YAChemT: Halfway point towards endpoint (solved)

Ricemarine

Lifer
Sep 10, 2004
10,507
0
0
Ok, so basically, my class had NaOH and H2SO4 to do a titration, with the known molarity of the NaOH being 0.09901 M. So we did a titration with 10 mL, and got about 29.9 mL of NaOH required to reach the endpoint of the 10 mL of H2SO4.

Obj: Find the concentration of Na + and SO4(2-) ions @ the halfway towards the endpoint.

So what I did

10 mL H2SO4 * 1L/1000mL * 0.148 mol H2SO4/L*** * 2 mol NaOH/1 mol H2SO4 = 0.00296 mol NaOH...

*** = Calculated after solving for the molarity of H+ and SO4 ions.

0.00296 mol NaOH/ ? * 2 mol Na+/2 mol NaOH * 1000 mL/L = ____ M.

So what am I supposed to divide by and what am I supposed to get the correct molarity?...

Thanks guys for your help.
 

bobert

Senior member
Dec 6, 2004
505
0
0
I haven't done chemistry in years, but if I'm reading this correctly I think you divide by the volume of NaOH you used to do the titration. the "mol NaOH" should cancel out each other and the 29.9 mL on bottom should cancel with the 1000mL to 1 L leaving you with the mol/L of Na+

so 0.00296*1000 / 29.9 should be your answer in mol/L

just curious, how old are you?
 

Ricemarine

Lifer
Sep 10, 2004
10,507
0
0
Originally posted by: bobert
I haven't done chemistry in years, but if I'm reading this correctly I think you divide by the volume of NaOH you used to do the titration. the "mol NaOH" should cancel out each other and the 29.9 mL on bottom should cancel with the 1000mL to 1 L leaving you with the mol/L of Na+

so 0.00296*1000 / 29.9 should be your answer in mol/L

just curious, how old are you?

If you divided 0.00296 * 1000 / 29.9, that would result in the starting molarity, thus it should actually be the endpoint as well since basically you used all of the 29.9 mL. The thing is, if you divide 0.00296 by 14.95 mL (half of the titrant used, NaOH), then you get twice the molarity. But from other collegue's, they get different answers, so I believe the molarity should still be under the starting/ending molarity. Clarify if I'm wrong.

I'm 17. Taking first class of the three part chem series.
 

bobert

Senior member
Dec 6, 2004
505
0
0
Originally posted by: Ricemarine
Originally posted by: bobert
I haven't done chemistry in years, but if I'm reading this correctly I think you divide by the volume of NaOH you used to do the titration. the "mol NaOH" should cancel out each other and the 29.9 mL on bottom should cancel with the 1000mL to 1 L leaving you with the mol/L of Na+

so 0.00296*1000 / 29.9 should be your answer in mol/L

just curious, how old are you?

If you divided 0.00296 * 1000 / 29.9, that would result in the starting molarity, thus it should actually be the endpoint as well since basically you used all of the 29.9 mL. The thing is, if you divide 0.00296 by 14.95 mL (half of the titrant used, NaOH), then you get twice the molarity. But from other collegue's, they get different answers, so I believe the molarity should still be under the starting/ending molarity. Clarify if I'm wrong.

I'm 17. Taking first class of the three part chem series.

Oh lol, I read the objective wrong.

bah nvm, i'll think a little longer

edit: are you looking for the equivalence point?
 

Ricemarine

Lifer
Sep 10, 2004
10,507
0
0
Originally posted by: bobert
Originally posted by: Ricemarine
Originally posted by: bobert
I haven't done chemistry in years, but if I'm reading this correctly I think you divide by the volume of NaOH you used to do the titration. the "mol NaOH" should cancel out each other and the 29.9 mL on bottom should cancel with the 1000mL to 1 L leaving you with the mol/L of Na+

so 0.00296*1000 / 29.9 should be your answer in mol/L

just curious, how old are you?

If you divided 0.00296 * 1000 / 29.9, that would result in the starting molarity, thus it should actually be the endpoint as well since basically you used all of the 29.9 mL. The thing is, if you divide 0.00296 by 14.95 mL (half of the titrant used, NaOH), then you get twice the molarity. But from other collegue's, they get different answers, so I believe the molarity should still be under the starting/ending molarity. Clarify if I'm wrong.

