Question

If 25.00 mL of 0.20 mol/L HCO2H(aq) is titrated with 0.20 mol/L NaOH(aq) (the titrant), determine the pH
a) befor titration began

b)after 10.00 mL of NaOH has been added

c) at the equivalence point

Answer 1

a) before titration begins. That is a 0.2 M solution of HCOOH.
HCOOH ==> H^+ + HCOO^-

(HCOOH) = 0.2-x
(H^+) = x
(HCOO^-) = x
Ka = (H^+)(HCOO^-)/(HCOOH)
Plug the above variables into Ka and solve for (H^+), then convert to pH.

b) after 10.00 mL NaOH has been added.
HCOOH + NaOH ==> HCOONa + H2O

mols HCOOH initially = M x L = ??
mols NaOH added = M x L = ??
mols HCOONa formed = ??

pH = pKa + log[(base)/(acid)]

c)at the equivalence point. We have the salt HCOONa + H2O at the equivalence point.
HCOO^- + HOH ==> HCOOH + OH^-
Kb = Kw/Ka = (HCOOH)(OH^-)/(HCOO^-)
(HCOOH) = x
(OH^-) = x
(HCOO^-) = 0.2 - x
Plug into Kb and solve for x = (OH^-), then convert to pH.

Post your work if you get stuck.
Check my thinking. Check my work. Check my arithmetic.

Answer 2

a) Ka = (H^+)(HCOO^-)/(HCOOH)
= (x)(x)/(0.2-x)
= (x^2) / 0.2 - x

Next step please...I Hate these problems :P:)