Titration Question

[DJtiesto]

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Hey all, just wondering if someone could help with this problem:

If 25 ml of 1.0 M acetic acid were to be titrated with .5M solution of NaOH, the equivalence point would be reached after what volume of NaOH is added?



I used M1V1=M2V2:

(25ml)(1.0M)=(.5M)V2

V2=50ml

50 ml is the total amount of the solution after the titration thus 25 ml must be added because 25 ml was there to begin with.

Can someone tell me what I am doing wrong? The answer is 50 ml.

There is a graph of the titration to go with this and the answer said to use the graph. It is easy to get the answer by analyzing the graph but I would like to know why this method does not work.

Thanks

 

[Phosphorus Ylide]

The equivalence point is when you have stoichiometrically equivalent amounts of acid and base. If you have X moles of HOAc, you need X moles of NaOH.

You have 25 mL of HOac x 0.1 M HOAc, so that means you have 2.5 mmol or 0.025 M of HOAc. You now need to add 0.025 mol of NaOH to reach the equivalence point.

You have 0.5 M NaOH. If you take 0.025 mol /0.5 M, you get 0.05L, which is equivalent to 50 mL.

You were right, you do need 50 mL of NaOH, but that must be added since there was none present initially. If you added 25 mL of NaOH, you would have .0125 moles of NaOH instead which is obviously not equivalent to 0.025 mol HOAc.



Hope this helps. Probably would have explained it better on a piece of paper and in person. 🙂

 

[wait4me]

Eq point is when mol acid=mol base

that is why you can use the equation MV=MV because that basically tells you how many mols are in each since (mol/L)(L) gives you the number of mols in each. Therefore, 50 mL would have to be added to give you the equivalence point.

If you were asked about the HALF eq point then I believe 25mL would be the correct answer

 

[DJtiesto]

Oooh... got it. There were some problems in the BR books that showed subtracting some initial amount of something but I can't find it now.

Regardless, it is clear now.

Thanks