What is the molarity of the acetic acid if 0.5ml of a vinegar solution has been titrated with the 0.101M solution of NaOH and the volume of the NaOH solution at the equivalence point is 5.0mL?
construct the balance reaction equation of Acetic acid and NaOH.
take note of the mole ratio
calculate the actual mole(n) of NaOH reacting with acetic acid using n=cv
use the mole ration to determine the mole of acetic acid from the reacted NaOH mole.
calculate the concentration of acetic acid by dividing its mole by its volume
hope that helps
To find the molarity of acetic acid in the vinegar solution, we can use the concept of titration.
In a titration, a known concentration of one substance (in this case, NaOH) is reacted with an unknown concentration of another substance (acetic acid). The point at which the reaction is complete is called the equivalence point.
To find the molarity of acetic acid, we need to use the balanced chemical equation for the reaction between acetic acid (CH₃COOH) and sodium hydroxide (NaOH):
CH₃COOH + NaOH -> CH₃COONa + H₂O
From the balanced equation, we can see that the mole ratio between acetic acid and sodium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
Given that the volume of the NaOH solution at the equivalence point is 5.0 mL (or 0.005 L), we can calculate the number of moles of NaOH using the equation:
moles of NaOH = Molarity × Volume (in liters)
moles of NaOH = 0.101 mol/L × 0.005 L
moles of NaOH = 0.000505 mol
Since the mole ratio between acetic acid and sodium hydroxide is 1:1, the number of moles of acetic acid present in the vinegar solution is also 0.000505 mol.
The equation for molarity is:
Molarity = Moles of solute / Volume of solution (in liters)
We are given that 0.5 mL of the vinegar solution was titrated. To convert this to liters, we divide by 1000:
Volume of solution = 0.5 mL / 1000 = 0.0005 L
Now we can calculate the molarity of the acetic acid:
Molarity = Moles of acetic acid / Volume of solution
Molarity = 0.000505 mol / 0.0005 L
Molarity = 1.01 M
Therefore, the molarity of the acetic acid in the vinegar solution is 1.01 M.