Can someone please tell me if I did this correctly?
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
----------------------
0.06705L
M=0.223713647M of NaOH
M1V1=M2V2
M1=M2V2
----
V1
M1=0.223713647M of NaOH * 0.06705L
---------
0.025L
M1= 0.600000001 M of HCl
0.600000001 M of HCl * 0.025L=
0.223713647 M of NaOH * 0.06705L
Brad, I tried to help yesterday by pointing out that these boards don't allow spacing; therfore, trying to draw division lines to show division is useless. If you want to divide a by b, you do it a/b = ?? and put parentheses around those parts that are together to avoid confusion. Also, I suggested you type the problem but I still don't see that. I see volume of both acid and base clear enough but I don't see a molarity, at least not one labeled molarity.
If you will type in L NaOH, L HCl, and molarity of one or the other we can help you.
Calculate the molarity of the HCl from the volumes of acid and base at equivalence point and the molarity of the NaOH.
L HCl=0.025
L NaOH=0.06705
NaOH moles M=0.015000
Does this help?
Calculate the molarity of the HCl from the volumes of acid and base and the molarity of the NaOH.
L HCl=0.025
L NaOH=0.06705
NaOH M=0.015000
Yes, very much, and thank you. I'm sorry to have taken so long to answer but I'm up to my neck in income tax time and some of my columns don't balance so.....
Here is the problem. Do it one of two ways, which ever is the easiest for you to do.
mols NaOH = L x M = 0.06705 L x 0.01500 = 0.00100575.
mols HCl = the same thing of 0.0010125.
The Molarity HCl = mols HCl/L HCl = 0.00100575/0.025 = 0.04023 M HCl.
The other way is to use the formula you were using.
M1V2 = M2V2 OR
MHClVHCl = MNaOHVNaOH.
MHCl*0.025 = 0.01500*0.06705
Solve for MHCl = 0.04023
Check my work. It's late and past my bed time.
You can see I made a typo in the post but I corrected it later.
mols NaOH = L x M = 0.06705 L x 0.01500 = 0.00100575.This is correct
mols HCl = the same thing of 0.0010125 This should read the same thing of 0.00100575.
The Molarity HCl = mols HCl/L HCl = 0.00100575/0.025 = 0.04023 M HCl.
The answer is correct.
Thank-you for your help. I did it the second way. The mistake I did make was that I divided when I should have multiplied. I appreciate your help in helping me find my error.
To calculate the molarity of the HCl, you can use the equation:
(Molarity of acid) x (Volume of acid) = (Molarity of base) x (Volume of added base)
In this case, you have the volumes of acid and base, and the molarity of the NaOH.
The volume of HCl is given as 25 mL, which is equivalent to 0.025 L (since 1 mL = 0.001 L).
The volume of NaOH is given as 67.05 mL, which is equivalent to 0.06705 L.
Next, you need to find the moles of NaOH. The equation for molarity is:
Molarity = Moles of solute / Liters of solution
In this case, you have 0.015000 moles of NaOH and 0.06705 L of solution. Plugging these values into the equation gives:
M = 0.015000 moles of NaOH / 0.06705 L
M = 0.223713647 M of NaOH
Next, you can use the formula M1V1 = M2V2 to find the molarity of HCl. Rearranging the formula gives:
M1 = M2V2 / V1
Plugging in the values, you get:
M1 = 0.223713647 M of NaOH * 0.06705 L / 0.025 L
M1 = 0.600000001 M of HCl
Therefore, the molarity of HCl is 0.600000001 M.
0.600000001 M of HCl * 0.025 L = 0.223713647 M of NaOH * 0.06705 L, so the equation is balanced.