0
$\begingroup$

Baking soda (sodium bicarbonate) + citric acid react when mixed with water to form some amount of carbon dioxide gas ($\ce{CO2}$). I am trying to determine what percentage of the products is carbon dioxide: in other words, how much does reaction make?

$\endgroup$

closed as off-topic by Jan, andselisk, ron, Tyberius, airhuff Oct 2 '17 at 17:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ It makes a lot of CO2. $\endgroup$ – Curt F. Aug 10 '15 at 3:16
  • 2
    $\begingroup$ OK, snark aside, the answer depends on how much citric acid and baking soda you use. What does "a lot" mean to you? "A little"? $\endgroup$ – Curt F. Aug 10 '15 at 3:17
  • $\begingroup$ I mean the percentage, a lot would be upwards of 90% of the reaction, a little would be like %10 $\endgroup$ – Luke Aug 10 '15 at 3:56
  • $\begingroup$ I edited, hope I was right about your intentions $\endgroup$ – Mithoron Aug 10 '15 at 12:55
6
$\begingroup$

I'm not sure what you mean by 90% and 10%. If you mean 90% yield (i.e. 90% or more of the bicarbonate ions in sodium bicarbonate will become carbon dioxide), then yes, that will happen if there is enough acid. How much is enough acid? We'll get to that in a minute. If you mean something else, like 90% of the mass of the components will become carbon dioxide, then that probably will not happen.

Let's look at why 90% of the mass of products is not carbon dioxide.

Bicarbonate anions react with protons from acid to produce carbonic acid, which decomposes to form carbon dioxide:

$$\ce{H+ + HCO3- -> H2CO3 -> H2O + CO2}$$

Let's start with a gram of sodium bicarbonate. How many bicarbonate ions are present? The molar mass of sodium bicarbonate is $80.0066\mathrm{\ g/mol}$, where a mole (mol) is a quantity equal to the number of carbon-12 atoms in 12 grams of carbon-12 or an Avogadro's number $(6.022\times 10^{23})$ of things. Atoms and molecules are very small, and the numbers of them in macro measurements are large.

So we have

$$\mathrm{1.00\ g \times \dfrac{mol}{80.0066\ g}= 0.0125\ mol}$$ $$\mathrm{0.0125\ mol \times \dfrac{6.022\times 10^{23}}{mol}=7.53\times 10^{21}\ molecules}$$

Let's stick with the mole unit since it gives us easier to conceptualize numbers.

The molar mass of carbon dioxide is $\mathrm{44.01\ g/mol}$, so if all of the bicarbonate ion in sodium bicarbonate is converted into carbon dioxide, we will get $\mathrm{0.0125\ mol}$ of carbon dioxide which is

$$\mathrm{0.0125\ mol \times \dfrac{44.01\ g}{mol}=0.550 \ g}$$

So 1.00 gram of sodium bicarbonate will product 0.550 grams of carbon dioxide. We have not even accounted for the citric acid yet. Citric acid is $\ce{H3C6H5O7}$, and it is tempting to think that some of those carbon and oxygen atoms in that molecule might become molecules of carbon dioxide. They can, but not in a reaction with sodium bicarbonate (acid-base). If you react citric acid with oxygen (combustion), then you can produce carbon dioxide.

How much acid?

Citric acid serves as a proton source - in fact it has three acidic protons. That means in theory, each molecule of citric acid can donate three protons to convert three bicarbonate anions into carbon dioxide:

$$\ce{H3C6H5O7 + 3HCO3- -> C6H5O7^3- + 3H2O + 3CO2}$$

However, those three acidic protons are not equal (and the reason for that would make a great follow-up question). Two of the protons are more acidic than carbonic acid, and the third one is slightly less acidic than carbonic acid. In theory, since carbonic acid decomposes into cabron dioxide, which escapes from solution, we can count on Le Châtelier's principle to pull even that third proton. However, in practice, we might not want to wait around. Let's use one equivalent of citric acid per two equivalents of sodium bicarbonate:

$$\ce{H3C6H5O7 + 2HCO3- -> HC6H5O7^2- + 2H2O + 2CO2}$$

So, to figure out how much citric acid (molar mass $\mathrm{192.12\ g/mol}$) we need, we have to do some calculations. Remember that 1 gram of sodium bicarboncate is 0.0125 moles, and we need 0.5 moles of citric acid for every mole of bicarbonate:

$$\mathrm{0.0125 \ mol\ bicarb\times \dfrac{1\ mol\ CA}{2\ mol\ bicarb}\times\dfrac{192.12\ g\ CA}{1\ mol\ CA}=1.20\ g\ citric\ acid}$$

$\endgroup$
  • $\begingroup$ Thanks Ben, that's all the chemistry (plus some) I forgot from 30 plus years ago. I sure lost it from lack of using it. Given your careful calculations, I go back to the kitchen from whence the question arose and find roughly two teaspoons of Citric Acid to a bit over three quarters of a teaspoon of Bicarbonate of Soda (Baking Soda) to get to your ratio of 1.20 g of Citric acid to 0.55 grams of Baking Soda. 2 teaspoons of Citric Acid weigh close to 8.5 grams. A little over 3/4 teaspoon of Baking Soda weighs in at 4.25 grams. $\endgroup$ – FredRock Oct 1 '17 at 20:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.