You should have told me what Ka values you are using. Not all texts have the same values although they are close. I'm using 1.77E-4 for formic acid and I will call that HF. I know that isn't formic acid and you know that, too, but it saves some space on the line. HAc is acetic acid and I'm using Ka for HAc of 1.8E-5.
............HF ==> H^+ + F^-
initial....0.185...0......0
change......-x......x.....x
equil.....0.185-x...x.....x
1.77E-4 = (x)(x)/(0.185-x)
I'm not going to do this step by step but this is the way you set it up. You should get an answer for x = 0.00572 if you assume 0.185-x = 0.185.
.............HAc ==> H^+ + Ac^-
initial......0.225....0.....0
change.......-x.......x......x
equil.....0.225-x.....x.......x
Ka = 1.8E-5 = (0.00572+x)(x)/(0.225-x) and solve for x
(Note:I suspect you didn't substitute the 0.00572 here. That's a common error.)
If I assume 0.00572+x = 0.00572 and 0.225-x = 0.225, then x = 7.2E-4
Then 0.00572 + 7.2E-4 = ? and -log of that is 2.19. Voila!.
That's all I did earlier in the day when I first responded to your post. You may ask what happens if we don't make those assumptions so here is what you get.
For formic acid, x = 0.00563 instead of 0.00572 (hardly worth talking about).
For acetic acid, x = 0.000644 (again, not much difference)
So 0.00563 + 0.000644 = 0.00627 and the pH =2.20
Let me know if you don't understand what I did. The only thing I've omitted is the algebra.