If 36.3 ml of 0.152M NaOH is required to neutralize 25.00 ml of an HC2H3O2 solution ,what is the molarity of the acetic acid?
To find the molarity of the acetic acid (HC2H3O2) solution, we can use the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction:
HC2H3O2 + NaOH → NaC2H3O2 + H2O
From the balanced equation, we can see that one mole of HC2H3O2 reacts with one mole of NaOH. Therefore, the number of moles of NaOH can be determined using the following equation:
moles of NaOH = Molarity of NaOH × Volume of NaOH (in liters)
Given that the volume of NaOH used is 36.3 ml (or 0.0363 liters) and the molarity of NaOH is 0.152M, we can calculate the number of moles of NaOH used:
moles of NaOH = 0.152M × 0.0363 L = 0.0055156 moles
Since the stoichiometry of the reaction indicates a 1:1 mole ratio between HC2H3O2 and NaOH, the number of moles of HC2H3O2 present in the reaction is also 0.0055156 moles.
To find the molarity of the acetic acid, we divide the moles of HC2H3O2 by its volume in liters. The volume of HC2H3O2 is given as 25.00 ml (or 0.025 L).
Molarity of HC2H3O2 = moles of HC2H3O2 / Volume of HC2H3O2 (in liters)
Molarity of HC2H3O2 = 0.0055156 moles / 0.025 L = 0.2206 M
Therefore, the molarity of the acetic acid (HC2H3O2) solution is 0.2206 M.
mols NaOH = M x L = ?
mols acetic acid = mols NaOH (write and balance the formula to see that).
M acid = mols acid/L acid.