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According to the Wikipedia article IUPAC nomenclature of inorganic chemistry,

he prefix bi- is a deprecated way of indicating the presence of a single hydrogen ion

A very common example is the commonplace 'bicarb of soda', or sodium bicarbonate (or using its correct chemical name sodium hydrogen carbonate). On the Wikipedia page for sodium bicarbonate there is a note that states that:

The prefix "bi" in "bicarbonate" comes from an outdated naming system and is based on the observation that there is two times as much carbonate ($\ce{CO3}$) in sodium bicarbonate ($\ce{NaHCO3}$) and other bicarbonates as in sodium carbonate ($\ce{Na2CO3}$) and other carbonates.

However, the statement above is not particularly clear, especially as carbonates are not the only compounds that used the 'bi' prefix, such as:

Bisulphite $\ce{HSO3-}$

Bisulphate $\ce{HSO4-}$

Why was the prefix 'bi' used to indicate the presence of a single hydrogen ion in compounds?

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According to the Online Etymology Dictionary, the word bicarbonate is derived as follows:

1814, bi-carbonate of potash, apparently coined by English chemist William Hyde Wollaston (1766-1828), from bi- + carbonate.

The question then becomes, what is the meaning of the prefix bi- here? We know it means two, but it also can mean double or twice. If that is the case, then sodium bicarbonate is sodium double carbonate.

Consider a given quantity of acid, let's say $\ce{HCl}$, that we are going to neutralize with sodium carbonate, $\ce{Na2CO3}$, and an equal quantity of acid that we are going to neutralize with sodium bicarbonate, $\ce{NaHCO3}$. We add the sodium carbonate to the acid, and as the neutralization reaction goes forward, it produces some quantity of carbon dioxide. The same process is repeated for sodium bicarbonate, and the carbon dioxide is again collected.

$\ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O}$

$\ce{NaHCO3 + HCl -> NaCl + CO2 + H2O}$

In the first reaction, one unit of sodium carbonate is able to neutralize two units of acid. In the second reaction, one unit of sodium bicarbonate can only neutralize one unit of acid. Another way of looking at this is that double the quantity of carbon dioxide is produced from the sodium bicarbonate neutralization. That is, the compound is a double carbonate.

Similar explanations make sense for bisulfate and bisulfite. The acid-base reactions that a sulfate or sulfite ion might undergo are already partially completed in the bi- versions of the compound, so it takes twice as much sulfate or sulfite to do the same job as the non-hydrogen form.

The hydrogen phosphates don't fit this pattern as nicely.

Additionally, if the cation in these salts is a relatively light weight one, then the required mass of the bicarbonate or bisulfate is substantially more than that of the non-hydrogen form of the compound, though it's not quite double. 53g of $\ce{Na2CO3}$ could neutralize one mole of $\ce{HCl}$, but it takes 84g of $\ce{NaHCO3}$ to do the same job. This is closer to accurate for the more massive sulfur and phosphorus based oxyanions.

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The use of "bi" for 'two" denotes that there are two different anions (e.g. $\ce{Na+, H+}$ in the examples), but it is confusing, because it could imply that there are two of a single anion, so its use is deprecated. Similarly, in English, biweekly can mean either twice a week or every two weeks.

The use of "di" for "two" is more consistent, i.e. it always means two of a single species, as in carbon dioxide, $\ce{CO2}$. [Perhaps "always" is a bit strong... anyone know of contrary examples?]

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  • $\begingroup$ Diphosphate, as in adenosine diphosphate (ADP) is the most common contrary example. "Diphosphate" is a side group composed of two esterified phosphates, $\mathrm{H_3P_2O_7—}$, whereas a bisphosphate compound such as fructose 1,6-bisphosphate bears two $\mathrm{H_2PO_4—}$ side groups, which may or may not be bound to the same atom. (The protonation state of all of these will of course depend on pH.) $\endgroup$ – hBy2Py Jun 1 '15 at 2:32
  • $\begingroup$ I'm curious, do you have a source for this answer? Everything I can find indicates that the answer suggested by santiago in the question is correct. $\endgroup$ – Jason Patterson Jun 1 '15 at 4:23

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