Posted by Brad on Tuesday, February 19, 2008 at 6:48pm.
You would have done much better to type in the problem before showing your solution. It's good to show your work, however, you have no data (at least none identifiable). For example, where did the 0.01500 mols come from? Finally, you have made a typo in the second sentence (the formula). You show a + sign and it should be a x sign. If you will post the problem someone will go through and check it for you. Thanks for using Jiskha.
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
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0.06705L
M=0.223713647M of NaOH
M1V1=M2V2
M1=M2V2
----
V1
M1=0.223713647M of NaOH * 0.06705L
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0.025L
M1= 0.600000001 M of HCl
0.600000001 M of HCl * 0.025L=
0.223713647 M of NaOH * 0.06705L