## anonymous one year ago 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?

1. Photon336

@Rushwr

2. anonymous

Yeah, I don't know how to do titrations very well so I thought I'd come here first #_#

3. Photon336

@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

4. Photon336

well for starters, $M _{1}V _{1} = M_{2}V_{2}$

5. Rushwr

Since its a diprotic weak acid it will have 2 Ka values. We can consider the neutralization of alanine here . |dw:1442265191959:dw|

6. Rushwr

|dw:1442265344770:dw|

7. Rushwr

Ka1> Ka2

8. Rushwr

So in the 1st end point we get 100% HA and in the 2nd end point we get 100% A^-

9. anonymous

Oh how do I know Ka1 is greater than Ka2?

10. anonymous

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?

11. Rushwr

That was just an example lol !!!! I can remember my teacher said that it was lyk that. :)

12. anonymous

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.

13. Rushwr

@woodward do u have the answers ! I'll just get the answer in my way and see if it tally ?

14. anonymous

The answer it gives is 300 g, I think you did it right probably just lost some multiple of 10 somewhere!

15. Rushwr

i took the concentration of NaOH as 1M not as 0.1 M

16. anonymous

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. $pH =pKa_1 + \log_{10}\frac{[HA]}{[H_2A^+]}$ $pH =pKa_2 + \log_{10}\frac{[A^-]}{[HA]}$

17. anonymous

18. Rushwr

it was not like that because we weren't given with pH values nah ! So basically what I did was stupid.

19. anonymous

I mean it looks like what you did got the right answer so I think I'm doing the stupid stuff not you!

20. Rushwr

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

21. Rushwr

22. Rushwr

@Photon336 What do u think ?

23. anonymous

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?

24. Photon336

so that was a polyprotc acid so wait you didn't have to use both of them?

25. anonymous

Thank you @Rushwr and @Photon336 !

26. Rushwr

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^-