Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: hell0joe on February 10, 2017, 12:58:06 AM
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This is a question that I came across that I don't fully understand the solution for:
"You have been supplied with 2M solution of carbonic acid (H2CO3) and a 6M solution of NaOH. What
volumes of these materials are required to prepare 1 L of a 50 mM buffer solution at pH 7 (pKa for
carbonic acid is 6.35)?"
My solution:
Using Henderson-Hasselbalch equation, the ratio of HCO3- to H2CO3 is 4.47:1. With a system of equations I solved for the concentration of each, getting [HCO3-] = 0.041 M and [H2CO3] = 0.009 M.
Then I stated that [HCO3-] = [NaOH] = 0.041 M.
Using the dilution formula I found that 6.8 mL of NaOH are required.
Part 2:
For the volume of H2CO3, I found using the same method of dilution that 4.5 mL of H2CO3 are required.
However, the solution states that 25 mL of H2CO3 are required, rather than 4.5 mL. This seems to suggest that [H2CO3] = 0.05 M, but this is clearly not the case...?
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Solubility of Co2 in water is ~0.05 M. 2 M solution is a nonsense.
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This seems to suggest that [H2CO3] = 0.05 M, but this is clearly not the case...?
Where does the HCO3- come from?
(And yes, AWK is right about the given concentration of H2CO3 being nonsensical, but that's not a reason why you are wrong.)
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It would come from the dissociation of H2CO3.
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Buffers always contain acid (or base) and its salt.
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It would come from the dissociation of H2CO3.
In some small amounts - yes. But most of it comes from other reaction.
What happens when you add strong base to the solution of an acid?
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One of the products would be HCO3-. I meant to say this instead of saying "dissociation".
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One of the products would be HCO3-.
Do you see now what happens to some of the H2CO3?
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One of the products would be HCO3-.
Do you see now what happens to some of the H2CO3?
Yes, it makes sense now. Sorry for the late reply!