Chemistry Help with Adding a Strong Acid to a Buffer-trying to find delta pH.?

Okay so here is the problem:



A beaker with 155 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 9.00 mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. What is the delta pH? (change in pH).





What I initially thought--

H C2H3O2 --%26gt; H+ %26amp; (C2H3O2)-1

I .005549____________.009951

C +.00419 ____________-.00419

E .009739____________.005761

(btw I got .00419 by adding .009 and .410 then dividing by 1000)



then I used this formula:

1.74x10^-5= x(.005761/.009739)

x=[H+]= 2.94148x10^-5

-log(2.94148x10^-5)= 4.5313

therefore the pH=4.53

since they were asking for the change of pH I subtracted

5.00-4.53 = .4686

and apparently that is INCORRECT....



ssooo.....I dont know what I did wrong

PLEASE help me, and give step my step instructions as well

Because I'm totally lost now...Chemistry Help with Adding a Strong Acid to a Buffer-trying to find delta pH.?
Let the initial concentrations be a and b(of acid and conjugate base res.)



5 = 4.76 + log(b/a)



So b/a =1.738.



And b + a = 0.1



Solving the two, b = 0.0365 and a = 0.0635



milliMoles of acid = 0.0635*155 = 9.8425



milliMoles of conjugate base = 5.6575



On addition of .410*9 = 3.69 milliMoles HCl, the acid = 9.8425 + 3.69= 13.5325 and base = 5.6575 - 3.69 = 1.9675.



So the pH = 4.76 + log(1.9675/13.5325) = 4.76 - 0.8375



So change in pH is 5 - 4.76 + 0.8375 = 1.0775(approx.)



There might have been some calculation mistake please check the calculations.