polyprotic titrations

[rainashley]

could someone please explain to me how to do this example? thank you

a 30mL NaOH solution (the unknown) is titrated with 60 mL H2SO4 solution (4x10-3M), calculate the original pH
The right answer is pH = 12.1
I know the formula is MV = MV
What is the correct way of doing it? What am I missing?
Thanks

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[rainashley]

could someone please walk me through how to do this question? thanks

 

[RogueUnicorn]

do the same question but instead of 60mL H2SO4, use 120mL HCl of the same concentration. can you do that?

 

[rainashley]

do the same question but instead of 60mL H2SO4, use 120mL HCl of the same concentration. can you do that?

why?

 

[premed101]

why?

because that is an equivalent question. He just made the polyprotic (with 2 protons ready to donate) into a monoprotic, but doubled the amount.

 

[rainashley]

because that is an equivalent question. He just made the polyprotic (with 2 protons ready to donate) into a monoprotic, but doubled the amount.

i still dont understand
how do you do my original problem?

 

[Isoprop]

Since NaOH is a strong base, it will pull off two protons from sulfuric acid. So every mole of sulfuric acid will react with 2 moles of NaOH.

Calculate how many moles of sulfuric acid you have, then figure out how many moles of NaOH reacted with it using the ratio above. Then calculate the molarity of the original 30 mL sample. From there, calculate the pH.

 

[rainashley]

could someone please write out the exact calculation
i need to see it to understand it
thanks

 

[RogueUnicorn]

an HCl - NaOH titration is something you should be able to do. we've pointed you in the right direction, make an effort.

 

[boaz]

What are the products of this reaction:

H2SO4+NaOH------> ???

 

[rctriplefresh5]

an HCl - NaOH titration is something you should be able to do. we've pointed you in the right direction, make an effort.

should be able to do??? based on what> if he doesnt understand it stop being a jackass. op i hate titrations, im at the end of my gen chem semester and they sucked,. i dont think you use the m1v1=m2v2 equation unless your dealing with dilutions.

 

[RogueUnicorn]

should be able to do??? based on what> if he doesnt understand it stop being a jackass. op i hate titrations, im at the end of my gen chem semester and they sucked,. i dont think you use the m1v1=m2v2 equation unless your dealing with dilutions.

based on it being basic general chemistry and something that can be learned from any textbook that the OP surely must have. nobody needs you to be their white knight. what the OP needs to learn is figuring thing out on his/her own; we've nudger him/her the right way. if the continued response is "can you just do the whole thing out" then the OP is either not trying hard enough or is too far behind the basic concepts to be well helped through a single Q.

 

[rainashley]

You need to get the molarity of NaOH to find the original pH
thus
60x10-3 L x 4x10-3 M / 30x10-3 L = 8x10-3 M
Thus the poH is about 3
so the pH is about 11
But the answer is over 12
WHY?
Thanks

 

[boaz]

You need to get the molarity of NaOH to find the original pH
thus
60x10-3 L x 4x10-3 M / 30x10-3 L = 8x10-3 M
Thus the poH is about 3
so the pH is about 11
But the answer is over 12
WHY?
Thanks

Let's take this step by step.

Why did you write that equation? Yes, M1V1=M2V2. How do you know this equation is true??? Please provide an explanation.

 

[Isoprop]

first, your calculation is wrong.

MV = MV really says the number of moles of acid should equal the number of moles of bae. But you have 2 moles of H reacting, so the concentration of H2SO4 should double.

Second, -log(8 x 10^-3) is closer to 2 not 3.

 

[RogueUnicorn]

nvm

 

[rainashley]

ok this is going nowhere
could someone just PLEASE explain to me what to do?
i dont want to waste anyones time
thanks

 

[rainashley]

first, your calculation is wrong.

MV = MV really says the number of moles of acid should equal the number of moles of bae. But you have 2 moles of H reacting, so the concentration of H2SO4 should double.

