Chemistry
posted by George .
A 50.0 mL sample of 0.12 M formic acid, HCOOH, a weak monoprotic acid, is titrated with 0.12 M NaOH. Calculate the pH at the following points in the titration. Ka of HCOOH = 1.8 multiplied by 104.
What is the pH when
50.0 mL NaOH is added
60.0 mL NaOH is added
70.0 mL NaOH is added

George, how much of this do you know how to do? This is a lot of work for me just to check your answers and I don't want to work these if you already know how. If you have worked these, post your work and I'll be happy to check your answers. I can do that much faster than I can type.

I'm actually not sure how to do any of it.

To do zero mL base.
HCOOH ==> H^+ + HCOO^
Prepare an ICE chart and substitute into the Ka expression.
Ka = (H^+)(HCOO^)/(HCOOH)
Solve for (H^+) and convert to pH.
For all of the others. The equation is
HCOOH + NaOH ==> HCOONa + H2O
1. First determine where the equivalence point is; that is, how many mL NaOH must be added to arrive at the equivalence point. For all mL BEFORE the equivalence point to the following:
2. Determine mole HCOOH you have initially. M x L = moles.
3. Add moles NaOH. M x L = moles
4. An ICE chart is the easiest to use but use information from 2 and 3 to determine the amount of salt (HCOONa) formed and the amount of the acid (HCOOH) remaining unreacted. Plug those values into the Ka expression and solve for H^+, then convert to pH.
At the equivalence point, the pH is determined by the hydrolysis of the salt, HCOONa. It's the HCOO^ part that is hydrolyzed.
HCOO^ + HOH ==> HCOOH + OH^
Set up an ICE chart for this and plug into the following.
Kb = (Kw/Ka) = (HCOOH)(OH^)/(HCOO^)
YOu know Kw and Ka (for formic acid), (HCOOH) = (OH^) = x and (HCOO^, the salt) you know from the equivalence point work you did. Solve for x = (OH^) and convert to pH.
For everything past the equivalence point the pH is determine by the excess of NaOH added past the e.p.
Post your work if you get stuck.