Answer to Question #109816 in Chemistry for Ale

Question

What is the molarity of acetic acid in vinegar?
If volume of vinegar sample is 5.00ml, molarity of NaOH is .25M, initial reading is .25ml and final reading is 17.50ml

Answer 1

Let's start by writing a balanced reaction equation:

NaOH + HAc $\rightarrow$ NaAc + H₂O.

As you see from the equation, 1 mole of sodium hydroxide reacts with 1 mole of acetic acid. Therefore, at the equivalence point, the amount of NaOH added is equal to the amount of acetic acid in the sample taken for the titration. Let's calculate the amount of NaOH from the concentration and the volume of the NaOH solution required for the titration:

$n(\text{NaOH}) = cV = 0.25 (\text{mol/L})\cdot(17.50 - 0.25)\cdot10^{-3}(\text{L}) = 4.31\cdot10^{-3}$ mol.

Then, the amount of HAc in the sample is:

$n(\text{HAc}) = n(\text{NaOH}) =4.31\cdot10^{-3}$ mol.

Finally, the concentration of HAc in the vinegar sample is:

$c(\text{HAc}) =\frac{n}{V} = \frac{4.31\cdot10^{-3} \text{mol}}{5.00\cdot10^{-3} \text{L}} = 0.863$ M

Answer: If volume of vinegar sample is 5.00ml, molarity of NaOH is .25M, initial reading is .25ml and final reading is 17.50ml, the molarity of acetic acid in vinegar is 0.863 M.