Question

1. Describe how you determined the pK_afor the plots involving a weak acid or base.

2. Why is the pH at the equivalence point different when a strong acid is titrated by strong base as opposed to when a weak acid is titrated with strong base?

3. Why is the titration curve nearly flat at the end of the titration, well past the equivalence point?

4. For the titrations involving weak acids or weak bases, at what pH value is the buffering capacity (resistance of pH change to the addition of both acid and base) of the solution at a maximum? Why is this?

5. Calculate the theoretical initial pH for 0.04 M acetic acid, HC₂H₃O₂.

Answer 1

1. Describe how you determined the pKa for the plots involving a weak acid or base.

the pKa = pH in the HALF-EQUIVALENCE point; since in this point the equation

pH = pKa + log(conjugate/acid)

and conjugate=Acid os log(1) = 0

then pH = pKa

2. Why is the pH at the equivalence point different when a strong acid is titrated by strong base as opposed to when a weak acid is titrated with strong base?

because strong acids/bases titration will yield a pH = neutral, whereas the pH of a storng acid with weak base, will show a conjguate, i.e. HB+ that will form B + H+ ions, that is, aslightly acidic (and basic for the weak acid)

3. Why is the titration curve nearly flat at the end of the titration, well past the equivalence point?

Because the pH changes are no longer going to change with respect to the large amount of volume, no conjguate is formed

4. For the titrations involving weak acids or weak bases, at what pH value is the buffering capacity (resistance of pH change to the addition of both acid and  base) of the solution at a maximum? Why is this?

the max capacity is when

pH = pKa

since then you have a max ratio os conjugate/acid

5. Calculate the theoretical initial pH for 0.04 M acetic acid, HC2H3O2.

Ka = [H+][A-]/[HA]

1.8*10^-5 = (x*X)(0.4-x)

x= 8.39*10^-4

pH =-log(8.39*10^-4) = 3.076