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July 26, 2014

July 26, 2014

Posted by **Tim** on Wednesday, February 27, 2008 at 4:40pm.

Calculate molarity of HCl from the volumes of acid and base at the equivalence point and the molarity of NaOH from the titration curve.

(M of Acid)x(v of acid)= (m of base)x(v of added base)

M1= 0.50M of NaOH x 0.050L/0.01505L

M1= 1.661129568 M of HCl

titration curve equivalence volume

Find the molarity of the acid with this equivalence volume.

(equivalence volume = 0.01505L)

What numbers do I use to find the molarity of the acid? Do I use the numbers I used in the first part (0.50M of NaOH and 0.05L of NaOH volume at the equivalence point) or do I use the Moles and Volume when the solution changed colors?

- chemistry -
**DrBob222**, Wednesday, February 27, 2008 at 5:08pmThere is a subtle difference in the questions and if your prof hasn't explained it the problems can be confusing, as you have just demonstrated. Here is the point of the two problem.

There is an equivalence point which is the point at which the EXACT number of mols of acid equals the EXACT number of mols of base. This is a theoretical number which is the point at which acid and base are EXACTLY neutralized. That is what your first problem is.

When you perform a titration, you must have some way to KNOW when you have reached the equivalence point. So you add an indicator to your solution and titrate away until the indicator changes color. This is not called the equivalence point; it is called the END POINT. You HOPE (with the emphasis on HOPE) that the indicator will tell you when you have reached the equivalence point. I notice in your problem that it calls what I have called the "end point" the "titration equivalence point." Same difference. So the first problem is to establish the theoretical equivalence point and the second problem is to establish the titration equivalence point (or end point) and then to see the difference in the two numbers. The difference is called the indicator error. With good choice of indicator and good skills as an analytical chemist, we can keep the indicator error very small although we can't do away with it entirely. I haven't answered your question directly but I think I have indirectly. Please let me know if there is anything on which you need me to elaborate.

- chemistry -
**Tim**, Wednesday, February 27, 2008 at 5:50pmSo if I understand you correctly the numbers I would need to use are the ones found at the end point or the titration equivalence point. My equation would then be:

0.015000M of NaOH x 0.01505L/0.01505L

- chemistry -
- chemistry -
**DrBob222**, Wednesday, February 27, 2008 at 6:11pmYes, I think you use 0.01505 for the second part. Four things I should mention.

1. It appears to me you used 0.01505 for the first part, too, from your first post. I don't think that's the idea of the problem.

2. It's hard to tell exactly what number means what since you didn't post any of the numbers EXCEPT in your first solution.

3. I don't think the values in your last post are correct (unless they came from other unposted data) but it's obvious to me that 0.01500 x 0.01505/0.01505 = 0.01500 but the molarity of NaOH is 0.500 M?

4. You have carried out the first calculation to far too many places. Did you have extra zeros on the 0.50 you didn't print. If so, and that number is 0.5000 and the buret readings are 0.01500 then you can have four significant figures.

- chemistry -
**Julie**, Wednesday, March 19, 2008 at 1:40pmOkay, I am doing a lab with titration as well. I have read everything here, and have a question for you as well... if that is okay? The hard part for me is that I don't have a prof. to talk to face to face since my whole class is on-line. Anyway...

Titrating an acid/base - 1 M. solution of NaOH into HCL. I have to Calculate the molarity of the HCl from the volumes of acid and base at the equivalence point and the molarity of the NaOH.

This is what I have:

Volume of Acid (HCL) at equivalence point: 25ml

Volume of Base (NaOH) at equivalence point: 15.05ml

Am I correct in thinking of it this way:

M of Acid x 025 = 0.01505 M of base x 0.01505 vol. of base added

Any insight you can give would be greatly appreciated - I feel like I am pulling my hair out trying to figure this out.

- chemistry -
**Julie**, Wednesday, March 19, 2008 at 1:42pmOkay, I am doing a lab with titration as well. I have read everything here, and have a question for you as well... if that is okay? The hard part for me is that I don't have a prof. to talk to face to face since my whole class is on-line. Anyway...

Titrating an acid/base - 1 M. solution of NaOH into HCL. I have to Calculate the molarity of the HCl from the volumes of acid and base at the equivalence point and the molarity of the NaOH.

This is what I have:

Volume of Acid (HCL) at equivalence point: 25ml

Volume of Base (NaOH) at equivalence point: 15.05ml

Am I correct in thinking of it this way:

M of Acid x 025 = 0.01505 M of base x 0.01505 vol. of base added

Any insight you can give would be greatly appreciated - I feel like I am pulling my hair out trying to figure this out.

- chemistry -
**Julie**, Wednesday, March 19, 2008 at 1:43pmTitrating an acid/base - 1 M. solution of NaOH into HCL. I have to Calculate the molarity of the HCl from the volumes of acid and base at the equivalence point and the molarity of the NaOH.

This is what I have:

Volume of Acid (HCL) at equivalence point: 25ml

Volume of Base (NaOH) at equivalence point: 15.05ml

Am I correct in thinking of it this way:

M of Acid x 025 = 0.01505 M of base x 0.01505 vol. of base added

Any insight you can give would be greatly appreciated - I feel like I am pulling my hair out trying to figure this out.

- chemistry -
**Julie**, Wednesday, March 19, 2008 at 1:43pmTitrating an acid/base - 1 M. solution of NaOH into HCL. I have to Calculate the molarity of the HCl from the volumes of acid and base at the equivalence point and the molarity of the NaOH.

This is what I have:

Volume of Acid (HCL) at equivalence point: 25ml

Volume of Base (NaOH) at equivalence point: 15.05ml

Am I correct in thinking of it this way:

M of Acid x 025 = 0.01505 M of base x 0.01505 vol. of base added

Any insight you can give would be greatly appreciated - I feel like I am pulling my hair out trying to figure this out.

- chemistry -
**Julie**, Wednesday, March 19, 2008 at 1:44pmsorry - i am having trouble posting the rest of my question - information

- chemistry -
**Julie**, Wednesday, March 19, 2008 at 1:44pmTitrating an acid/base - 1 M. solution of NaOH into HCL. I have to Calculate the molarity of the HCl from the volumes of acid and base at the equivalence point and the molarity of the NaOH.

This is what I have:

Volume of Acid (HCL) at equivalence point: 25ml

Volume of Base (NaOH) at equivalence point: 15.05ml

Am I correct in thinking of it this way:

M of Acid x 025 = 0.01505 M of base x 0.01505 vol. of base added

Any insight you can give would be greatly appreciated - I feel like I am pulling my hair out trying to figure this out.

- chemistry -
**Julie**, Wednesday, March 19, 2008 at 1:45pmThis is what I have:

Volume of Acid (HCL) at equivalence point: 25ml

Volume of Base (NaOH) at equivalence point: 15.05ml

Am I correct in thinking of it this way:

M of Acid x 025 = 0.01505 M of base x 0.01505 vol. of base added

Any insight you can give would be greatly appreciated - I feel like I am pulling my hair out trying to figure this out.

- yuzn djbicwt -
**yuzn djbicwt**, Sunday, February 1, 2009 at 5:22pmgpnkd tvios lwaseo itgdqwa sxkhwnav emksq kdqepwtay

- chemistry -
**Kyle**, Monday, July 20, 2009 at 4:22pmJulie, do you still need help?

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