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
To calculate the molarity of the HCl, we can use the equation:
M1V1 = M2V2
where M1 is the molarity of the HCl, V1 is the volume of the HCl, M2 is the molarity of the NaOH, and V2 is the volume of the NaOH.
Given:
HCl volume = 25 mL = 0.025 L
NaOH volume = 67.05 mL = 0.06705 L
Molarity of NaOH = 0.223713647 M
Substituting the values into the equation, we have:
M1 * 0.025 L = 0.223713647 M * 0.06705 L
Solving for M1:
M1 = (0.223713647 M * 0.06705 L) / 0.025 L
M1 = 0.600000001 M
Therefore, the molarity of the HCl is approximately 0.600000001 M.
To calculate the molarity of HCl, we need to use the equation M1V1 = M2V2, where M1 is the molarity of HCl, V1 is the volume of HCl, M2 is the molarity of NaOH, and V2 is the volume of NaOH added.
Given the following values:
- Volume of HCl = 25 mL = 0.025 L
- Volume of NaOH added = 67.05 mL = 0.06705 L
- Molarity of NaOH = 0.223713647 M
First, we calculate the moles of NaOH by using the formula:
Moles = Molarity * Volume
Moles of NaOH = 0.223713647 M * 0.06705 L = 0.015000 moles
Next, we can calculate the molarity of HCl using the formula:
Molarity = Moles / Volume
Molarity of HCl = 0.015000 moles / 0.025 L = 0.600000001 M
So, the molarity of HCl is approximately 0.600000001 M.