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?
Yes, you are close to getting the correct molarity. Let's go through the calculations step by step to confirm your result.
First, let's calculate the molarity of NaOH:
Molarity (M) = moles of solute / liters of solution
You provided the moles of NaOH (0.015000 moles) and the volume of NaOH (0.06705 L). Therefore, we can calculate the molarity of NaOH as follows:
M(NaOH) = 0.015000 moles / 0.06705 L ≈ 0.223713647 M
Now, let's calculate the molarity of HCl using the equation you provided:
M1V1 = M2V2
We have the molarity of NaOH (M2) as 0.223713647 M, the volume of NaOH (V2) as 0.06705 L, and the volume of HCl (V1) as 0.025 L.
M1 = M2 * V2 / V1
M1 = 0.223713647 M * 0.06705 L / 0.025 L ≈ 0.600000001 M
Based on your calculations, the molarity of HCl is approximately 0.600000001 M. Therefore, your answer seems to be correct.
Just keep in mind that due to rounding errors, there may be slight discrepancies in the calculated values. Nevertheless, you have arrived at the correct approach to determine the molarity of HCl and NaOH.