A 0.600 g sample of a pure, weak diprotic acid gives end points at 20.0 mL and 40.0 mL when it is titrated with 0.100 M NaOH. What is the molar mass of the weak acid?

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A 0.600 g sample of a pure, weak diprotic acid gives end points at 20.0 mL and 40.0 mL when it is titrated with 0.100 M NaOH. What is the molar mass of the weak acid?

Chemistry
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Yeah, I don't know how to do titrations very well so I thought I'd come here first #_#
@Woodward i'm not good at these either.. do you know @cuanchi that guy is amazing at these types of questions. I'm going to look this over myself

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well for starters, \[M _{1}V _{1} = M_{2}V_{2}\]
Since its a diprotic weak acid it will have 2 Ka values. We can consider the neutralization of alanine here . |dw:1442265191959:dw|
|dw:1442265344770:dw|
Ka1> Ka2
So in the 1st end point we get 100% HA and in the 2nd end point we get 100% A^-
Oh how do I know Ka1 is greater than Ka2?
Actually maybe this question is too hard for me right now I don't really even remember what an endpoint is, I think for a single protic acid it's always going to happen at pH=7 right?
That was just an example lol !!!! I can remember my teacher said that it was lyk that. :)
Wait I don't think I'm using my brain, I can write out the equilibrium constant equations (Henderson-Hasselbach) One sec I'll see how far I get, thanks @Rushwr I think the chemical equations are helping for me to visualize this better. I'm going to try to write out the equations in a sec and see what I can figure out.
@woodward do u have the answers ! I'll just get the answer in my way and see if it tally ?
The answer it gives is 300 g, I think you did it right probably just lost some multiple of 10 somewhere!
i took the concentration of NaOH as 1M not as 0.1 M
I'm looking at these equations, not sure how or why if this is true \(pH = pKa\) at the endpoint? I know there are some things like this but I don't know why or what they are. \[pH =pKa_1 + \log_{10}\frac{[HA]}{[H_2A^+]}\] \[pH =pKa_2 + \log_{10}\frac{[A^-]}{[HA]}\]
or is this the wrong direction, how'd you get your answer?
it was not like that because we weren't given with pH values nah ! So basically what I did was stupid.
I mean it looks like what you did got the right answer so I think I'm doing the stupid stuff not you!
if we luk at the equations, as i told the 1st end point is 100% HA I took the moles of acid = moles of HA HA + NaOH --------> NaA + H2O So the molar ratio is 1:1 So I found the moles of NaOH used until the 1st end point That would be = 0.1M x 20 x 10^-3 So the no. of moles of acid = 0.1 x 20 x 10^-3 No. of moles = mass divided by molar mass we know the mass used and the moles so molar mass = 0.6 divided by 0.1 x 20 x 10^-3 = 300g
I dnt think I made sense ! Sorry about that !
@Photon336 What do u think ?
Ahhh! Ok this makes sense, we didn't even have to use the 40 mL they just gave that to try to throw me off then?
so that was a polyprotc acid so wait you didn't have to use both of them?
Thank you @Rushwr and @Photon336 !
If we used 40ml then that's the end point where 100% A^- right so by then we have released 2 hydrogen atoms so we have to divide that by 2 ! SO wwe will get the same answer there !!!! H2A----> H^+ + HA^- HA^- -------> H^+ + A^-

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