Second, -log(8 x 10^-3) is closer to 2 not 3.

why double? please explain
so whats the calculation?
thanks

 

[rainashley]

first, your calculation is wrong.

MV = MV really says the number of moles of acid should equal the number of moles of bae. But you have 2 moles of H reacting, so the concentration of H2SO4 should double.

Second, -log(8 x 10^-3) is closer to 2 not 3.

why double? could you explain this to me?
so whats the real way of doing it? calculation?

 

[boaz]

You will be able to figure out the right way of doing it if you answer my question. See above.

 

[Isoprop]

why double? please explain
so whats the calculation?
thanks

we double the concentration because effectlvely, there is 1 mole of sulfuiric acid that is going to react with 2 mole of NaOH. The sulfuiric acid is producing 2 moles of protons to react. So every 1 M of sulfuric acid produces 2 M of H+ to react with the NaOH.

Redo your calculation remembering to double the H concentration. Then properly take the -log. Double check on your calculator. Then see if you get the right answer.

 

[rainashley]

we double the concentration because effectlvely, there is 1 mole of sulfuiric acid that is going to react with 2 mole of NaOH. The sulfuiric acid is producing 2 moles of protons to react. So every 1 M of sulfuric acid produces 2 M of H+ to react with the NaOH.

Redo your calculation remembering to double the H concentration. Then properly take the -log. Double check on your calculator. Then see if you get the right answer.

Could you please elaborate on this: The sulfuiric acid is producing 2 moles of protons to react. So every 1 M of sulfuric acid produces 2 M of H+ to react with the NaOH.
Thanks

 

[rainashley]

What are the products of this reaction:

H2SO4+NaOH------> ???

NaSO4 + H3O
ok so?

 

[Isoprop]

You really need to brush up on your basic general chemistry.

The balanced equation for this reaction is:

H2SO4 + 2NaOH --> Na2SO4 + 2H2O

For every mole of H2SO4, it reacts with 2 moles of NaOH.

 

[rainashley]

You really need to brush up on your basic general chemistry.

The balanced equation for this reaction is:

H2SO4 + 2NaOH --> Na2SO4 + 2H2O

For every mole of H2SO4, it reacts with 2 moles of NaOH.

ok, so what happens to the other H?

 

[rainashley]

Could you please elaborate on this: The sulfuiric acid is producing 2 moles of protons to react. So every 1 M of sulfuric acid produces 2 M of H+ to react with the NaOH.
Thanks

 

[Isoprop]

H2SO4 will react with 1 OH- to produce HSO4- and water.
Then the HSO4- will react with with another OH- to produce SO4-2 and water.

So both hydrogens come off H2SO4 to react with NaOH.

 

[determinedone]

You found the error!!! GOOD JOB!!! I hope you didn't think you'd actually be able to do all of that on the MCAT...time is of the essence!!! One more thing, you've got to understand what's going on CONCEPTUALLY and not absolutely QUANTITATIVELY (at least up til calculating pH).

Now for my REAL advice...REVIEW Acids/Bases in your g.chem textbook and supplement with your MCAT (physical sciences section) prep book, and most importantly PRACTICE problems that are very similar to this question and to make sure you fully understand Acids/Bases in general...you will see it again on Test Day!!! Best wishes to you!!!

 

[rainashley]

Greetings RainAshley:

M of -OH = N of -OH/moles of -OH = (16 X 10^-3)/(2)= 8X 10^-3 M -OH

i dont think youre supposed to divide by 2 here
whyd you divide by 2?

 

[determinedone]

Quote:
Originally Posted by determinedone
Greetings RainAshley:

M of -OH = N of -OH/moles of -OH = (16 X 10^-3)/(2)= 8X 10^-3 M -OH


i dont think youre supposed to divide by 2 here
whyd you divide by 2?

