# Chemistry

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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 10-4.
What is the pH when

• Chemistry -

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.

• Chemistry -

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

• Chemistry -

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.

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