Adding a strong acid to a buffer

A beaker with 195 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 0.1 M. A student adds 8.80 mL of a 0.320 \it M \rm HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76. I have such a hard time with this stuff PLEASE HELP!!!!Adding a strong acid to a buffer
the total mmoles of th base and the acid form must total to 195mL x 0.1M = 19.5 mmoles

pH = pKa + log base /acid



5 = 4.76 + log base /19.5 - base

0.24 = log base/19.5 -base

1.737 = base/19.5-base solving for base 33.887=2.737Base

base = 12.38 mmoles acid =19.5 -12,38 = 7.12 mmoles



check



pH = 4.76 + log 12.38/7.12 = 4.76 + 0.24 = 5



adding 8.8 mL of 0.32 M acid is adding

8.8mL x 0.32M = 2.816 mmoles adding 2.816 mmoles of acid will reduce the amount of the base in the numerator by 2.816 and add that same amount to the denominator so now

the pH = 4.76 + log (12.38-2.816)/7.12 + 2.816 = 4.76 + - 0.0166 = 4.74Adding a strong acid to a buffer
First you need to find the equilibrium concentrations of acetic acid and acetate. Use this equation:

pH = pKa + log(base/acid)



Let x be the concentration of acid at equilibrium.

So .1 - x is the equilibrium concentration of acetate



5.00 = 4.76 + log[(.1 - x)/x]

x = .0365M



Now we know the concentrations of all the parts of the buffer:

Acetic acid is .0365M

Acetate is .0635M

Hydronium is 10^-5M (remember that the pH = -log[hydronium]



Now multiply the concentrations by .195L to find out how many moles of each there are.

Acetic acid is 7.1175x10^-3mol

Acetate is .01238mol

Hydronium is 1.95x10^-6



Now add the total moles of hydronium from the HCl

Since HCl is a strong acid, it completely dissolves, so the total moles of HCl equals the total moles of hydronium it yields.

.320M x .0088L = 2.816x10^-3mol

1.95x10^-6 + 2.816x10^-3mol = 2.818x10^-3mol



Next find the new concentrations of all the reactants. Remember the new volume is .195 + .0088 = .2038L

Acetic acid .0349M

Acetate .0607M

Hydronium .0138M

Now use the pKa to find the Ka

4.76 = -logKa

Ka = 1.7378x10^-5

This solution isn't at equilibrium. But we know that a certain amount of acetate and hydronium will react to form acetic acid until it is. Let that amount be x.

1.7378x10^-5 = (.0138 - x)(.0607 - x)/(.0349 + x)

x = .013782M



.0607 - .01378 = .04692M

-log.04692 = 1.33