posted by Gagan .
calculate the pH at 0ml at the quivalence point and at 40ml in a titration of 25 ml of .120M formic acid with .105M NaOH
(kA of formic acid= 1.8x10^-4)
The key to this type problem is to recognize what is in the solution at the point of calculation.
0 mL must be pure formic acid.
The equivalence point is the salt, sodium formate. so the pH is determined by the hydrolsis (the Kb) of the salt.
At 40.00 mL, (past the equivalence point), it is essentially NaOH.
Does this help get you started? If you have questions, post your work to that point and explain what you don't understand.