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Topic: Quantitatively determining the composition of NaHCO3/ Na2CO3 solution  (Read 4094 times)

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Offline licamine

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Given a  400 mL of a 4.82% (by mass) NaOH solution with a density of 1.05 g/cm3 through which 0.400 mol of CO2 are bubbled, how does one quantitatively determine the composition of the final solution?

So I know the reactions which take place are:

NaOH + CO2 --> NaHCO3
2NaOH + CO2 --> Na2CO3

I figured the number of NaOH moles would be 0.506, and from there on I'm not sure how to proceed. I tried thinking that, for all of the CO2 to react, the mole percentage of each reactant could be given by:

0.400 = 0.506x + 0.253y
x + y = 1

In which x is the NaHCO3 percentage and y is the Na2CO3 percentage (which I got to given the stoichometric proportions 1NaOH : 1CO2 : 1NaHCO3 and 2NaOH : 1CO2 : 1Na2CO3), but the result I got to doesn't match the expected one.

Answer:
The resulting solution contains 5.53% NaHCO3 and 2.64% Na2CO3 by mass.

Thank you very much!

Offline orthoformate

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I'm assuming when they say bubbled, they mean CO2 enters the solution and does not leave.

Do you know the equilibrium constant for the first reaction? (carbon dioxide to sodium bicarbonate)

The second reaction you wrote down is not correct, do you see why?

Offline AWK

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I'm assuming when they say bubbled, they mean CO2 enters the solution and does not leave.
In undergraduate courses this is always assumed.

Quote
Do you know the equilibrium constant for the first reaction? (carbon dioxide to sodium bicarbonate)
When the above is assumed the equilibrium constant is unimportant.

Quote
The second reaction you wrote down is not correct, do you see why?
Though this equation is not completely balanced, it is sufficient to calculate correctly its stoichiometry.

Quote from: licamine
0.400 = 0.506x + 0.253y
x + y = 1
Set both equations using moles of CO2 and moles of NAOH, then calculate masses and eventually percentages (if needed).
AWK

Offline licamine

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Quote
Set both equations using moles of CO2 and moles of NAOH, then calculate masses and eventually percentages (if needed).

Thank you! I'm sorry but I still can't seem to get it right... I still don't see what's wrong with my first equations. If all of the 0.506mol of NaOH were to react with CO2 yielding Na2CO3, 0.253mol CO2 would be spent, given the 2:1 molar proportion. If all of it were to react with CO2 yielding NaHCO3, 0.506mol CO2 would be spent, given the 1:1 proportion. Thus, for 0.400mol CO2 to be spent, I thought that the percentage of each one could be given by the equations I set. 

Offline orthoformate

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Is this an OK way to answer the question:

The NaOH is at a concentration of 1.265 M and the CO2 is at a concentration of 1.0 M.

NaOH + CO2 --> NaHCO3
2NaOH + CO2 --> Na2CO3 +H2O

Since CO2 is the limiting reactant, this must be true: Na2CO3 + NaHCO3 = 1M

You can express the carbonates in terms of NaOH consumed: Na2CO3=1/2x & NaHCO3=x

1/2x + x = 1M

x=2/3

Na2CO3=0.33 Mol/L*0.4 L=0.133 mol*105 g/mol=13.96 g
NaHCO3=0.66 Mol/L*0.4 L=0.264 mol*84 g/mol= 22.17 g

these weight %'s are approximately correct



Offline AWK

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Yo have 0.400 mole of CO2 and 0.5061 mole NaOH.
System of equation according to your chemical equations is:
x+y = 0.400 (x concerns Na2CO3, y - NAHCO3)
2x+y = 0.5061
This system can be solved even without calculator and computer and final results (now with SCIENTIFIC CALCULATOR FOR CHEMISTS) are:
11.245g (2.68%) of Na2CO3
and
24.689 g (5.88 %) NaHCO3.
(Note - exact masses of elements within this calculator are used but the same percentage you will get with 84 and 106)

Solve system of equations and check my results.
AWK

Offline orthoformate

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I solved your system:

x+y=0.4
2x+y=0.5061
y=0.5061-2x
x+0.5061-2x=0.4
x=0.1061mol
y=0.2939mol

0.1061mol*106g/mol=11.245 g
0.2939mol*84g/mol=24.689 g

Offline AWK

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My percentage does not take into acount the mass of absorbed CO2 and contains about 5 % error.
Corrected data are 2.57 and 5.64 % respectively.
AWK

Offline AWK

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I solved your system:

x+y=0.4
2x+y=0.5061
y=0.5061-2x
x+0.5061-2x=0.4
x=0.1061mol
y=0.2939mol

0.1061mol*106g/mol=11.245 g
0.2939mol*84g/mol=24.689 g
Better, for x - from equation 2 subtract equation 1.
AWK

Offline orthoformate

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I set up an experiment to mimic this problem.

To a clean dry amber glass jar was added 33.6 g of sodium bicarbonate (0.4 mol) and 400 mL of DI water containing 0.106 mol of NaOH. The mixture was stirred until dissolution was observed. The pH was measured with a calibrated pH probe, giving a reading of 9.47. The ratio of CO3 to CO3H was determined to be 0.14 based on the pKa of NaCO3H value 10.32.

Do you think this experiment accurately mimics the problem we were working on? I removed 0.4 mols from the original 0.506 mols of NaOH because I was adding 0.4 mols of NaHCO3 instead of 0.4 mol of CO2. Since we were assuming all NaOH reacted, I figured this was just a different way to get to the same ending position.

Offline AWK

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Re: Quantitatively determining the composition of NaHCO3/ Na2CO3 solution
« Reply #10 on: July 27, 2016, 03:39:00 PM »
At this ionic strenght pKa2 of H2CO3 is ~9.7
AWK

Offline orthoformate

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Re: Quantitatively determining the composition of NaHCO3/ Na2CO3 solution
« Reply #11 on: July 27, 2016, 03:56:54 PM »
At this ionic strenght pKa2 of H2CO3 is ~9.7

That changes the ratio to 3/5, slightly different than the calculated value of ~1/3. How can we account for this discrepancy?

It makes sense to me that you must consider all the components in the mixture, assume that the system is completely closed and isolated, and calculate where the components must be at equilibrium. Does this make sense.


Offline AWK

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Re: Quantitatively determining the composition of NaHCO3/ Na2CO3 solution
« Reply #12 on: July 27, 2016, 04:16:49 PM »
pH is a measure of activity, not concentration - you may use concentration at ionic strenght ~0.01 (when acivity coefficient of H3O+ is about 0.98 - then difference between calculated and measured pH is between 0.01 and 0.02).
Prepare 0.1 M phospate buffer with calculated pH 7.2 (1:1) - you measured pH will be close to 6.7.
AWK

Offline licamine

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Re: Quantitatively determining the composition of NaHCO3/ Na2CO3 solution
« Reply #13 on: July 27, 2016, 04:45:44 PM »
Thank you all for the replies! I finally got it :)

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