Chemical Reactions

[Asian]

For a Double Replacement Reaction, I don't understand how or when we use the "Rule for the Solubility of Salts in Water"

 

Also.......

 

How can you easily predict what state a compound is in, in any chemical reaction.

 

Lastly.......

 

How can you easily tell the difference between a Single Replacement Reaction and a Synthesis/Combination/Composition Reaction because i find it difficult to tell when a synthesis reaction is in the following form "element+compound---------> More Complex Compound" because a single replacement reaction is in the similar form when given the reactants.

 

-Thanks for Any Help

[RyanJ]

1. I don't understand the question.

 

2. Generally you can't unless you know the information before hand. It depends on lots of factors, some you can predict and some are derived by experimentation. For example sometimes a reaction in aqueous solution will produce its products in solution and sometimes as a precipitate.

 

3. I don't understand the use of the term "Replacement reaction" in this context. Are you referring to when one, more reactive compound displaces another from its compound?

 

In general if you are then for a displacement reaction we have this as the general formula:

 

[ce]XY + Z -> ZY + X[/ce]

 

If Z is more reactive than X then it will displace the element and take its place in the compound.

 

I can't think of a way to explain a synthesis type reaction with a general equation because there are so many possible variants and forms it can take.

[Asian]

for #1 there are a list of solubility rules for a double replacement reaction. i dont quite understand when to apply them in that type of reaction.

[encipher]

For a Double Replacement Reaction, I don't understand how or when we use the "Rule for the Solubility of Salts in Water"

 

The solubility rules for salts in water are in use all the time. Whenever two aqueous (or aqueous and solid) mix, Depending on the solubility of the products,we can determine if there will be a precipitate or not.

 

For example:

 

[CE]Ba(NO3)2 + Na2SO4 -> BaSO4 + NaNO3[/CE]

 

In order for us to write a net ionic equation for that reaction we must know what the solubility of the products are, in this case we know that all sodium salts are soluble ( as a general rule) and most nitrates are soluble, therefore NaNO3 will remain aqueous. However, not all barium salts are soluble, and if BaSO4 is looked up on the solubility table, you will realize that it is not a soluble salt. Therefore the net ionic equation would be as follows:

 

[CE]Ba_{(aq)}^2+ + SO4_{(aq)}^2- -> BaSO4_{(s)} [/CE]

 

Hope this helps

[RyanJ]

Well I don't understand the context in which this is used. Solubility wouldn't normally affect the reaction per se. There are reactions where the products of the displacement would form salts that were insoluble and would thus precipitate out, a great example of this is solutions of potassium iodide and lead nitrate are mixed, you get a bright yellow precipitate.

 

[ce]2KI + Pb(NO3)2 -> PbI2 + 2KNO3[/ce]

 

This is because the solubility of lead iodide is really low in water so it precipitates out. These are pretty simple to predict normally based on periodic trends and group trends too.

[encipher]

How can you easily tell the difference between a Single Replacement Reaction and a Synthesis/Combination/Composition Reaction because i find it difficult to tell when a synthesis reaction is in the following form "element+compound---------> More Complex Compound" because a single replacement reaction is in the similar form when given the reactants.

 

As a general rule, a synthesis reaction would have ONE product, as opposed to a single replacement which would have generally two products (depending on the replacement thats occurring)

[RyanJ]

As a general rule, a synthesis reaction would have ONE product, as opposed to a single replacement which would have generally two products (depending on the replacement thats occurring)

 

This is true for most simple reactions but for more complex ones this is rarely the case even though the rule still holds for general reference its worth pointing out its not 100% reliable.

[encipher]

Yes, which is why i said it is a general rule. Judging from the questions 'Asian' is asking, it would seem he is still taking very basic chemistry, and I doubt anything more complex than that would appear since it was that way when i took basic chemistry for the first time. And he also stated synthesis/combination... which to my knowledge in general chemistry refers to a chemical bonding between two reactants to form a single product.

[RyanJ]

Yes, which is why i said it is a general rule. Judging from the questions 'Asian' is asking, it would seem he is still taking very basic chemistry, and I doubt anything more complex than that would appear since it was that way when i took basic chemistry for the first time.

 

Yea, I wasn't disagreeing with you, just thought that point should be clarified in case something a little more complex crops up

 

@Asian: Normally you can tell the reaction type without much thought, just look at a few examples and you should see a pattern - this pattern is pretty easy to spot when you know what your looking for

[Asian]

so if it is soluable, then the product will be a solid??? and if it is a precipitate it would be a solid? im not really sure what a precipitate is. One more thing we mainly use the solubility rules to write the net ionic equation?

 

-thanks for the help guys

[encipher]

I just explained the use of solubility to determine net ioinic equations in my previous reply to your question. A precipitate is a solid that you cannot dissolve in a solvent. Meaning, something that is not a precipitate, will remain aqueous in a solution.

[Asian]

I think I understand this now. If something is soluable based on the "rules" than we know that it dissolves in water and it must be aqueous. And whichever compound is not soluable than it is in the solid state (which is the precipitate). And usually in a double replacement reaction the general formula for the states is (aq)+(aq)------>(aq)+(s) where the solid is the preciptate.

