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### Topic: Neutralization reaction between NaOH and H2CO3  (Read 56585 times)

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#### jdgf25

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##### Neutralization reaction between NaOH and H2CO3
« on: April 07, 2008, 07:48:14 PM »
I'm a bit confused with this reaction... is it either

NaOH + H2CO3 -> NaHCO3 + H2O

or

2NaOH + H2CO3 -> Na2CO3 + 2H2O

??

#### Rabn

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #1 on: April 07, 2008, 07:54:47 PM »
Not to confuse you further, but it's actually both...or rather a mix of the two.  Consider a few things and get back to us.

1) Is H2CO3 a strong or weak acid? How does this effect the reaction?

2) What are the spectator ions in this reaction if there are any? How does this effect reaction equation?

#### jdgf25

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #2 on: April 07, 2008, 08:30:37 PM »
Well this is the problem, I have to figure out the concentration of H2CO3 in a titration that we did in class...
this was the process:
1)  pour 300 mL of tap water into a large beaker. add 5-6 drops of phenolphthalein indicator solution, swirl to mix well
2) pour 80 mL of the solution from 1) into an erlenmeyer flask
3) get a person to exhale through a straw into the flask for 20-30 secs
4) measure the amount of CO2 exhaled at this time. titrate the solution in the erlenmeyer flask as follows
a) fill the burette with 0.010M NaOH
measure initial and final volume of the NaOH solution in the burette

so these are the results..

6 mL of NaOH were needed to neutralize the H2CO3 in the water.

then..
n(H2CO3) = n(NaOH)

m(H2CO3) x v(H2CO3) = m(NaOH) x v(NaOH)

but i need the mole ratio
i dont know if i should use 1:1 or 2:1
We did this in biology class, and i took chemistry a looong time ago so i barely remember any of this

#### Rabn

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #3 on: April 07, 2008, 09:05:15 PM »
Well, Let's begin by writing out the system of reaction equations:

CO2 + H2O -> H2CO3

H2CO3 <-> HCO3- + H+

HCO3- + H+ + NaOH -> HCO3- + H2O + Na+

HCO3- + NaOH -> CO32- + H2O + Na+

write the equilibrium equations out for these reactions, make sure you use the proper Ka values, then use this system of equations to solve for H2CO3.  That there should put you on the right track. Try it out, if you get hung up, post what you've done and your ?'s and we'll work from there.

#### jdgf25

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #4 on: April 07, 2008, 10:08:26 PM »
How would i solve the equations if i only have the concentration of NaOH?

I'm sorry if im making stupid questions! i forgot all this

Thanks for your help by the way.

#### Rabn

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #5 on: April 07, 2008, 10:13:44 PM »
Let's look at this another way.  Look at the reaction equations I gave you.  How many equivalents of NaOH does it take to neutralize one equivalent of carbonic acid?

#### jdgf25

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #6 on: April 07, 2008, 10:25:30 PM »
It would take 2 equivalents of NaOH to neutralize carbonic acid..I think

#### Rabn

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #7 on: April 07, 2008, 10:36:03 PM »
According to the reaction equations it would take 2 equivalents, that is correct.  We have to keep equilibrium in mind though; what pH does phenolphthalein indicate when it changes color?

#### jdgf25

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #8 on: April 07, 2008, 10:38:01 PM »
I believe it's 8.3

#### Borek

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #9 on: April 08, 2008, 03:31:15 AM »
Almost OK 8.3 is where the color change starts, but before it becomes apparent you need to go a little bit higher with the pH. Color change ends at around 9.8. You end your titration somewhere in between, around 9 hopefully.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Rabn

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #10 on: April 08, 2008, 04:06:38 AM »
Thanks for jumping in Borek.

I guess here is where I have a question for you jdgf25. Are you expected to get very precise results or just general results?  If you need very precise results you need to use what Borek reported as well as the Ka values for carbonic acid and equilibrium equations. If it is a low-ish level bio course you probably need just general results.

#### Borek

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #11 on: April 08, 2008, 06:02:28 AM »
IMHO - judging from the titration curve its is not possible to get meaningfull results.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### jdgf25

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #12 on: April 08, 2008, 09:48:49 AM »
Yes we are not supposed to know a whole lot about equilibriums, since it is not part of the course. So no, we don't need very accurate results...
the purpose of the experiment is to estimate the rate of cellular respiration. So first i would have to figure out the concentration of H2CO3 and then  the concentration of CO2 in the water and then the amount of CO2 and then the rate of CO2 production.
So this is what I've done so far..I don't think it's right but still...

Concentration of NaOH - 0.010M
Volume of NaOH - 0.006 L
Concentration of H2CO3 - Unknown
Volume of H2CO3 - 0.080 L

at endpoint

n(carbonic acid) = n(sodium hydroxide)
m(carbonic acid) x v = m(NaOH) x v
m(carbonic acid) = 0.010 x 0.006 / 0.080
m(carbonic acid) = 7.5 x 10-4

and then i dont know how to get the concentration of carbon dioxide..
how to set up the equilibrium equations

#### jdgf25

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #13 on: April 08, 2008, 09:56:01 AM »
Do i use

Ka = [CO2][H2O] / [H2CO3]

or
Ka = [CO32-][H+]2 / [H2CO3]

or
Ka = [H+][HCO3] / [H2CO3]

#### Rabn

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##### Re: Neutralization reaction between NaOH and H2CO3
« Reply #14 on: April 08, 2008, 03:35:54 PM »
Find out how many moles of NaOH you added. That is also the moles of OH- you added.  If it takes 2 equivalents of NaOH to neutralize 1 equivalent of H2CO3, how many moles of H2CO3 were there?