Ionic equations

Can someone help me on how to write equations? For example: Sodium carbonate solution with nitric acid and copper oxide with HCL. I know the balanced equations but where do I go next? How do I know which ions to keep in and which are spectators?

Scroll to see replies

Reply 1

MathsNerd1

Original post by Year11guy

Moved to Chemistry section :smile:

Reply 2

Decerto

Original post by Year11guy

After having the balanced equation, split the ionic compounds into their ions. Then you removed the spectator ions and rewrite the ionic equation.

Reply 3

Year11guy

ok so:
CuO+2HCl--->CuCl2+H2O
Now split the ionic compounds into ions:
Cu+2 + O2- + 2HCl---> Cu2+ + 2Cl- + H2O
Is this right? How do I tell which the spectators are?

Reply 4

QuantumSuicide

Original post by Year11guy

ok so:
CuO+2HCl--->CuCl2+H2O
Now split the ionic compounds into ions:
Cu+2 + O2- + 2HCl---> Cu2+ + 2Cl- + H2O
Is this right? How do I tell which the spectators are?

Looks good except for HCl. You have to split HCl into 2H+ and 2Cl-.
Spectator ions appear on both sides of the equation. So here you can see that 2Cl- is a spectator ion because it appears on both sides of the equation

Posted from TSR Mobile

Reply 5

Year11guy

Original post by QuantumSuicide

Looks good except for HCl. You have split HCl into 2H+ and 2Cl-.
Spectator ions appear on both sides of the equation. So here you can see that 2Cl- is a spectator ion because it appears on both sides of the equation

Posted from TSR Mobile

I thought HCl was covalent?

Reply 6

username1042883

Original post by Year11guy

ok so:
CuO+2HCl--->CuCl2+H2O
Now split the ionic compounds into ions:
Cu+2 + O2- + 2HCl---> Cu2+ + 2Cl- + H2O
Is this right? How do I tell which the spectators are?

They'll appear on both sides of the reaction, like Cu2+ does.
Shouldn't HCl be split up into H+ and Cl-?

Reply 7

Year11guy

Original post by charlesn202

They'll appear on both sides of the reaction, like Cu2+ does.
Shouldn't HCl be split up into H+ and Cl-?

I didn't know it was ionic, I thought it was covalent.

Reply 8

QuantumSuicide

Original post by Year11guy

I thought HCl was covalent?

Yep, HCl is covalent with a permanent dipole (electrons are attracted towards the Cl more strongly because Cl is more electronagtive). But i think in this case HCl is aqueous and aqueous species always separate into their ions.

This is what i've been taught. Maybe someone has a proper explanation.

Posted from TSR Mobile

Reply 9

username1042883

Original post by Year11guy

I didn't know it was ionic, I thought it was covalent.

HCl is covalent in gas form but when it's "dissolved" (I'm sure that's not the right word) in water the proton is released. That's one of the definitions of acidity anyways.

Reply 10

Year11guy

Original post by charlesn202

HCl is covalent in gas form but when it's "dissolved" (I'm sure that's not the right word) in water the proton is released. That's one of the definitions of acidity anyways.

Ah yes, how could I forget. It's the dissociation that makes acids acidic, am I right?

Reply 11

QuantumSuicide

Original post by Year11guy

Ah yes, how could I forget. It's the dissociation that makes acids acidic, am I right?

Acids are proton donors. Here the HCl dissociates into H+ and Cl-

H2O acts as the base here:
H2O + H+ ---->H3O+

Posted from TSR Mobile

Reply 12

Year11guy

Back to the question:

Cu+2 + O2- + 2H+ 2Cl- l---> Cu2+ + 2Cl- + H2O

The final equation after omitting Cu and Cl is:

O2- + 2H+ ------> H2O(l)

Reply 13

QuantumSuicide

Original post by Year11guy

Back to the question:

Cu+2 + O2- + 2H+ 2Cl- l---> Cu2+ + 2Cl- + H2O

The final equation after omitting Cu and Cl is:

O2- + 2H+ ------> H2O(l)

Yeh fam. That's kool

Posted from TSR Mobile

Reply 14

GDN

in my understanding only the dissolved ionic substances can be shown as separate ions - since copper oxide is insoluble in water it should not be shown as separate ions in the construction of your ionic equation so the e ionic equation is

CuO + 2H⁺ --> Cu²⁺ + H₂O

only the Cl^{^-} ions are spectator ions

Reply 15

Year11guy

Original post by QuantumSuicide

Yeh fam. That's kool

Posted from TSR Mobile

Ok, so if I try the next one by myself, the reaction of Sodium Carbonate with nitric acid.

Na+ + CO3(2-) + H+ + NO3- ------> Na+ + NO3- + H2O(l) + CO2.

Remove the spectator ions:

CO3(2-) + 2H+ ------> H2O(l) + CO2(g)

Is this correct, or are water and carbon dioxide ionic now too?

Reply 16

QuantumSuicide

Original post by GDN

Im no expert but i know this isnt right. Dont ask me why tho lol

Posted from TSR Mobile

Reply 17

Year11guy

Original post by QuantumSuicide

Im no expert but i know this isnt right. Dont ask me why tho lol

Posted from TSR Mobile

Can you check the other one I just posted

Original post by GDN

The copper oxide is still involved in the reaction so it still separates into ions which may be why you're mistaken. I'm just going by what QuantumSuicide said as they said you're incorrect