hi again, thanks to the previous help i got the first part of the question. the second part builds on the first part as we used that stock solution for titration of the next. we are now given:
Concentration of Standard Acid Soultion (M) = 0.100 M
Volume of Standard Acid Solution (mL) = 25.00 mL
Volume oof Stock Solution of NaOH (mL) = 18.65 Ml
Concentration of Stock Solution of NaOH (M) = 0.108 M (from previous question asked)
Now it is asking for the concentration of the stock solution.
Would you just do this:
0.100M X 0.025 L = ?? mols
and then ?? mols / 18.65 mL?
is that how to do it or is there more processes involved?
You're on the right track and you ALMOST have it.
Would you just do this:
0.100M X 0.025 L = ?? mols
exactly right.
and then ?? mols / 18.65 mL?
remember definition of M = mols/Liter. Therefore, you must change 18.65 mL to liters. Otherwise, you have it down ok but see the note in italics that follows.
Technically you can't work the problem nor solve for the molarity of the solution without an equation. You may have one that was given to you and you just didn't copy it into this post. If you know what the acid is (HCl for example and I think that was the previous post) then you can write it
NaOH + HCl ==> NaCl + H2O.
The idea here, and what I'm cautioning you against, is that we assumed above that the mols of the acid and the mols of the base were the same and that works as long as the mole ratio in the equation is 1:1 as it is for HCl and NaOH. BUT, if you have an equation in which the ratio is NOT 1:1, for example,
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O THEN
you must convert using the equation from mols of one reactant to mols of the other. In this case you were given standard acid of 0.100 M. Let's supposer that was H2SO4 and you have the equation I wrote above. Then 0.100 x 0.025 L = 0.0025 mols H2SO4.
Then 0.0025 mols H2SO4 x (2 mol NaOH/1 molH2SO4) = 0.0050 mols NaOH.
Then M NaOH = mol NaOH/L = 0.0050 mols/0.01865 L = ??molarity NaOH.
that did make sense, and my little mistake was the conversion from mL to L. i referenced into some of my material and i believe the standard acid is probably H2SO4 so i will try it out and see my result.
sorry to keep bugging you, but if it makes you feel better i am understanding it more.
what if you have the same numbers, but you are trying to find the concentration of an UNKNOWN acid. all you have is:
Volume of Unknown Acid
Volume of Stock Solution
Concentration of Stock solution..
how would you determine that? same process?
To find the concentration of the stock solution of NaOH, you can use the principle of dilution. Here's how you can solve it step-by-step:
1. Convert the volume of the standard acid solution from milliliters to liters:
Volume of standard acid solution = 25.00 mL = 0.025 L
2. Calculate the number of moles (mols) of the standard acid solution:
Number of moles = Concentration × Volume
Number of moles = 0.100 M × 0.025 L
3. Determine the number of moles of NaOH used in the titration.
Since NaOH and the acid react in a 1:1 ratio, the number of moles of NaOH is equal to the number of moles of the standard acid solution.
4. Next, you need to find the volume of the stock solution of NaOH that is used in the titration. To do this, use the concept of molarity:
Molarity = Moles / Volume
Rearranging the equation, Volume = Moles / Molarity
Volume = Number of moles / Concentration
Volume = Number of moles / 0.108 M
5. Substitute the known values into the equation to calculate the volume of the stock solution:
Volume of stock solution = Number of moles / 0.108 M
6. Convert the volume of the stock solution to milliliters:
Volume of stock solution = (Volume of stock solution in L) × 1000 mL/L
7. Lastly, calculate the concentration of the stock solution of NaOH:
Concentration = Number of moles / Volume of stock solution
So, to answer your question, there are a few more steps involved than just dividing the number of moles by the volume of the stock solution. Make sure to follow the steps outlined above to get the correct answer.