Adding Sodium Oxalate to hard/soft/distilled water.

[Esoremada]

Homework Statement

Im supposed to predict what will happen if I add Na₂C₂O₄ to hard water, soft water, or distilled water. Hard water has Mg²⁺ and Ca²⁺ ions dissolved in it from calcium bicarbonate/carbonate and magnesium bicarbonate/carbonate.

It's a theoretical lab where we predict what would happen if we did the lab.

A student added 1 mL of hard water to one test tube, 1mL of soft water to another, and 1 mL of distilled water to a third. The hard water sample contained Ca²⁺ and Mg²⁺. The soft water contained lower concentrations of these ions.

Into each test tube, the student put two drops of 0.1 mol/L sodium oxalate solution. Then the student mixed the contents.

We didn't learn about measurements in class yet, I don't know what 0.1 mol/L means, so I'm assuming the volumes are irrelevant?

The Attempt at a Solution

Hard water:

MgCO_{3 (aq)} + Na₂C₂O_{4 (aq)} + H₂O_(l) -> MgC₂O_{4 (s)} + Na₂CO_{3 (aq)} + H₂O_(l)

(am I supposed to write H₂O in the chemical equation?)

A double displacement reaction would occur because MgC₂O₄ is a solid so a precipitate would be formed.

Soft water:

I'm not sure what to do here.

Distilled water:

Na₂C₂O_{4 (aq)} + H₂O _(l) -> 2NaOH _(aq) + H₂C₂O_{4 (aq)}

Does a reaction occur here? How do I know when an acid can decompose? It seems like it could turn in to

Na₂C₂O_{4 (aq)} + H2O-> ₂NaOH + H₂O _(l) + CO_{2 (g)} + CO _(g)


I don't completely understand how to know if a reaction occurs in a double displacement reaction with aq elements as products that are bases/acids

 

[Borek]

We didn't learn about measurements in class yet, I don't know what 0.1 mol/L means, so I'm assuming the volumes are irrelevant?

0.1 mol/L is a concentration, most likely it is irrelevant.

MgCO_{3 (aq)} + Na₂C₂O_{4 (aq)} + H₂O_(l) -> MgC₂O_{4 (s)} + Na₂CO_{3 (aq)} + H₂O_(l)

(am I supposed to write H₂O in the chemical equation?)

No, you shouldn't put water in the reaction when it is just a solvent.

Reaction is correct, but you should add another, separate reaction for calcium.

Technically these should be net ionic reactions, but if you don't know what these are, don't worry.

Soft water:

I'm not sure what to do here.

Honestly - I am not sure what to do too. Soft water contains both Mg²⁺ and Ca²⁺ as well, just in much lower concentration. Whether they will precipitate or not depends on the concentration of ions and solubility (more precisely - solubility product), so there is no simple answer to that.

Distilled water:

Na₂C₂O_{4 (aq)} + H₂O _(l) -> 2NaOH _(aq) + H₂C₂O_{4 (aq)}

Your other reaction is completely off, this one makes partial sense - but if you have just started to learn chemistry and you know nothing about acid/base equilibrium and hydrolysis, it is probably better to say nothing happens.

 

[Esoremada]

0.1 mol/L is a concentration, most likely it is irrelevant.

No, you shouldn't put water in the reaction when it is just a solvent.

Reaction is correct, but you should add another, separate reaction for calcium.

Technically these should be net ionic reactions, but if you don't know what these are, don't worry.

Honestly - I am not sure what to do too. Soft water contains both Mg²⁺ and Ca²⁺ as well, just in much lower concentration. Whether they will precipitate or not depends on the concentration of ions and solubility (more precisely - solubility product), so there is no simple answer to that.

Your other reaction is completely off, this one makes partial sense - but if you have just started to learn chemistry and you know nothing about acid/base equilibrium and hydrolysis, it is probably better to say nothing happens.

Thanks for the answer. I was taught that to predict a double displacement reaction you make the equation and a reaction occurred:

-if there's a gas liquid or solid in the products
-if both products are aq and one is a weak acid or base (can break down)

So if one product was H₂CO₃ it would break down into carbon dioxide and water meaning a reaction occurred.


Also how bout this

Na₂C₂O_{4 (aq)} + H₂O _(l) -> Na₂O _(aq) + H₂C₂O_{4 (aq)}


Would the oxalic acid break down, or would no reaction occur?


edit, meant Na₂O

 

[Borek]

No such thing as NaO(aq). First, it is Na₂O if anything, second, Na₂O reacts immediately with water, so it can't exist as (aq).

Rules you have learned are not without a merit, but they are very simplified, so they can be misleading in more complicated cases.

 

[DeeNos]

Adding sodium oxalate to hard water will result in a double displacement reaction, forming a precipitate of magnesium oxalate (MgC2O4) and sodium carbonate (Na2CO3). The reaction equation you wrote is correct, and it is not necessary to include H2O as it is a solvent and does not participate in the reaction.

In soft water, the lower concentration of calcium and magnesium ions may not be enough to form a significant amount of precipitate. However, a small amount of magnesium oxalate may still form.

In distilled water, no reaction will occur as there are no ions present to react with sodium oxalate. The equation you wrote for this reaction is also correct, and no further decomposition would occur.

In general, acids can decompose when they are heated or when they react with a base. In this case, the sodium oxalate is a salt of a weak acid (oxalic acid) and a strong base (sodium hydroxide), so it is unlikely to decompose unless heated to high temperatures. Additionally, the reaction with water to form sodium hydroxide and oxalic acid is a reversible reaction, so it would not produce significant amounts of carbon dioxide and carbon monoxide.