According to the problem we are adding a 0.1 M solution of KOH to a 0.1 M solution of HCOOH. So, we are titrating a weak acid using a strong base.
The reaction that describes this titration is:
HCOOH (aq) + KOH (aq) <----> HCOOK (aq) + H₂O (l)
The reaction is 1 to 1, since all the coefficients are 1.
The equivalence point occurs when equal number of moles of acid reacts with equal number of moles of base.
Let's find the initial number of moles of metanoic acid that we had:
Initial number of moles of HCOOH = 10.00 mL * 0.1 mmol/mL
Initial number of moles of HCOOH = 1.00 mmol
In the equilibrium point we need the same number of moles of the strong base.
Moles of KOH =
