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
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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
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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