Asked by 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 of a solution
M=0.015000 moles of NaOH
0.06705L
M1V1=M2V2
M1V1=M2V2
V1
M1=M2V2
V1
M1=0>223713647 M of NaOH * 0.06705L
0.025L
M1=0.600000001M of HCl
0.600000001 M of HCl* 0.025L =
0.223713647 M of NaOH * 0.06705L
Am I even close?
(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 a solution
M=0.015000 moles of NaOH
0.06705L
M1V1=M2V2
M1V1=M2V2
V1
M1=M2V2
V1
M1=0>223713647 M of NaOH * 0.06705L
0.025L
M1=0.600000001M of HCl
0.600000001 M of HCl* 0.025L =
0.223713647 M of NaOH * 0.06705L
Am I even close?
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