[licamine]

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!

[orthoformate]

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?

[AWK]

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

In undergraduate courses this is always assumed.

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.

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.

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).

[licamine]

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. 

[orthoformate]

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

[AWK]

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.

[orthoformate]

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

[AWK]

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]

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.

[orthoformate]

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.

[AWK]

At this ionic strenght pKa₂ of H₂CO₃ is ~9.7

[orthoformate]

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.

[AWK]

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.

[licamine]

Thank you all for the replies! I finally got it