Change in pH Question!!?

Question

Ok so my teacher really confused me when going over this in class! Please help, I have no clue! Please show your steps. Any help appreciated!

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.77-M HCl to 400. mL of each of the following solutions.
Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.
a) water
pH before mixing =
pH after mixing=
pH change =

b) 0.193 M C2H3O21-
pH before mixing =
pH after mixing=
pH change =

c) 0.193 M HC2H3O2
pH before mixing =
pH after mixing=
pH change =

d) a buffer solution that is 0.193 M in each C2H3O21- and HC2H3O2
pH before mixing =
pH after mixing=
pH change =
Link to acid base table: http://www.webassign.net/wertz/caqs/html/resource/acidbase_pka.html

Answer 1

I will help you on some, but you need to figure out the rest, especially all the pH change. OK?
For ilustration purpose, let us assume that there is a state after mixing but before reaction. The new HCl concentration is:
10.0*2.77/(10.0+400) = 0.0676 (M)
This data will be used in each case.

a) You should know pH=7.00 for pure water. hence:
pH(before) = 7.00
HCl completely dissociate in the solution to H+ and Cl-. Thus
pH(after) = -log([H+]) = -log(0.0676)

b) pKa for acetic acid (HC2H3O2) is 4.76. That means pKb for C2H3O2- is 14 - 4.76=9.24 and Kb = 10^(-9.24) for the following reaction:
C2H3O2- + H2O <==> HC2H3O2 + OH-.
[OH-] = sqrt(0.193*10^(-9.24)) = 1.054x10^-5 (M)
pH(before) = 14+log([OH-]) = 14+log(1.054x10^-5) =

The new C2H3O2- concentration after mixing is:
400*0.193/(10.0+400) = 0.188 (M)
Thus after nuetralization, we have:
[HC2H3O2] = 0.0676M
[C2H3O2-] = 0.188M-0.0676M = 0.121M
Since pKa for acetic acid (HC2H3O2) is 4.76:
HC2H3O2 <==> H+ + C2H3O2-, pKa = 4.76
pH(after) = pKa + log([C2H3O2-]/[HC2H3O2])
= pKa + log(0.121/0.0676) =

c) If you now fully understand Part (b), we can go fast:
[H+] = sqrt(0.193*10^-4.76) = 0.00183M
pH(before) = -log([H+]) = -log(0.00183) =

You may roughly approximate pH(after) as that in Part (a). To be more regorous, we do the following:
Let [C2H3O2-] simply be [X].
So for reaction:
HC2H3O2 <==> C2H3O2- + H+
0.188-[X]------------[X]-------0.0676+[X]
Ka = 10^-4.76 = [X]*(0.0676+[X])/ (0.188-[X])
Solve this quadratic equation to get [X], and [H+] = 0.0676+[X]. Finally use: pH = -log([H+]).

d) Try to do it yourself.