[blightcutter]

Just getting into calculating these titrations. I have a 50mL solution that contains 0.10M HClO₄ and 0.08M HCO₂H (Ka = 1.8x10^-4) that was titrated with a solution of 0.20M KOH. I need to calculate the pH of the solution after the addition of 25.0mL of the base solution (KOH).

How do I want to approach this?;
0.18M(0.050L)=
0.20M(0.025L)=
That will give me moles...Then do I just use the Ka to calculate the final pH? This one has me confused for some reason because I can't seem to get the right answer to come out (which is pH=2.5)


Another question I have is on determining an unknown amount of something by titrating. In this example a 0.6334g sample that contains an unknown amount of HgO (MW = 216.6g/mol) was dissolved in water that contained KI.

so, HgO+4I^-+H₂O  HgI₄^-+2OH^-

In order to determine the amount of OH^- present the solution was titrated with 0.1178M HCl. The end point was reached after the addition of 42.6 mL of the acid solution. What percent of the original sample is HgO?

So far I have the moles of the acid = 0.1178M(0.0426L)
And the titration is finished so that means that there is no more OH left. If that even matters. Anyway, the answer is 85.8%.

I'll keep trying these but any help you guys can provide would be nice.

[blightcutter]

 
Ok, so for the HgO problem here is what I did to get the answer,
First of all I forgot about the stoichiometric ratio...   Ya, makes a a lot of sense now.

           (0.0426 x 0.1178 x (1/2) ) x 216.6 g HgO
%HgO =                           .6334 g sample 

x 100 = 85.8%

...the first one is still giving me problems though           

[AWK]

Write down a balanced reactions for neutralization of HClO₄ and HCOOH separately, then calculate volumes of 0.20 M KOH neded for each acid. Note - neutralize a strong acid first.
There are a few possibilities:
1.HCLO₄ is more then KOH
2. HCLO₄ is exactly enough for neutralization of KOH - then you have a weak acid problem
3. HCOOH is neutralized partially  - then you have a buffer problem
4. The whole amount of HCOOH is neutralized - then you have a hydrolysis problem
5. You have an excess of KOH - then you have a strong base problem

[blightcutter]

Thanks for the input AWK. I can see from the problem that HCOOH is my strong acid, also looking over the volumes I see that the titration has reached the first equivalence point (after titration). so,
After the moles HCOOH are calculated (0.004M) I can say,
[H₃O⁺] = SQRT(1.8x10^-4*(0.004M/0.075L))

Then pH=-log [H₃O⁺]
        pH=2.5

Yay!...unless I totally missed something.