What volume of a 0.2 N NaoH solution do you expect to use in order to titrate 25ml (sample out of 250ml) of a 0.200 N oxalic acid solution?
heeeelp!
Equation is
H2C2O4 + 2NaOH==> 2H2O + Na2C2O4
I assume that is 25.00 mL H2C2O4.
milliequivalents(meq) NaOH = meq H2C2O4
25 mL x 0.2 = ??mL x 0.200
Do you suppose 25.00 mL.
Most schools don't teach normality now, but notice how easy it is to use normality. There are no mols and then convert mols H2C2O4 to mols NaOH, then mols/L = M for NaOH. It is all done for us by using normality. Neat!!
wow...thanks DrBob, I'll try it!
The problem will come when the question does NOT provide normality but provides molarity. Then you must convert M to N yourself OR work it the long way using mols. But note that whereas mL x N = mL x N (ALWAYS, ALWAYS, ALWAYS), mL x M /=/ (is not equal) mL x M EXCEPT when you want to work a dilution problem. For a stoichiometry problem, however, mL x M = mL x M is true ONLY when the reacting ratios are 1:1 (or 2:2 or 3:3 etc).
Thanks...I didn't know that
You're welcome! I'm here to help. Normality (N) is a concept that is not commonly taught in schools nowadays, but it can be quite useful in certain situations. In this case, the equation you provided is balanced, so the reacting ratios are 1:1.
To calculate the volume of the NaOH solution needed to titrate the oxalic acid solution, you can use the formula:
mL(NaOH) x N(NaOH) = mL(oxalic acid) x N(oxalic acid)
Plugging in the given values, we have:
25 mL x 0.2 N = X mL x 0.2 N
To solve for X, we need to isolate it on one side of the equation. We can do this by dividing both sides of the equation by 0.2 N:
X mL = (25 mL x 0.2 N) / 0.2 N
Simplifying, we find:
X mL = 25 mL
So, you would need to use 25 mL of the 0.2 N NaOH solution in order to titrate 25 mL of the 0.200 N oxalic acid solution.
I hope this explanation helps! Let me know if you have any further questions.