Ionic reaction

[akcapr]

In ionic reactions a new ionic compund forms, sometimes a percipetate or just a solid. I want to know why a reaction happens and sometimes doesnt. For example: Silver Nitrate and Sodium Chloride react to make new ionic compounds- silver chloride and Sodium nitrate. But how is it determined if reaction will happen? Are the produced compunds somehow more favorable to make than the starting 2? If you mix NaCl and Kno3 no new compound is formed- why do some react and some dont?

 

EDIT: I guess Nacl and KNO3 do react, but the point being why some dont and some do.

[RyanJ]

In ionic reactions a new ionic compund forms, sometimes a percipetate or just a solid. I want to know why a reaction happens and sometimes doesnt. For example: Silver Nitrate and Sodium Chloride react to make new ionic compounds- silver chloride and Sodium nitrate. But how is it determined if reaction will happen? Are the produced compunds somehow more favorable to make than the starting 2? If you mix NaCl and Kno3 no new compound is formed- why do some react and some dont?

 

 

I think you'll get [ce]KCl[/ce] and [ce]NaNO_3[/ce] - I think the element with the highest electropositivity will displace the one with the lower electropositivity. Maybe it also has something to do with the new stability of the ions in the exchange but I'm not too shure about that

 

These type of reactions normaly occur in th following form:

 

[math]\ce{A + BX -> AX + B}[/math] is A is more reactive than B.

 

E.G.

 

[math]\ce{CaCl_2 + K -> KCl_2 + Ca}[/math]

 

Cheers,

 

Ryan Jones

[xeluc]

Before I say anything, I just want you to know that this is what I think is going on, so maybe you should wait till someone agrees or corrects me before you take this...

 

Yeah, that works for single replacement reactions, but double replacement reactions are a little more difficult. I know that if you have two solutions and mix them together and one or even two of the four possible compounds (2 of which being what you started out with of course) is insoluble, a reaction will occur. Ex: CuCl2 (aq) + NaOH (aq) --> Cu(OH)2 + NaCl (aq) . I believe that when all possible compounds are soluble, then maybe there wouldn't BE a reaction? the way I see it, you just have lots of Ions floating around, so really if you mix 2KCl (aq) + 2NaOH (aq) and you then evaporate the water, you might get KCl + NaCl + NaOH + KOH. So it's like an Equilibrium. This of course only applies to aqeous solutions. If you have solid reactants, then I supose you'd figure out each Ions tendency to steal away a different Ion.

Like..

Na = Good at displacing

Cu = Not good.

F = I think good at displacing, It's sure reactive..

OH = Dunno, but it's not as good as F

 

So I THINK NaOH + CuF would cause a double replacement reaction. I'll need someone to second this though...

[akcapr]

Also, I'm trying to figure out what the products of Sodium bIcarbonate + Iron (III) sulfate are. Some iron hydroxide surely.

[woelen]

If you have a solution of ionic compounds, then you really have separate ions in solution.

 

Suppose you take 1 mol of KNO3 and 1 mol of NaCl and you dissolve both solids in 1 liter of water and label this beaker 'A'.

Suppose you take 1 mol of NaNO3 and 1 mol of KCl and you dissolve both solids in 1 liter of water and label this beaker 'B'.

 

Now, you give both beakers, labeled 'A' and 'B' to a super-chemistry-expert and you ask him/her which beaker contains KNO3+NaCl and which contains NaNO3+KCl, then that expert will have NOT A SINGLE WAY to determine which was put in 'A' and which was put in 'B'. The expert only can observe that the beakers contain Na(+), K(+), NO3(-) and Cl(-) ions.

 

So, solutions of ionic compounds really have separate ions in solution and not entities like NaNO3 and so on.

 

Many ionic compounds have limited solubility, which is specified by means of a number Ksp (solubility product), which for an ionic compound [math]C_nA_m[/math] (C=cation, e.g. Na(+), A is anion, e.g. NO3(-)) equals

 

[math]Ksp = max(\[A^{m+}\]^n\[b^{n-}\]^m)[/math]

 

As soon as the product of the ions in solution exceeds Ksp for the given compound, then it precipitates from solution, such that the concentration becomes equal to Ksp. So, Ksp tells in some sense how much of a compound can be in solution.

 

When two compounds with large Ksp (e.g. NaCl and AgNO3) are mixed in a single solution, then because of the fact that all ions exist separately (see first part of this lengthy post), the following four Ksp's come into play:

 

Ksp(NaCl)

Ksp(AgCl)

Ksp(NaNO3)

Ksp(AgNO3)

 

Ksp(AgCl) is much smaller than all other Ksp's and you can expect AgCl to precipitate out of solution.

 

So, a reaction occurs, when the product of ionic concentrations of one of the possible combinations of ions exceeds the value for the Ksp for that compound.

 

So, when CuSO4 and NaOH are mixed, then the following Ksp's come into play:

 

Ksp(CuSO4) = max([Cu(2+)]*[sO4(2-)])

Ksp(Cu(OH)2) = max([Cu(2+)]*[OH(-)]*[OH(-)])

Ksp(Na2SO4) = max([Na(+)]*[Na(+)]*[sO4(2-)])

Ksp(NaOH) = max([Na(+)]*[OH(-)])

 

Here, the value of [Cu(2+)]*[OH(-)]*[OH(-)] easily exceeds the maximum allowed value, because Ksp(Cu(OH)2) is very small. So, a precipitate of copper hydroxide is formed.

 

As Xeluc states, when two solutions of ionic compounds are mixed and none of the Ksp's is exceeded for any of the four combinations, then no reaction occurs. The four different ions simply coexist in the solution.

