Question

A weak acid was HA (pKa=5.00) was titrated with 1.00M KOH. The acid solution had a volume of 100ml and a molarity of 0.100M. Find the pH at the following volumes of base added Vb= 0, 1, 5, 9, 9.9, 10, 10.1 and 12.

Answer 1

Without any KOH added, this weak acid undergoes self dissociation:
HA ? H+ + A-, pKa = 5.00
hence [H+] ? [A-]
Ka = 10^-5.00 = [H+]*[A-]/[HA] ? [H+]^2/0.100
pH = -log([H+]) = -0.5*log(Ka*0.100) = 3.00

2. I would not solve all the pH for you and leave you nothing to practice. Instead, I will just arbitrarily pick one case, say, 5ml base added, and you SHOULD solve the rest (base added: 1, 9, and 9.9) by following exactly the same method.
To start with, we have: 0.100M*0.1000L = 0.01 mol of HA. By adding 5ml of KOH, that is 1.00M*0.005L = 0.005 mol KOH, the following reaction takes place:
HA + KOH == KA + H2O, to use up 0.005 mol of HA. We thus have 0.005 mol of HA left and have 0.005 mol A- (from KA) formed. Now:
Ka = 10^-5.00 = [H+]*[A-]/[HA] = [H+]*0.005/0.005 = [H+]
pH = -log([H+]) = 5.00

3. When added KOH is 10ml, or 1.00M*0.01L = 0.01 mol KOH. The reaction between KOH and HA will use up all HA to form 0.01 mol A-. A- may react with water:
A- + H2O ? HA + OH-
that is to say, [HA] is approximately the same as [OH-]. Hence:
Ka = 10^-5.00 = [H+]*[A-]/[HA]
= [H+]*[OH-]*[A-]/[HA]*[OH-] = Kw*[A-]/[OH-]^2
Thus [OH-]^2 = (Kw/Ka)[A-], and
pH = 14+log([OH-]) = 14+0.5*log((10^-14/10^-5)*0.01) = 8.5

4. Again for cases 10.1 and 12ml of KOH, I would just arbitrarily pick one case, say, 12ml base added, and you SHOULD solve the rest (base added: 10.1ml) by following exactly the same method.
To start with, we have 0.01 mol of HA. By adding 12ml of KOH, that is 1.00M*0.012L = 0.012 mol KOH, the reaction:
HA + KOH == KA + H2O, uses up all HA, and still leave 0.002 mol KOH un-neutralized. Now:
pH = 14+log([OH-]) = 14+log(0.002) = 11.3