 

Also is this a correct statement? something that is a preciptate is insoluable.

 

But my last question is how you came up with the net ionic equation on the previous post, im just a little confused on that....

 

 

- Thanks so much i understand this so much better

[RyanJ]

I think I understand this now. If something is soluable based on the "rules" than we know that it dissolves in water and it must be aqueous. And whichever compound is not soluable than it is in the solid state (which is the precipitate). And usually in a double replacement reaction the general formula for the states is (aq)+(aq)------>(aq)+(s) where the solid is the preciptate.

 

Some double displacement reactions do produce a solid precipitate but not all so this may work in general but its not always true.

 

But my last question is how you came up with the net ionic equation on the previous post, im just a little confused on that....

 

Net ionic equations are normally pretty simple. We know that an ionic solid dissolves in water because of the polarisation in the sigma bond caused by bonding hydrogen too a strongly electronegative element like Oxygen, Nitrogen, etc. This means that water molecules have an area of slight negative charge over the oxygen atom and a slight positive one over the hydrogens. Add an ionic solid too water and the water molecules will surround each individual ion, pulling it from its lattice by breaking the electrostatic attractions between the ions in the solid and replacing them with hydrogen bonds with the water. When we do an ionic equation we only deal with the ions we are interested in, any other ions in solution are simple ignored.

 

For example, if we were doing the reaction of hydrochloric acid [ce]HCl_{(aq)}[/ce] with magnesium oxide [ce]MgOH_{(aq)}[/ce] we would do this:

 

[ce]2Cl^{-} + Mg^{+} -> MgCl2[/ce]

 

As you can see we ignore the hydroxyl ion and the oxonium ion. Those are ions that are involved but we generally ignore them as the main produce of the reaction is shown.

[woelen]

Ryan, I have to disagree with your last example of reaction with magnesium hydroxide and hydrochloric acid. It is the hydroxide and oxonium ions which do react, the Mg(2+) and Cl(-) ions are just spectator ions.

 

OH(-) + H3O(+) --> 2H2O (+ a lot of heat)

 

When you add hydrochloric acid to solid Mg(OH)2, then you obtain Mg(2+) and Cl(-) in solution and some water is formed. On evaporation you obtain MgCl2.

[RyanJ]

Ryan, I have to disagree with your last example of reaction with magnesium hydroxide and hydrochloric acid. It is the hydroxide and oxonium ions which do react, the Mg(2+) and Cl(-) ions are just spectator ions.

 

OH(-) + H3O(+) --> 2H2O (+ a lot of heat)

 

When you add hydrochloric acid to solid Mg(OH)2, then you obtain Mg(2+) and Cl(-) in solution and some water is formed. On evaporation you obtain MgCl2.

 

Point taken Woelen, we were just told to only show for the ions for the expected product and ignore the oxonium ions and hydroxyl ions. I've no idea why because I agree with you (and would write the full equation anyway) but that's just what we were told

 

Maybe because in reactions like this the (acid + base) water is always formed so it can generally be ignored.

[woelen]

Maybe because in reactions like this the (acid + base) water is always formed so it can generally be ignored.

It can only be ignored in aqueous solution. In non-aqueous solution the water, formed in such reactions can be quite disturbing.

As an example, I tried to make sodium nitronate from nitromethane and sodium hydroxide in methanol solvent. This did not work, because the nitronate ion is destroyed quickly by the water, formed in this reaction. Only the use of sodium methoxide as base does the job without problem.

[RyanJ]

It can only be ignored in aqueous solution. In non-aqueous solution the water, formed in such reactions can be quite disturbing.

As an example, I tried to make sodium nitronate from nitromethane and sodium hydroxide in methanol solvent. This did not work, because the nitronate ion is destroyed quickly by the water, formed in this reaction. Only the use of sodium methoxide as base does the job without problem.

 

Your point is well taken Woelen, thanks for pointing it out! Seeing the level of the question I'd guess it was in aqueous solution thankfully

[jdurg]

When I was instructed on writing net ionic equations and learning solubility rules, I was ALWAYS told to write out each and every compound as an aqueous compound. Do this on both sides of the equation. Then, when you have all your ions written out, use your solubility rules to pair up insoluble ions. As an example, let's say you are mixing Barium Nitrate and Sodium Sulfate. I would write everything out like the following:

 

2Na+ + SO4(-2) + Ba+2 + 2NO3(-) => 2Na+ + SO4(-2) + Ba+2 + 2NO3(-)

 

Now based upon solubility rules we know that sodium salts and nitrate salts are soluble. So we can remove them from the equation:

 

SO4(-2) + Ba+2 => SO4(-2) + Ba+2

 

Barium sulfate is one of those compounds that we know is not soluble, so as a result we can rewrite our above equation as follows:

 

Ba+2 + SO4(-2) => BaSO4(s)

 

There. You know have a net ionic equation. This is a fairly simple method to learn about writing net ionic equations as well as learning which compounds are soluble and which are not.