# Show that the molarity of ethanoic acid in the original vinegar is 0.937 M.A 50.0 mL sample of vinegar was diluted to 250.0 mL in a volumetric flask. A 20.00 mL aliquot of this solution required...

Show that the molarity of ethanoic acid in the original vinegar is 0.937 M.

A 50.0 mL sample of vinegar was diluted to 250.0 mL in a volumetric flask. A 20.00 mL aliquot of this solution required the addition of 27.98 mL of 0.134 M sodium hydroxide solution in order to be neutralised.

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Let's trace this problem **backwards** to find the solution:

We know that 20.00 mL of the diluted solution required 27.98 mL of 0.134 M NaOH be added to neutralize the acid. Ethanoic acid is a monoprotic acid (see link below, keep in mind Ethanoic Acid = Acetic Acid), so the number of moles of NaOH added will be equivalent to the number of moles of ethanoic acid in the aliquot.

So we'll find the number of moles NaOH added. We do this by multiplying the concentration of NaOH by the volume of solution added to the aliquot (**remember, because M is in moles/L, we need to convert mL to L by dividing by 1000**):

Moles NaOH = (27.98 /1000) L * 0.134 moles NaOH/L

Moles NaOH = 3.75*10^-3 moles

Remember, the moles of NaOH is equivalent to the moles of ethanoic acid in the aliquot. Therefore, the number of moles in the aliquot are:

Moles (Ethanoic Acid in Aliquot): 3.75*10^-3 moles

Now, let's figure out how many moles of ethanoic acid were in the diluted solution. We know that 20.00 mL of solution contained 3.75*10^-3 moles of ethanoic acid. So, we set up a proportion (EA = Ethanoic Acid):

**Total moles/Total volume = Aliquot moles/Aliquot volume**

Total Moles EA/250.0 mL = 3.75*10^-3 moles EA/20.00 mL

Multiplying both sides by 250.0 mL...

Total Moles EA = (3.75*10^-3 moles EA/20.00 mL) * 250.0 mL

Total Moles EA = 4.688*10^-2 moles EA

Now that we have the total number of moles in our diluted solution, we can determine the total number of moles in our original solution:

**Moles in original solution = Moles in diluted solution**

Moles EA in original solution = 4.688*10^-2 moles

Bam! Easy! This is because when we make the dilute solution from the original solution, we didn't add or remove any of the ethanoic acid!

Now, to complete the problem, we simply go back to the definition of concentration: number of moles/volume (remember, volume for concentration is in liters!):

Original Concentration of EA = Original Moles of EA/Original Volume (in L)

Original concentration of EA = 4.688*10^-2 moles/(50.0 mL /1000)

**Original concentration of EA = 0.9375 M**

And there you have it. All you do is figure out how much material you started with, then you figure out its concentration. Remember, when you get problems involving a series of steps, and you're trying to figure out a quantity that you had in the beginning, you generally work backwards to solve the problem. Good luck!

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