Homework Help: Chemistry

Posted by Brad on Wednesday, February 20, 2008 at 9:22pm.

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 - DrBob222, Wednesday, February 20, 2008 at 10:00pm
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 - Brad, Wednesday, February 20, 2008 at 10:27pm
  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 - Brad, Thursday, February 21, 2008 at 12:34am
    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 - DrBob222, Thursday, February 21, 2008 at 1:19am
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
M_HClV_HCl = M_NaOHV_NaOH.
M_HCl*0.025 = 0.01500*0.06705
Solve for M_HCl = 0.04023
Check my work. It's late and past my bed time.
- Chemistry - DrBob222, Thursday, February 21, 2008 at 1:23am
  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 - Brad, Thursday, February 21, 2008 at 2:10pm
  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.

Answer this Question

Related Questions

Chemistry - Calculate the molarity of the HCl from the volumes of acid and base ...
chemistry - I am confused on how to do the last question. Calculate molarity of ...
Brad - My problem is calculate the molarity of HCl and NaOH (molarity of acid)x(...
chemistry - My problem is calculate the molarity of HCl and NaOH. (molarity of ...
chemistry - The following data were collected at the endpoint of a titration ...
Chemistry - Acid Base Titration - When you complete a acid base titration of ...
chemistry - the following titration data were collected: a 10 mL portion of a ...
Chem - A tank car containing concentrated HCL (molarity 12.1) derails and spills...
Chemistry - 1) You are given solutions of HCl and NaOH and must determine their ...
chemistry - Could someone help me with this please? In chemistry, we did a lab ...

For Further Reading

First Name:
School Subject:
Answer: