Chemistry with buffers?

A beaker with 100 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 7.30 ml of a 0.360 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Chemistry with buffers?
pH = pKa + log (base/acid)



5.00 = 4.76 = log (x/1)



I want the ratio of base to acid which is why I used x/1



log x = 0.24



x = 1.74



my base:salt ratio is 1.74:1



let x = moles of acid



1.74x + x = 0.01 total moles of solute



x = 0.00365 moles (this is the acid)

base = 0.00635 mol



Add strong acid, it protonates the base



(0.360 mol/L) (0.00730 L) = 0.002628 mol of strong acid added



base: 0.00635 - 0.002628 = 0.003722

acid: 0.00365 + 0.002628 = 0.006278



pH = 4.76 + log(0.003722/0.006278)



pH = 4.53



Note: I used moles in that last H-H expression rather than using the new volume of 107.3 mL to get molarities. Since there would be a 0.1073 L in the numerator and the denominator, I just skipped that step.



Good problem. Check my math carefully and talk the solution through with your study group. Best wishes.



Update:



http://www.chemteam.info/AcidBase/Hender



problem #7