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July 30, 2015

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

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