A 25.0 mL sample of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 ✕ 10-5) is titrated with 0.100 M KOH solution.
A) 8.0ml
B) 12.5 ml
C) 20.0 ml
D) 25.0 ml
E) 25.1 ml
F) 28.0 ml
and the question is? If it is the volume of KOH, then
NoralityAcid*VolumeAcid=NormalityBase*VolumeBAse
.1*25ml=.1*Vb
vb=25ml
To determine the volume of KOH solution required to reach the equivalence point in the titration, you need to use the concept of stoichiometry and the balanced chemical equation for the reaction between propanoic acid (HC3H5O2) and KOH.
1. Write the balanced chemical equation for the reaction:
HC3H5O2 + KOH → KC3H5O2 + H2O
2. Determine the stoichiometric ratio between HC3H5O2 and KOH from the balanced chemical equation. In this case, 1 mole of HC3H5O2 reacts with 1 mole of KOH.
3. Calculate the number of moles of HC3H5O2 in the 25.0 mL sample using its molarity:
moles of HC3H5O2 = volume (L) × molarity (mol/L)
moles of HC3H5O2 = 0.025 L × 0.100 mol/L
moles of HC3H5O2 = 0.0025 mol
4. Since the stoichiometric ratio is 1:1 between HC3H5O2 and KOH, the number of moles of KOH required to react with the HC3H5O2 is also 0.0025 mol.
5. Calculate the volume of 0.100 M KOH solution required to react with 0.0025 moles of KOH using its molarity:
volume of KOH solution (L) = moles of KOH / molarity of KOH
volume of KOH solution (L) = 0.0025 mol / 0.100 mol/L
volume of KOH solution (L) = 0.025 L
6. Convert the volume of KOH solution to mL:
volume of KOH solution (mL) = 0.025 L × 1000 mL/L
volume of KOH solution (mL) = 25.0 mL
Therefore, the volume of 0.100 M KOH solution required to reach the equivalence point is 25.0 mL. So the correct answer is (D) 25.0 ml.