Calculate the pH of a 100mL solution of 0.10M acetic acid ka=1.8x10(-5)

Question

i calculated the pH to be 2.87

from the square root of
(1.8x10(-5) * 0.1, the negative log gives me the pH.

The next question wants you to calculate the pH with 50mL HCl added. I found the pH to be 2.78 by just adding .05 to the bottom [HA] value are these correct so far?

Answer 1

2.87 is correct for the first one.
For #2, you don't provide the (HCl). Although (HCl) is not listed; I suspect your answer of 2.78 is not right. Assume (HCl) is 0.1M and you add 50 mL so that is 5 millimoles/150 mL = 0.0333M HCl.
Since HCl is 100% ionized (and acetic acid is only about 1% ionized) most of the H^+ will be contributed by HCl and pH = 1.48.
If (HCl) is 0.01M, we add 0.5 millimole HCl and that in 150 mL = 0.00333 which is 2.48 for pH.