# Question 395a2

Aug 23, 2015

The answer is (D) glacial ethanoic acid + sodium carbonate.

#### Explanation:

The way I see it, the answer is actually provided in the question.

Notice that test tube (C) contains calcioum carbonate chips, which is another way of saying the you're reacting glacial ethanoic acid and solid calcium carbonate.

Test tube (D) does not specify that you're dealing with solid sodium carbonate, which could mean that you're actually mxiing glacial ethanoic acid with a solution of sodium carbonate - this is your culprit!

First of all, glacial ethanoic acid means pure ethanoic acid. SImply put, you're not dealing with a solution of ethanoic acid (think vinegar), you're dealing with pure ethanoic acid.

Let's start with test tube (A). The reaction between the glacial ethanoic acid and the pH paper will not take place because ethanoic acid is not really an acid unless it is in aqueous solution.

Since pure ethanoic acid contains only ethanoic acid molecules, which implies of course that no water molecules are present, all the ethanoic acid molecules will remain undissociated.

The same principle applies to test tube (B). Magnesium will react with ethanoic acid solutions because the presence of water molecules allow some of the ethanoic acid molecules to dissociate.

In pure ethanoic acid the molecules are undissociated, which is why the reaction that usually forms magnesium ethanoate, "Mg"("CH"_3"COO")_2, and hydrogen gas, ${\text{H}}_{2}$, will not take place.

For test tube (C), you react solid calcium carbonate with glacial ethanoic acid. Once again, the fact that you have no water molecules present to allow for some of the ethanoic acid molecules to donate their acidic protons will prevent the reaction from taking place.

In aqueous solution, ethanoic acid reacts with calcium carbonate to form calcium ethanoate, "Ca"("CH"_3"COO")_2, carbon dioxide, ${\text{CO}}_{2}$, and water.

This is exactly what will happen in test tube (D). If indeed you're reacting glacial acetic acid with a sodium carbonate solution, the reaction will produce sodium ethanoate, $\text{NaCH"_3"COO}$, carbon dioxide, and water.

That happens because the sodium carbonate solution contains water molecules, which means that a part of the ethanoate molecules will ionize and release their acidic protons.

This will provide enough protons to the reaction to allow for the formation of carbon dioxide and water.

The balanced chemical equation for this reaction is

$2 {\text{CH"_3"COOH"_text((aq]) + "Na"_2"CO"_text(3(aq]) -> 2"NaCH"_3"COO"_text((aq]) + "CO"_text(2(aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

In essence, you can think about this reaction in terms of the protons coming from the ethanoic acid

overbrace(2"H"_text((aq])^(+))^(color(red)(stackrel("coming from")("ethanoic acid ionization"))) + "CO"_text(3(aq])^(2-) -> "CO"_text(2(aq]) + "H"_2"O"_text((l])#