To get molarity from normality you must divide the normality of the unknown acid or base equivalents by the number of moles of the acid or base equivalents. You are dividing by 2 here because, according to the balanced reaction, there are 2 -OH ions total that dissociate, since you now have 2 moles of the base so, now 2 moles of -OH ions dissociate. So, that's why you have to divide the normality by 2.

Hope this helps!

 

[rainashley]

I am sorry, I still dont understand...
could someone PLEASE explain clearly why you MULTIPLY by 2 and ALSO DIVIDE by 2? Why both? Specifically?
So sorry
thanks a lot
i reallyyyy appreciate it

 

[BerkReviewTeach]

i dont think youre supposed to divide by 2 here
whyd you divide by 2?

All of the coaxing from the previous posters such as bleagrh and isoprop paid off. Your are right, you don't divide by 2 there. You recognized determinedone's error.

So if you believe that the OH- concentration is 16 x 10exp-3 before titration begins, all you need to do is take the pOH of that solution. A few posters have already pointed out the math:

pOH = -log(1.6 x 10exp-2) = 2 - log 1.6 = 2 - (about 0.2 or so) = 1.8. If pOH = 1.8, then pH = 12.2.

You can do this.

 

[rainashley]

All of the coaxing from the previous posters such as bleagrh and isoprop paid off. Your are right, you don't divide by 2 there. You recognized determinedone's error.

So if you believe that the OH- concentration is 16 x 10exp-3 before titration begins, all you need to do is take the pOH of that solution. A few posters have already pointed out the math:

pOH = -log(1.6 x 10exp-2) = 2 - log 1.6 = 2 - (about 0.2 or so) = 1.8. If pOH = 1.8, then pH = 12.2.

You can do this.

Now I am MORE confused... what goes on exactly... could someone PEASE tell me and CLARIFY what is going on exactly? Why do you ONLY multiply by 2? Specifically?
Thanks

 

[Pulsar]

The number of moles of H+ have to equal the number of moles of -OH. Since H2SO4 has two acidic hydrogens, it will contribute two H+'s for every molecule of H2SO4.

# of moles of H+ = (2 mol H+ / 1 mol H2SO4) * (4*10^-3 mol H2SO4 / 1000 mL H2SO4) * (60 mL H2SO4) = 4.80 * 10^-4 moles H+

This means there must be 4.80 * 10^-4 moles of -OH, since we said H+ = -OH.

The concentration of OH is the number of moles of OH (4.80*10^-4) divided by the volume (30 mL).

Concentration OH = mol OH / volume = 4.80*10^-4 / 0.030 L = 0.016M

pOH = -log(OH concentration) = -log(0.016M) = 1.796

pH + pOH = 14 => pH = 14-pOH = 14-1.796 = 12.2

 

[determinedone]

Now I am MORE confused... what goes on exactly... could someone PEASE tell me and CLARIFY what is going on exactly? Why do you ONLY multiply by 2? Specifically?
Thanks

Oh, dear!!! Sorry about that, but I'm glad you found the error!!!
GOOD JOB!!! I hope you didn't think you'd actually be able to do all of that on the MCAT...time is of the essence!!! One more thing, you've got to understand what's going on CONCEPTUALLY and not absolutely QUANTITATIVELY (at least until calculating pH).

Now for the best advice I can give you at this point...REVIEW Acids/Bases in your g.chem textbook and supplement with your MCAT (physical sciences section) prep book, and most importantly PRACTICE problems that are very similar to this question and make sure you fully understand Acids/Bases in general...you will see it again on Test Day!!! Best wishes to you!!!

 

[RogueUnicorn]

OP, you need a tutor, or at least a refresher class.

 

[rainashley]

OP, you need a tutor, or at least a refresher class.

No, what I need is for someone to explain this to me CORRECTLY
I am getting different answers from everyone... and the answers are not clear
could someone please tell me the correct quantitative and qualitive way of going about this? the correct way?
please
i really appreciate it

 

[RogueUnicorn]

lol ok. good luck then.