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Someone introduced $\ce{CO2}$ into saturated $\ce{Na2CO3}$ solution, and observed that white precipitate is produced. He therefore claims that $\ce{NaHCO3}$ has less solubility than $\ce{Na2CO3}$.

His claim is of course true, but does his experiment suffice to support this claim? The solubility of a solid substance refers to the mass of a substance that dissolves when it reaches saturation in $100~\mathrm g$ of water.

The related reaction is $$\ce{Na2CO3 + CO2 + H2O}=\ce{2NaHCO3}\downarrow.$$

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    $\begingroup$ My doubts include: 1. The reaction uses up water; 2. The mass of $\ce{NaHCO3}$ produced is greater than the mass of $\ce{Na2CO3}$ reacted. $\endgroup$
    – youthdoo
    Jan 15 at 10:25
  • $\begingroup$ The essential info is the stage when the precipitation occurs. If it occurs at very beginning, then yes, it would be proof it is less soluble. If it occurs when most of Na2CO3 reacted, it would not be the proof. $\endgroup$
    – Poutnik
    Jan 17 at 4:54

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The solubility of $\ce{NaHCO3}$ is $0.82$ M ($69$ g/L) at $0$°C, $1.95$ M ($164$ g/L) at $60$°C, and the salt start being decomposed at higher temperatures.

As a comparison, the solubility of $\ce{Na2CO3}$ is rather strange. The stable hydrate contains $10$ $\ce{H2O}$ at $\pu{T}$ < $30$°C, and only $1$ $\ce{H2O}$ at $\pu{T}$ > $30$°C. Expressed in moles, the solubility does not depend on the number of $\ce{H2O}$ crystallized with $\ce{Na2CO3}$. This is why we will reason with molar units.

The decahydrate $\ce{Na2CO3·10 H2O}$ has a solubility of $0.245$ $\pu{M}$ ($70$ g/L) at $0$°C, and $1.357$ $\pu{M}$ ( $388$ g/L) at $30$°C. At T > $30$°C, the solubility of the monohydrate $\ce{Na2CO3·H2O}$ is higher but it decreases with T. It is $4.075$ $\pu{M}$ ($505$ g/L) at $30$°C, and $455$ g/L ($3.66$ M) at $100$°C).

As a consequence, if we compare their molar concentrations, $\ce{NaHCO3}$ is more soluble than $\ce{Na2CO3}$ at $0$°$\pu{C}$. This tendency is reversed at $30$°C, where $\ce{NaHCO3}$ has a solubility $1.32$ $\pu{M}$, which is slightly lower than the solubility of $\ce{Na2CO3·10H2O}$ which is $1.357$ $\pu{M.}$ Above $30$°C, $\ce{Na2CO3·H2O}$ is much more soluble than $\ce{NaHCO3}$.

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  • $\begingroup$ But the question is not if it is less or more soluble but if the precipitation is the sign of the lower solubility. Even if solubility of NaHCO3 was by 50% higher than of Na2CO3, it would still start precipitation if reacted stoichiometrically, because of higher mass and slighly less of water. $\endgroup$
    – Poutnik
    Jan 15 at 16:36
  • $\begingroup$ @Poutnik. I don't understand what you say. In my opinion, precipitation is the sign of a lower solubility, on the molar scale. Higher mass and small amount of water do not play the role of solubility. Precipitation happens if the solubility is low. That's all. Except in cases of supersaturation. $\endgroup$
    – Maurice
    Jan 15 at 16:48
  • $\begingroup$ Solubility is evaluated on mass scale, not amount scale. Na2CO3 has from the solubility table on Wikipedia value about 21 g/100 mL. If reacted stoichiometrically with CO2, NaHCO3 would precipitate if having solubility around 35 g/100 mL or less (has less than 10 g/100 mL). So precipitation is NOT a selective sign of lower solubility, as it would precipitate as well if it was somewhat more soluble than Na2CO3. $\endgroup$
    – Poutnik
    Jan 15 at 16:52
  • $\begingroup$ Yes. Precipitation IS a sign of low solubility. It is even the best sign of a low solubility. Or maybe you don't like the word "precipitation". Would you prefer another word ? $\endgroup$
    – Maurice
    Jan 15 at 16:57
  • $\begingroup$ In my opinion, precipitation is the sign of a lower solubility, on the molar scale. - That assumes 1:1 molar ratio. But 1 mol of Na2CO3 forms 2 mol of NaHCO3. So hypothetically and illustratively, if molar solubility would have been 0.2 mol/100 mL for Na2CO3 and 0.3 mol/100 mL for NaHCO3, it would have precipitated even with the higher molar solubility of NaHCO3, as 0.4 mol NaHCO3/100 mL would not stay dissolved ( not speaking about water bonded in HCO3-) $\endgroup$
    – Poutnik
    Jan 15 at 17:31

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