0.0535 M
[CH₃COOH] = {0.0355 × (200/206.2) + 0.300 × (6.20/206.2)} M = 0.0435 M

Consider the dissociation of CH₃COOH again:
CH₃COOH() ⇌ CH₃COO⁻(aq) + H⁺(aq) …… pKa = 4.740

Henderson-hasselbalch equation :
pH = pKa + log([CH₃COO⁻]/[ CH₃COOH]) = 4.740 + log(0.0535/0.0435) = 4.83

From the Henderson Hasselbalch equation
pH = pKa + log [salt]/[acid]
Before any addition you have 5.000 = 4.740 + log [salt]/[acid] and solving for [salt]/[acid]...
log [salt]/[acid] = 0.2600
[salt]/[acid] = 1.81
[salt] + [acid] = 0.1 so [salt] = 1.81[acid]
[acid] = 0.0356 M
[salt] = 0.0644 M
After addition of HCl:
6.20 ml x 0.300 mmole/ml = 1.86 mmoles HCl added
Final volume = 200 ml + 6.20 ml = 206.2 ml = 0.2062 L
Initial HAc = 0.0356 mol/L x 0.2 L = 0.00712 moles
Initial Ac- = 0.0644 mol/L x 0.2 L = 0.01288 moles
Final HAc = 0.00712 mol + 0.00186 mol = 0.00898 moles
Final Ac- = 0.01288 mol - 0.00186 mol = 0.01102 moles
Final [Ac-] = 0.01102 mol/0.2062 L = 0.0534 M
Final [HAc] = 0.00898 mol/0.2062 L = 0.0435 M
pH = pKa + log [Ac-]/[HAc]
pH = 4.740 + log (0.0534/0.0435) = 4.740 + 0.0891
pH = 4.829