I'm 17. Taking first class of the three part chem series.

Oh lol, I read the objective wrong.

bah nvm, i'll think a little longer

edit: are you looking for the equivalence point?

Nope. PH is not a factor in this case, since we're never sure if the pH is exactly 7.00.
 

Paperdoc

Platinum Member
Aug 17, 2006
2,292
270
126
First you solve for the actual concentration of the original H2SO4 solution, which you did.
Now, half way through the titration process, you have exactly as much SO4-- ions as you started with and ended with. All the titration process does is add both Na+ ions and OH- ions until the OH_ and H+ ions are equal. (Oh yeah, it also adds water and thus impacts the volume of the solution.) It does absolutely nothing to the SO4-- ions. BUT you must remember that it takes TWO moles of NaOH to neutralize ONE mole of H2SO4, right? So, although you know the number of moles of OH- ions it took to reach neutrality (and hence, the same number of moles of H+ ions involved), there are only HALF as many SO4-- ions present.

Now, for the Na+ ion quantity half way to neutrality: that is simply half the number of moles of NaOH solution it took to get all the way. And since there is one mole of Na+ per mole of OH-, that's easy.

Now, watch out here. Each of the calculations above gets you to the point of knowing how many moles of SO4-- and Na+ are present at the half-way point. However, you also need to know the VOLUME of the solution in order to calculate the concentrations of each. For these conditions it is safe to assume additive volumes. So take the volume at half-neutralized to be the volume of the original H2SO4 (10.00 ml) plus half the final volume (29.9 / 2 = 14.95 ml) of NaOH solution. Now you have a volume and numbers of moles present for each of Na+ and SO4--. Calculate molarity for each.
 

dainthomas

Lifer
Dec 7, 2004
14,588
3,414
136
Originally posted by: Ricemarine
Ok, so basically, my class had NaOH and H2SO4 to do a titration, with the known molarity of the NaOH being 0.09901 M. So we did a titration with 10 mL, and got about 29.9 mL of NaOH required to reach the endpoint of the 10 mL of H2SO4.

Obj: Find the concentration of Na + and SO4(2-) ions @ the halfway towards the endpoint.

So what I did

10 mL H2SO4 * 1L/1000mL * 0.148 mol H2SO4/L*** * 2 mol NaOH/1 mol H2SO4 = 0.00296 mol NaOH...

*** = Calculated after solving for the molarity of H+ and SO4 ions.

0.00296 mol NaOH/ ? * 2 mol Na+/2 mol NaOH * 1000 mL/L = ____ M.

So what am I supposed to divide by and what am I supposed to get the correct molarity?...

42
 

Ricemarine

Lifer
Sep 10, 2004
10,507
0
0
Originally posted by: Paperdoc
First you solve for the actual concentration of the original H2SO4 solution, which you did.
Now, half way through the titration process, you have exactly as much SO4-- ions as you started with and ended with. All the titration process does is add both Na+ ions and OH- ions until the OH_ and H+ ions are equal. (Oh yeah, it also adds water and thus impacts the volume of the solution.) It does absolutely nothing to the SO4-- ions. BUT you must remember that it takes TWO moles of NaOH to neutralize ONE mole of H2SO4, right? So, although you know the number of moles of OH- ions it took to reach neutrality (and hence, the same number of moles of H+ ions involved), there are only HALF as many SO4-- ions present.

Now, for the Na+ ion quantity half way to neutrality: that is simply half the number of moles of NaOH solution it took to get all the way. And since there is one mole of Na+ per mole of OH-, that's easy.

Now, watch out here. Each of the calculations above gets you to the point of knowing how many moles of SO4-- and Na+ are present at the half-way point. However, you also need to know the VOLUME of the solution in order to calculate the concentrations of each. For these conditions it is safe to assume additive volumes. So take the volume at half-neutralized to be the volume of the original H2SO4 (10.00 ml) plus half the final volume (29.9 / 2 = 14.95 ml) of NaOH solution. Now you have a volume and numbers of moles present for each of Na+ and SO4--. Calculate molarity for each.

Too bad my lab report was due uhhhh... 4 hours ago.
Thanks though :)...