[akcapr]

Ahh thank you woelen. So if I understand correctly, if the solubility of one of the possible ion combinations has poor solubility then it will be formed, along with the correspondign byproduct.

[woelen]

Ahh thank you woelen. So if I understand correctly, if the solubility of one of the possible ion combinations has poor solubility then it will be formed, along with the correspondign byproduct.

Yes, you could say it like this. Easy to remember. The Ksp-things, however, also give you some quantitive powers. There are large tables with Ksp values for many salts. With such a table nearby you can predict which combinations give a reaction and which not.

 

An example of such a table is this:

 

http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm

[akcapr]

thx. Also woelen, do you know what the products of Sodium bIcarbonate + Iron (III) sulfate are?

[woelen]

thx. Also woelen, do you know what the products of Sodium bIcarbonate + Iron (III) sulfate are?

This is a difficult one. In order to understand this, I need to introduce the concept of hydrolysis. This cannot be solved by using the simple Ksp-story as explained before.

 

All metal ions, when dissolved in water, form aqua-complexes. When iron (III) sulfate is dissolved, then in water, you get Fe(3+) ions. However, these ions do not exist as bare ions, in fact, water molecules form a complex with metal ions.

 

The Fe(3+) ion happens to form hexaaqua complexes (many metals form hexaaqua complexes, but some form tetraaqua complexes or even other numbers).

 

So, a solution of ferric sulfate contains ions

 

[Fe(H2O)6](3+), nice formulation: [math][Fe(H_2O)_6]^{3+}[/math]

 

These aqua complexes have a central metal core, the oxygen atoms of the water-ligands are pointing towards the metal core and the two hydrogen atoms are pointing outwards. The charge of the ion as a whole is not simply located on the metal core, but it is distributed over the whole hexaaqua-metal ion, that is why I write [Fe(H2O)6](3+), with the (3+) charge distributed over all atoms between []. For many metals, the story ends here. The water molecules are called ligands of a complex. Complexes can have other ligands, e.g. NH3, CN(-), Cl(-), NO(+), but here the ligand is H2O.

 

For iron (III), another effect comes into play. The charge of the total ion is distributed, such that a relatively large part is on the H-atoms, pointing outwards. This makes it fairly easy to loose a H(+) atom. Hence, the hexaaqua-iron (III) ion is an acid!

 

[math][Fe(H_2O)_6]^{3+} <-> [Fe(H_2O)_5(OH)]^{2+} + H^+[/math]

 

The ion can even loose a second and a third proton! If three protons are lost, hydrous ferric hydroxide is formed:

 

[math][Fe(H_2O)_3(OH)_3][/math]

 

This phenomenon that a metal salt solution is acidic is called hydrolysis. It is an extremely important concept and one of the key-concepts for understanding reactions in inorganic chemistry. So, a water-ligand can change into a hydroxo-ligand by splitting off a proton.

 

This is what you'll observe when sodium bicarbonate is added. The H(+) from the acidic hexaaqua iron (III) ion is taken up by the bicarbonate ion, CO2 bubbles out of sulution and water is formed. Finally iron (III) hydroxide is precipitated.

 

--------------------------------------------------------------

 

Many metals hydrolyse in solution. Ferric ions do to quite some extent. Real hexaaqua iron (III) ions are almost colorless. The yellow/brown color of ferric solutions are due to the pentaaqua-hydroxo iron (III) ion.

 

A metal like bismuth hydrolyses much more. Even in 1 mol/l acid, it still hydrolyses so much, that bismuth hydroxide is precipitated or in the presence of nitrate, basic bismuth nitrate (the Ksp of Bi(3+)/OH(-)/NO3(-) is quite low).

 

If you want to see what color real iron (III) has, without hydroxo-ligands, look here at the section for oxidation state +3:

 

http://woelen.scheikunde.net/science/chem/solutions/fe.html

 

Here some misconceptions about the color of ferric salts and their solutions are covered.

[jdurg]

Actually woelen, if you have two beakers, one with NaNO3 and KCl, and another with KNO3 and NaCl, there is a way to find out which is which. If you SLOWLY evaporate the water out of the solution, eventually you'll reach a point where the Ksp of the highly soluble salts come into play. As a result, the less soluble salt will begin to ppt out of solution. Since the Ksp's of NaNO3, KCl, NaCl, and KNO3 are not exactly the same, they can all be fractionally crystalized out of solution. It's a big P.I.T.A. to do that, but it can be done.

[woelen]

Actually woelen, if you have two beakers, one with NaNO3 and KCl, and another with KNO3 and NaCl, there is a way to find out which is which. If you SLOWLY evaporate the water out of the solution, eventually you'll reach a point where the Ksp of the highly soluble salts come into play. As a result, the less soluble salt will begin to ppt out of solution. Since the Ksp's of NaNO3, KCl, NaCl, and KNO3 are not exactly the same, they can all be fractionally crystalized out of solution. It's a big P.I.T.A. to do that, but it can be done.

No, I do not agree with you. If in one beaker, you have dissolved 1 mol each of NaNO3 and KCl and in the other you have dissolved 1 mol each of KNO3 and NaCl, then in both beakers you will have 1 mol of each of the following ions: Na(+), K(+), NO3(-), Cl(-). If then you slowly evaporate the solutions, in both cases the same salt first crystallizes, because both beakers contain the same ions in the same concentrations.

[jdurg]

Ahhh. My bad. I didn't read that it was equal amounts added to each beaker. Oops. (Didn't have the time to read every paragraph. )