Acid-Base Titration question?

Question

A 10.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH3COOH, requires 16.9 mL of 0.500 M NaOH to reach the endpoint in a titration. What is the molarity of the acetic acid solution?

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

I understand how to do this type of problem; however, the "endpoint in a titration" is a bit confusing to me. Some help please. It would be great if you can provide the answer, so I can double check to make sure I am doing the problem correct. Thank you!

Answer 1

The endpoint of the titration is the volume of NaOH where your pH jumps from acidic to basic, that is where your curve crosses pH=7, or another way to say it: you've put the amount of base corresponding to the original amount of acid.
If it is 16.9mL for a 0.5M solution of NaOH, and you had 10mL of acid, I would say that you had a .845M solution of vinegar.
16.9*0.5/10 = 0.845

Answer 2

The "endpoint of a titration" is where the moles of acid = moles of base. When you actually titrate, say, a sample of acid with a sample of base, you know the volume of acid being titrated, the molarity (moles/L volume) of the base used for the titration, and the end point volume for the base (the amount of base you have added to neutralize the moles of acid in the given volume of acid that has been titrated). So the endpoint volume in this case is the volume of base that has been added (if you are titrating with base, that is, adding the base drop by drop or whatever into the acid). It is at this volume that the moles of base you have added (in the form of moles/L concentration of base) equals the moles of acid present in your specific volume sample, which means you can calculate the molarity of your unknown acid.

Note that the above explanation also works if you are titrating a known molarity acid into an unknown molarity of base.


For your problem, you just use MaVa = MbVb, where Ma = molarity of acid, Va = volume of acid, and Mb and Vb are for the base. This should give you 0.845 mol/L or 0.845 M for the concentration of your acetic acid.

Hope you understand. Good luck!

Answer 3

At the endpoint of the titration there is essentially no acetic acid left to titrate.

To solve: ( Vacid * M acid) = (V base * M base)
Just substitute and solve for Macid

Answer 4

hardship-free Acids: Ammonia Hydrochloric Acid (HCl) Lemon Juice espresso Orange Juice (Ascorbic acid) hardship-free Bases: Bleach Baking Soda (sodium bicarbonate) milk of magnesia NaOH (sodium hydroxide) if i did not list sufficient, i say sorry. i do not recognize how wide you needed me to flow. in case you look on the chemical formula of acids and bases, keep in mind that maximum acids have an H+ team (a large hydrogen ion) and maximum bases have an OH- team (also familiar as a unfavourable hydroxide ion) for instance, hydrochloric acid is made up of H+ and Cl-, forming HCl although, sodium hydroxide (NaOH) is made up of an Na+ ion, and the OH- ion, making it a base. desire this helped!

Answer 5

endpoint in a titration means equilibrium
In this case MV of acid = MV of base (changes for things involving normality)
so (.01)M = (.0169)(.5)
M=0.845

(I changed the volume to L, but you really don't need to.)