chemistry

I am confused on how to do the last question.

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?

  1. 👍 0
  2. 👎 0
  3. 👁 100
asked by Tim
  1. There 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.

    1. 👍 0
    2. 👎 0
  2. So 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

    1. 👍 0
    2. 👎 0
    posted by Tim
  3. Yes, 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.

    1. 👍 0
    2. 👎 0
  4. Okay, 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.

    1. 👍 0
    2. 👎 0
    posted by Julie
  5. 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.

    1. 👍 0
    2. 👎 0
    posted by Julie
  6. sorry - i am having trouble posting the rest of my question - information

    1. 👍 0
    2. 👎 0
    posted by Julie
  7. 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.

    1. 👍 0
    2. 👎 0
    posted by Julie
  8. 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.

    1. 👍 0
    2. 👎 0
    posted by Julie
  9. gpnkd tvios lwaseo itgdqwa sxkhwnav emksq kdqepwtay

    1. 👍 0
    2. 👎 0
  10. Julie, do you still need help?

    1. 👍 0
    2. 👎 0
    posted by Kyle

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chemistry (molarity) (sorta an emergency)

    I'm doing a titration lab and writing a lab report where I'm sort of stuck on how exactly to find the concentration of HCl. These are the following info I have from the lab, Equation: HCl(aq)+NaOH(aq)-> H2O(l)+NaCl(aq) -Calculated

    asked by Ray on November 13, 2016
  2. Chemistry

    A sample of 0.0020 mole of HCl is sissolved in water t omake a 2000ml solution. Calculate the molarity of the HCl solution, the (H30+) and the pH. For a strong acide such as HCl the (h30+) is the same as the molarity of the HCl

    asked by Susie on December 1, 2010
  3. Chem.

    Sorry if I confused you. I miss-wrote my question. My question was: How can I determine the concentration of HCl remaining in the flask. I calculated the # of moles of HCl neutralized by the tablet which was around 0.014 M. I

    asked by Help!! on July 7, 2015
  4. chemistry structures

    what is the stucture for heptanoic acid i need a to draw it CH3CH2CH2CH2CH2CH2COOH http://en.wikipedia.org/wiki/Enanthic_acid Molarity Calculate the molarity of 9.25 mol HCL in 2.25L of solution. mols/L = Molarity

    asked by kim on April 4, 2007
  5. Chemistry

    Hi, I posted yesterday with this lab question. "Calculate the molar amounts of NaOH used in the reaction with the HCl solution and with the HC2H3O2. I think I got the answer to that with the help that I received. I had 5 mL of

    asked by Rachel on April 22, 2007
  6. Chemistry

    0.0020 mole of HCl is dossolved in water to make a 2000ml solution. Calculate the molarity of the HCl solution, the H30, and the pH. The H30 is the same as the molarity of the HCl solution.

    asked by Susie on December 1, 2010
  7. Chemistry

    How do i determine the exact molarity of .2 M of HCL? The molarity of NaOH Used was .2010M used in the titration. And the volume used 10.33 mL used to reach equivalence point. What would be the Molarity of HCL?

    asked by Katie on March 22, 2011
  8. ap chemistry

    please explain titration problems. I'm a total noob at this and am trying to answer some prelab questions. Examples: 1. How many mL of a 0.800 M NaOH solution is needed to just neutralize 40 mL of a 0.600 M HCl solution? 2. You

    asked by chris on October 2, 2008
  9. Chemistry

    Can someone please tell me if I did this correctly? Calculate the molarity of the HCl from the volumes of acid and base and the molarity of the NaOH. Use the following equation: (molarity of acid)x(volume of acid)=(molarity of

    asked by Brad on February 20, 2008
  10. Chemistry

    I have a question about buffers. Part A So it starts with 20ml 0.1 sodium acetate and 25ml 0.1 acetic acid. Calculate ph of buffer is 4.74 because the acid and conjugate base have the same molarity correct? So the Ph is just pKa

    asked by Rio on March 31, 2010

More Similar Questions