Chemistry

posted by .

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 base)x(volume of added base)

equivalance point=number of mL of base to an acid

1mL=0.001L

HCl volume=25 mL=0.025L
NaOH=67.05mL=0.06705L

Molarity=Moles of solute
---------------
Liters of a solution

M=0.015000 moles of NaOH
----------------------
0.06705L

M=0.223713647M of NaOH

M1V1=M2V2

M1=M2V2
----
V1

M1=0.223713647M of NaOH * 0.06705L
---------
0.025L

M1= 0.600000001 M of HCl

0.600000001 M of HCl * 0.025L=
0.223713647 M of NaOH * 0.06705L

  • Chemistry -

    Brad, I tried to help yesterday by pointing out that these boards don't allow spacing; therfore, trying to draw division lines to show division is useless. If you want to divide a by b, you do it a/b = ?? and put parentheses around those parts that are together to avoid confusion. Also, I suggested you type the problem but I still don't see that. I see volume of both acid and base clear enough but I don't see a molarity, at least not one labeled molarity.
    If you will type in L NaOH, L HCl, and molarity of one or the other we can help you.

  • Chemistry -

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

    L HCl=0.025
    L NaOH=0.06705

    NaOH moles M=0.015000

  • Chemistry-Dr. Bob -

    Does this help?

    Calculate the molarity of the HCl from the volumes of acid and base and the molarity of the NaOH.

    L HCl=0.025
    L NaOH=0.06705

    NaOH M=0.015000

  • Chemistry -

    Yes, very much, and thank you. I'm sorry to have taken so long to answer but I'm up to my neck in income tax time and some of my columns don't balance so.....
    Here is the problem. Do it one of two ways, which ever is the easiest for you to do.
    mols NaOH = L x M = 0.06705 L x 0.01500 = 0.00100575.
    mols HCl = the same thing of 0.0010125.
    The Molarity HCl = mols HCl/L HCl = 0.00100575/0.025 = 0.04023 M HCl.

    The other way is to use the formula you were using.
    M1V2 = M2V2 OR
    MHClVHCl = MNaOHVNaOH.
    MHCl*0.025 = 0.01500*0.06705
    Solve for MHCl = 0.04023
    Check my work. It's late and past my bed time.

  • Chemistry -

    You can see I made a typo in the post but I corrected it later.
    mols NaOH = L x M = 0.06705 L x 0.01500 = 0.00100575.This is correct
    mols HCl = the same thing of 0.0010125 This should read the same thing of 0.00100575.
    The Molarity HCl = mols HCl/L HCl = 0.00100575/0.025 = 0.04023 M HCl.
    The answer is correct.

  • Chemistry-Dr. Bob -

    Thank-you for your help. I did it the second way. The mistake I did make was that I divided when I should have multiplied. I appreciate your help in helping me find my error.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    Could someone help me with this please? In chemistry, we did a lab in which he put brands of commerical vinegar in a microplate. We put in 10 drops of each vinegar in each plate thinger then we added 1 drop of phenolphthalein and then
  2. Brad

    My problem is calculate the molarity of HCl and NaOH (molarity of acid)x(volume of acid)=(molarity of base)+(volume of added base) HCl=acid NaOH=base HCl volume = 25mL=0.025L NaOH=67.05mL=0.06705 molarity = moles of solute Liters of …
  3. chemistry

    My problem is calculate the molarity of HCl and NaOH. (molarity of acid)x(volume of acid)=(molarity of base)+(volume of added base) HCl=acid NaOH=base HCl volume = 25mL=0.025L NaOH=67.05mL=0.06705 molarity = moles of solute Liters …
  4. Chemistry

    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 base)x(volume of added base) equivalance point=number of …
  5. 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 …
  6. chemistry

    The following data were collected at the endpoint of a titration performed to find the molarity of an HCl solution: Volume of acid (HCl) used = 14.4 mL; Volume of base (NaOH) used = 22.4 mL; Molarity of standard base (NaOH) = 0.200 …
  7. chemistry

    1. if 15.0mL of 4.5 M NaOH are diluted with water to a volume of 500mL, what is the molarity of the resulting solution?
  8. Chemistry 12

    Hi, I did the acid-base titration lab with HCl and 0.5M NaOH. I need to calculate the moles of NaOH from molarity of NaOH and the average volume used. My English isn't good but is it asking me to find the moles of NaOH using the molarity …
  9. chemistry

    Calculate the Molarity of the unknown acid if 25.0 mL of the acid reacts with 17.50 mL of 0.14 M NaOH solution. Assume that NaOH and acid react in 1:! ratio. Show your work. (Hint: Find the moles of NaOH from the given molarity and …
  10. 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 …

More Similar Questions