Ionic Equations General Help!

[Aquiche]

I am a first year AS chem student and missed the lesson on how to write ionic equations on a general basis...you know...the questions where they ask you to write the full equation and thenthe ionic? Well, i can not do it for the life in me!

 

I was wondering if there is a technique you can apply to all equations to reach the ionic equation or something along those terms.

 

PLEASE HELP....ANYBODY!!

[YT2095]

what part exactly are you having problems with?

 

is it the Electron count and charge? identifying the diff between Ionic and another type of bond?

[Aquiche]

If i try and use an example maybe it will become clearer.

 

Okay, so...say in an exam i was given the full equation:

 

2Na + 2HCl --> 2NaCl + H2

 

and had to convert it into a balanced ionic equation...where would i start? My friend once mentioned that if you eleminate the elements that are the same on both sides the equation that is left will be the ionic equation...is she right?

 

thank you for replying.

[Cap'n Refsmmat]

You need to split up each compound into its ions (assuming it's aqueous).

 

You'd split this:

 

[ce]2Na + 2HCl -> 2NaCl + H2[/ce]

 

into this:

 

[ce]2Na + 2H+ + 2Cl- -> 2Na+ + 2Cl- + H2[/ce]

 

and then eliminate "spectator ions" -- those that stay the same on both sides.

[Aquiche]

Oh wow, that has made it a lot easier...so...simply split the equation into it's seperate ions and eliminate. Can this be applied to the majority of equations?

[thedarkshade]

Your first need to know what is the oxidation number of the elements. For example, why do we write [ce]H2SO4[/ce] and not [ce]HSO4[/ce]. Well it turns out that there is a reason. You know that the sum of oxidation numbers in a compound must be zero. H is 1+, S is +6 (though it can change), and O is -2. So in [ce]H2SO4[/ce] we have 2*(+1) + (+6) + 4*(-2) and all that equals 0. In [ce]HSO4[/ce] the sum would give us -1, that's no good, that only happens when it forms salts.

 

Let's take a reaction, a neutralization reaction

 

[ce]2NaOH + H2SO4 -> Na2SO4 + 2H2O[/ce]

 

you know that [ce]NaOH[/ce] decomposes in [math]Na^+[/math] and [math]OH^{-2}[/math] and [ce]H2SO4[/ce] in [math]Na^+[/math] and [math]SO_4^{-2}[/math]. Now the [math]OH^{-2}[/math] combines with [math]H^+[/math] forming water and [math]Na^+[/math] with [math]SO_4^{-2}[/math] forming the salt, sodium sulphate.

[Aquiche]

Cool, i must say that i am really glad i joined this site...very good help form what appear to be very intellectual people. My mock exams are this week in Foundation Module 1, Chains and Rings and how far how fast...I did not realise the thing about the ON's. Thanks for the advice. Much appreciated.

[Fuzzwood]

Keep in mind however that not every salt dissociates. CaCO₃ doesnt dissolve in water and therefore should be written as CaCO₃ (s).

[thedarkshade]

Keep in mind however that not every salt dissociates. CaCO₃ doesnt dissolve in water and therefore should be written as CaCO₃ (s).

It might not dissociate in a particular solvent, but that doesn't mean it's dissociative.