Question

What is the molarity of the acid when 25.0 mL of 0.215 M KOH was titrated with a weak acid to the equivalence point?

a) 0.215 M
b) 0.430 M
c) 0.1075 M
d) 0.3225 M

Answer 1

The molarity of the weak acid is 0.215 M. This was calculated by determining the moles of KOH used in the titration and assuming a 1:1 molar ratio between KOH and the acid, as no different ratio was given. Therefore, the molarity of the acid is the same as the molarity of KOH.

Step-by-step explanation:

To find the molarity of the weak acid when 25.0 mL of 0.215 M KOH is titrated to the equivalence point, we must first consider the mole-to-mole reaction between the acid and the base. According to the concept of titration and the stoichiometry of a neutralization reaction, we know that at the equivalence point, the moles of acid will equal the moles of base.

Here's how we calculate it:

• Calculate the number of moles of KOH: moles = molarity × volume in liters = 0.215 M × 0.025 L = 0.005375 moles.
• Assume a 1:1 molar ratio between KOH and the acid because the question does not specify otherwise.
• Find the molarity of the acid: molarity = moles / volume = 0.005375 moles / 0.025 L = 0.215 M.

Therefore, the molarity of the acid is also 0.215 M, which matches the molarity of KOH. So, the correct answer is (a) 0.215 M.