How do you solve this chemistry problem? Need HELP?

Can someone please solve this chemistry problem for me? I really need help?

• A beaker with 1.90×10^2 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.40mL of a 0.480 M HCl solution to the beaker. How much will the pH change? The pka of acetic acid is 4.760. and the answer is not -0.202 or -0.492 thank you so so so much if you can help me, I really appreciate it!

• Answer:

  First we need to figure out how mnay moles of each species there are before adding the acid. Buffer problems use the henderson-Hasselbalch equation: pH = pKa + log ([base]/[acid]) . . .where [base] = [acetate] and [acid] = [acetic acid]. 5.000 = 4.760 + log ([acetate]/[acetic acid]) 0.240 = log ([acetate]/[acetic acid]) ([acetate]/[acetic acid]) = 10^0.240 = 1.74 [acetate] = 1.74[acetic acid] We also know that [acetate] + [acetic acid] = 0.100 . . .substituting 1.74[acetic acid] for [acetate] gives us 1.74[acetic acid] + [avcetic acid] = 0.100 2.74[acetic acid] = 0.100 [acetic acid] = 0.0365 M . .and [acetate] = 0.100 - 0.0365 = 0.0635 M. In 190 mL of solution, initial moles acetate = M acetate x L of solution = (0.0635)(0.190) = 0.0121 moles acetate initial moles acetic acid = M acetic acid x L of solution = (0.0365)(0.190) = 0.00694 moles acetic acid moles HCl to be added = M HCl x L HCl = (0.480)(0.00840) = 0.00403 moles HCl HCl will react with the base part of the buffer (acetate) to produce acetic acid: Moles . . . . . . . . . .acetate + HCl ==> acetic acid + Cl- initial . . . . . . . . . .0.0121 . . .0.00403 . . . .0.00694 . . . .0 change . . . . . . .-0.00403 . .-0.00403 . . . .+0.00403 . .+0.00403 final . . . . . . . . . . .0.0081 . . . .0 . . . . . . . .0.0110 . . .0.00403 So after adding the HCl we now have [acetate] = 0.0081 moles / 0.198 L = 0.0409 M [acetic acid] = 0.0110 moles / 0.198 L = 0.0556 M pH = 4.760 + log (0.0081 / 0.0110) = 4.760 - 0.133 = 4.627 So the pH dropped by 0.133 units.