Chemistry Help, please??

HCl is a strong acid which completely dissociates in aqueous solution to form H₃O⁺ ions.
Before the dissociation of CH₃COOH, the initial concentrations of the solution mixture :
[CH₃COOH]ₒ = (0.84 M) × [2.0 / (2.0 + 0.2)] = 0.764 M
[H₃O⁺]ₒ = (0.01 M) × [0.2 / (2.0 + 0.2)] = 0.000909 M

_______ CH₃COOH(aq) + H₂O(l) ⇌ CH₃COOH⁻(aq) + H₃O⁺(aq) …… Ka = 1.8 × 10⁻⁵
Initial: ___ 0.764 M ______________ 0 M ______ 9.09×10⁻⁴ M
Change: ___ -y M _______________ +y M _______ +y M
At eqm: _ (0.764 - y) M ____________ y M ___ (9.09×10⁻⁴ + y) M

Ka = [CH₃COO⁻] [H₃O⁺] / [CH₃COOH]
y (9.09×10⁻⁴ + y) / (0.764 - y) = 1.8 × 10⁻⁵
(9.09×10⁻⁴)y + y² = 1.38×10⁻⁵ - (1.8 × 10⁻⁵)y
y² + (9.27×10⁻⁴)y - 1.38×10⁻⁵ = 0
y = 3.28×10⁻³

pH = -log(9.09×10⁻⁴ + 3.28×10⁻³) = 2.38 ≈ 2.4 (to 2 sig. fig.)

Infinitesimal change in pH.....

What is the pH of the acetic acid?
CH3COOH(aq) <==> H+ + CH3COO^- .............. Ka = 1.8x10^-5

Ka = [H+][CH3COO-] / [CH3COOH]
1.8x10^-5 = x² / (0.84-x)
-- some algebra goes here --
x = 0.00354
[H+] = 0.00354
pH = -log[H+] = -log(0.00354) = 2.45

My gut feeling is that the change in pH will be less than the precision to which you can write the pH, which is to two decimal places.

Diluting the acetic acid with 0.2 mL of water...
2.0 mL x (0.84 mol / L) / 2.2 mL = 0.764 M

Ka = [H+][CH3COO-] / [CH3COOH]
1.8x10^-5 = x² / (0.764-x)
-- some algebra goes here --
x = 0.00338
[H+] = 0.00338
pH = -log[H+] = -log(0.00338) = 2.47

Adding HCl along with the extra water...
0.0002L x (0.010 mol H+ / 1L) = 2.0x10^-6 mol H+ added
This is indeed insignificant. Any minute change will be swamped by the lack of precision in calculating the pH.

Therefore, we can say that the final pH will be 2.47.

The numbers you have given in the problem are at best given to two significant digits. The Ka of acids are good to only two significant digits. Therefore, the pH is expressed to two significant digits, and those are the digits to the right of the decimal. When taking the log of a number, only the digits to the right of the decimal reflect the precision of the original number.