Funny Question

[dat_student]

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We have 3 M H+ and 1 M [CO3]2- in a solution. What is the molarity of the solution?

H2CO3 -> H+ + HCO3- -> H+ + [CO3]2-

 

[seoul]

We have 3 M H+ and 1 M [CO3]2- in a solution. What is the molarity of the solution?

H2CO3 -> H+ + HCO3- -> H+ + [CO3]2-

Is the answer 1 M?

 

[seoul]

I think [H2CO3]=4.46 x 10^18 M would be more likely if I understand your question correctly...

 

[dat_student]

I think [H2CO3]=4.46 x 10^18 M would be more likely if I understand your question correctly...

No, How can one have 3 M H+ and 1 M [CO3 ]2- and get 446000000000000000000 M solution????!!!!!!!!

1 M is also wrong.

Hint: "funny question"

 

[seoul]

No, How can one have 3 M H+ and 1 M [CO3 ]2- and get 446000000000000000000 M solution????!!!!!!!!

1 M is also wrong.

Hint: "funny question"

Question is not clear (whether 3 M H+ and 1 M [CO3]2- serve as starting materials or something already in equilibrium with the parent acid) ...

What's your own reasoning?

 

[dat_student]

Question is not clear (whether 3 M H+ and 1 M [CO3]2- serve as starting materials or something already in equilibrium with the parent acid) ...

What's your own reasoning?

I'd like to give everybody a chance to think about it. I'll post something [at some point in the future].

 

[fannian]

We have 3 M H+ and 1 M [CO3]2- in a solution. What is the molarity of the solution?

H2CO3 -> H+ + HCO3- -> H+ + [CO3]2-

Would this be on the DAT

 

[Flipper405]

Just going to toss out a guess: 4 M.

 

[Flipper405]

Where's the answer? 😀

 

[Comet208]

Where's the answer? 😀

He is still figuring it out

 

[fannian]

Where's the answer? 😀

it was a hoax

 

[dat_student]

He is still figuring it out

There is nothing to figure out

 

[fannian]

There is nothing to figure out

 

[Flipper405]

OK, OK, I'm dumb. Someone want to explain this to me? 😕

 

[Flipper405]

seriously ... what's so "funny" about this question? Seems like a typical chemistry question.

 

[RaiderNation]

I'd say that the answer doesn't exist. My reasoning is that all chemistry is fake. It's a hoax that a bunch of science and math nerds came up with. It's all complete crap.

 

[Richnator]

I second that, haha!!!

 

[joonkimdds]

We have 3 M H+ and 1 M [CO3]2- in a solution. What is the molarity of the solution?

H2CO3 -> H+ + HCO3- -> H+ + [CO3]2-


all we gotta do is find molarity of H and CO3 then add them up.
It's that simple.
I think it's very funny, because I would laugh so bad if this shows up on DAT, since this is piece of cake.

 

[dat_student]

We have 3 M H+ and 1 M [CO3]2- in a solution. What is the molarity of the solution?

H2CO3 -> H+ + HCO3- -> H+ + [CO3]2-


all we gotta do is find molarity of H and CO3 then add them up.
It's that simple.
I think it's very funny, because I would laugh so bad if this shows up on DAT, since this is piece of cake.

No. Wrong answer [don't laugh too much] 😉

 

[Flipper405]

Is it even possible to have two different molarities in one solution? This question is really weird. I'm not going to post anything else because it's just frustrating me now. 😡

 

[dat_student]

Is it even possible to have two different molarities in one solution? This question is really weird. I'm not going to post anything else because it's just frustrating me now. 😡

It's possible. You can assume that 2H+ combine with 1 [CO3]2-. In that case, you'll be left with 1 M H+ and 1 M H2CO3 which has a low dissociation constant.

 

[mccarth2]

We have 3 M H+ and 1 M [CO3]2- in a solution. What is the molarity of the solution?

H2CO3 -> H+ + HCO3- -> H+ + [CO3]2-

This is totally just a guess but:

H+ is a strong acid, and (CO3)2- is a very strong base. In solution, equilibrium will shift towards the formation of the weaker conj. acid and base which would be HCO-, but since H+ is in excess H2CO3 will form. The formation of H2CO3 will use up two moles of H+, there will still be 1 mol of H+ left and 1 mol of the newly formed product H2CO3. If they are already in solution together as a 1L soln then the molarity would be 2M, because 2 moles of solute in 1L of solution.

 

[dat_student]

This is totally just a guess but:

H+ is a strong acid, and (CO3)2- is a very strong base. In solution, equilibrium will shift towards the formation of the weaker conj. acid and base which would be HCO-, but since H+ is in excess H2CO3 will form. The formation of H2CO3 will use up two moles of H+, there will still be 1 mol of H+ left and 1 mol of the newly formed product H2CO3. If they are already in solution together as a 1L soln then the molarity would be 2M, because 2 moles of solute in 1L of solution.

No, if that were the case the question wouldn't be funny, bb in 1-2 hr(s)

 

[joonkimdds]

No, if that were the case the question wouldn't be funny, bb in 1-2 hr(s)

We have 3 M H+ and 1 M [CO3]2- in a solution. What is the molarity of the solution?

H2CO3 -> H+ + HCO3- -> H+ + [CO3]2-


first of all....there are couple of errors in this problem
the equation has two arrows , it should only have 1 arrow. I've never seen an equation with two arrows. that's just wrong.

2nd of all.. u said 3M, what does M mean?

 

[Victoria1999]

We have 3 M H+ and 1 M [CO3]2- in a solution. What is the molarity of the solution?

H2CO3 -> H+ + HCO3- -> H+ + [CO3]2-

H2CO3 is going to dissociate very fast to H2O and CO2. This acid is not stable at all.

 

[Flipper405]

bump