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$\ce{Na2CO3}$ and $\ce{NaHCO3}$ have similar reactions. For example, they can react with acid to produce $\ce{CO2}$.
So my question is:

Can we use a chemical test to distinguish between $\ce{Na2CO3}$ and $\ce{NaHCO3}$?


I know that $\ce{NaHCO3}$ can be thermal decomposed into $\ce{Na2CO3}$ via this reaction: $2\ce{NaHCO3->} \ce{Na2CO3} + \ce{CO2} + \ce{H2O}$
However, I am not sure if thermal decomposition is a chemical test.

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    $\begingroup$ You should give more info about your sample. Is solid? Is a disolved in water? In such case, do you know the concentration? I can not imagine something more direct than just check the pH of a solution. $\endgroup$ – user1420303 Jul 18 '16 at 17:00
  • $\begingroup$ The pH of Na2CO3 at 1mM = 10.52; Molar mass of Na2CO3: 105.9888 g/mol. The pH of NaHCO3 at 1mM = 8.27; Molar mass of NaHCO3: 84.007 g/mol. $\endgroup$ – Ben Welborn Jul 18 '16 at 17:12
  • $\begingroup$ You could also test for evaporating water (for instance, by having the the gaseous products go through some waterfree copper-(II)-sulfate, which would turn blue in presence of water). This requires, of course, assurance that it can only be, in fact, sodium carbonate or sodium hydrocarbonate. $\endgroup$ – Zubo Jul 18 '16 at 17:58
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As you mentioned in your post, thermal decomposition and measurement of the resulting gas would be one way to distinguish what you have. I would call this a physical, rather than chemical method, however.

This kind of thermal decomposition isn't something most labs would consider for two reasons:

  1. Both chemicals are cheap. They're generally bought by the kilo and looking at what we get charged for them the company is basically only charging us for the cost of the containers and shipping.

  2. Thermal decomposition and gas measurement is, nowadays at least, a fairly specialised technique. Most labs would struggle to piece together the required equipment and even if they could find it, would likely be reluctant to do so.

That said, your question seems purely academic rather than practical, so lets consider a possible chemical method.

Sodium carbonate is well known for its use in inorganic analysis, as it reacts with certain metals to give carbonate salts of well defined colour.

The test can distinguish between copper (Cu), iron (Fe), and calcium (Ca), zinc (Zn) or lead (Pb). Sodium carbonate solution is added to the salt of the metal. A blue precipitate indicates Cu2+ ion. A dirty green precipitate indicates Fe2+ ion. A yellow-brown precipitate indicates Fe3+ ion. A white precipitate indicates Ca2+, Zn2+, or Pb2+ ion. (Source: Wikipedia)

IIRC the test doesn't work with sodium bicarbonate, and as such you could use the reactions with a cheap metal such as iron to determine whether you had sodium bicarbonate or sodium carbonate.

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Heating salt is an important test in determining the anion in the salt by checking the gas evolved. This test is called dry heating test and is a important part of qualitative analysis of salt. But this test goes under physical examination of salt (checking color, appearence etc.).

One important chemical test to differentiate carbonate and bicarbonate is $\ce{MgSO4}$ test.

$$\ce{Na2CO3 + Mg^2+ -> MgCO3↓(white) + Na+}$$

$$\ce{NaHCO3 + Mg^2+ -> MgCO3↓(white) + H2O + CO2 ^}$$

When carbonate salt is reacted with $\ce{MgSO4}$, a white salt precipited while bicarbonate salt on reacting with $\ce{MgSO4}$, not only a white salt precipited but also a brisk effervescence occur. The gas is detected by passing it through lime water.

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depends on your equipment. You can estimate the amount of $\ce{NaHCO3}$ by taking a known amount of sample, heating it to $\sim\pu{300 °C}$, cooling it to room temperature and measuring the weight again. Zero loss means 100% $\ce{Na2CO3}$. Loss of $37\%$ of weight means $100\%$ $\ce{NaHCO3}$. The dependence is linear within $0-37\%$ weight loss.

Better way is to react a gram of material with $\ce{HCl}$, send forming $\ce{CO2}$ through saturated $\ce{Ba(OH)2}$ ($\ce{Ca(OH)2}$ can be used, but wouldn't work as well), collect, wash dry and weight the precipitate $\ce{BaCO3}$. Both experiments tell you how much carbonate anion you have in a sample. You have to assume that $\ce{NaHCO3}$ and $\ce{Na2CO3}$ were the only components in the mix.

You can also evaluate the ratio based on pH of the resulting solution if you know how to calculate pH for buffers.

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  • $\begingroup$ Feel free to comment your downvotes. $\endgroup$ – sixtytrees Jul 18 '16 at 17:05
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    $\begingroup$ You should try formatting your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance. $\endgroup$ – bon Jul 18 '16 at 18:08
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Dissolve a sample in water to a known concentration and measure the pH. Carbonate ion is a stronger base than bicarbonate ion and the pH will behave accordingly.

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