The data below was collected while performing a titration of acetic acid (in vinegar) to determine its concentration using sodium hydroxide. 150 ml of vinegar with unknown molarity 45.5 ml of .5 M NaOH was added to reach equivalence point Find the molarity of the vinegar.

Let's call acetic acid HAc.

HAc | NaOH ==> NaAc + H2O

mols NaOH = M x L = ?
mols HAc = mols NaOH
M HAc = mols HAc/L HAc = ?

For a 1:1 rxn ratio of monoprotic acid reacting with a GpIA strong base use ...

(Molarity x Volume)acid = (Molarity x Volume)base

Molarity Acid
= [(M x V)base]/(Vol Acid)
= [(0.5M)(45.5 ml)]/(150 ml)
= 0.16M HOAc

To find the molarity of the vinegar, we can use the concept of stoichiometry in a titration. The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is as follows:

CH3COOH + NaOH -> CH3COONa + H2O

In this reaction, one mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of sodium acetate and one mole of water.

Given the volume and molarity of the sodium hydroxide used (45.5 ml of 0.5 M NaOH), we can calculate the number of moles of sodium hydroxide used:

moles of NaOH = volume of NaOH (in liters) x molarity of NaOH
moles of NaOH = 0.0455 L x 0.5 mol/L
moles of NaOH = 0.02275 mol

From the balanced chemical equation, we can see that the molar ratio between acetic acid and sodium hydroxide is 1:1. This means that the number of moles of acetic acid is also 0.02275 mol.

Now, let's calculate the volume of vinegar used in the titration. We know that the vinegar and sodium hydroxide react in a 1:1 ratio, so the volume of vinegar used is equal to the volume of sodium hydroxide used.

volume of vinegar = volume of NaOH
volume of vinegar = 45.5 ml

Finally, to find the molarity of the vinegar, we divide the number of moles of acetic acid by the volume of vinegar used (in liters):

molarity of vinegar = moles of acetic acid / volume of vinegar (in liters)
molarity of vinegar = 0.02275 mol / 0.0455 L
molarity of vinegar = 0.5 M

Therefore, the molarity of the vinegar is 0.5 M.