Posted by 2 DrBob222 from DAN on Thursday, March 1, 2012 at 4:46pm.
Well, it turns out that way and to be a little honest about it I was surprised. But here are two ways to do it.
Let's assume the volume of the acid is ANYTHING but for the first case we'll take 50 mL. The 0.1M x 50mL = 0.08MNaOH x ?M. Solve for ?M NaOH which is the volume it will take for the NaOH to titrate to the equivalence point. I get 62.5 mL. So the product is 0.1 x 50mL = millimoles and that is dissolved in 50+62.5 mL = 112.5 mL, then 5mmols/112.4 mL -= 0.0444. Let's try another, say 100 mL of the 0.1M acid.
100 mL x 0.1M = 0.08M x ?mL NaOH and
?mL NaOH = 125 mL NaOH.
mmoles salt formed = 100 x 0.1 = 10 mmols and the volume is 100 + 125 = 225 mL. Then M = mmols/mL = 10/225 = 0.0444. Convinced? Try ANY volume of the acid and it will come out 0.04444444. :-). Do it with two or three other numbers to see for yourself. You can do it more esoterically this way.
0.1M*x mL = mmols of salt formed.
The ratio of the volume base to acid = base/acid = 0.1M/0.08M = 1.25 which means the volume of the base is always 1.25 times that of the acid.
Thus the salt will be 0.1x millimoles and the volume will be x from the acid + 1.25 from the base or total of 1 + 1.25 = 2.25. Then M = mmols/mL = (0.1x/2.25x) = 0.0444. :-).
It would be interesting to know if the prof expects you to use 0.0444 or to assume some other number. 0.0444 M is what should be used.
and did u get 625ml for NaOH just by guessin??
He got 62.5ml by making up a certain amount of volume for the 0.1M solution and setting it equal to the 0.08 NaOH( which is used to titrate the solution) and determined the amount of NaOH in milliliters needed until the solution hit its endpoint
basically he used...
M1 x V1 = M2 x V2
I used a 62.5 mL but I don't know where I have a 625 mL.
OK. Alison explained where the 62.5 mL came from. No, I didn't guess.
0.1 x 50 mL = 0.08 x ?mL
?mL = (0.1 x 50)/0.08 = 62.5 mL.Voila!
Note also that 62.5 really is 1.25 x volume acid used = 1.25 x 50 = 62.5
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