I need to write a net ionic equation...help?

Question

I need to do this using the Anion from Sodium Hydroxide (notsure what part of it is the anion, nor the charge) which is: NaOH-

And the Cation from Ammonium Chloride NH4ClO2 (not sure which part is the cation (positive)

Thanks~

And if you could name the final product (like in words) that would be AWESOME

Answer 1

Sodium hydroxide  ( NaOH )  ----- >  in solution  ------ >  Na +  ( Sodium Cation )  and  OH -  ( Hydroxide Anion )    

Ammonium Chloride  ( NH4Cl )  ----- >  in solution  ------ >  NH4 +  ( Ammonium Cation )  and  Cl -  ( Chloride Anion )    

[Na +]  +  [OH -]  +  [NH4 +]  +  [Cl -]  ------ >  NH3  +  H2O  +  [NA +]  +  [Cl - ]        

[Na +] ,  [Cl -]  remain unchanged after reaction   ( and can be ignored on both side of net equation )

So,  Net  Ionic Equation will be . . . .

[OH -]  +  [NH4 +]  ------ >  NH3  +  H2O    

Final products are { Ammonia ( NH3 Gas) } and Water  ( H2O )    

Please  also  bring  my  following  answer  to  the  question  before   your  question  ( Na2CO3 Titration with HCl )  [[[ Please try to find that question and then put my answer ( under your name )  -- Thanks ]]]

Dear Sarah E ,

Sorry for my previous answer . That was a " Fake " Answer .

I gave that answer to test whether people will believe ( since it also contains some correct facts ) .

The following answer will be a real and correct one.

The 1/2 V Equivalent Point 1 ( at 9.90 / 2 => 4.95 mL ) is the point where half of Carbonate ion was consumed and equal amount of Bicarbonate ion was generated. ====> [ CO3 ion ] = [ HCO3 ion ]

At this 1/2 V Equivalent Point 1 ; pH of solution will be equal to pKa2 of Carbonic acid.

The 1/2 V Equivalent Point 2 ( at 19.40 - 9.90 / 2 => 4.75 + 9.90 => 14.65 mL ) is the point where half of Bicarbonate ion was consumed and equal amount of Carbonic acid was generated. ====> [ HCO3 ion ] = [ H2CO3 ]

At this 1/2 V Equivalent Point 2 ; pH of solution will be equal to pKa1 of Carbonic acid.

Answer 2

Well, if u are trying to do an NET  ionic equation that just means the balanced equation in which there are no spectator ions. What that mean's is that since NaOH is a strong base, it automatically break up into Na+ and OH- ions. So since the Na+ ions are not going to do anything, they're spectators and are not put into the NET ionic equation.

So, now to the equation. Since your mixing the two, and one happens to be a strong base, this is a single replacement. so the OH- is going to switch places with the ClO2- ion. And that should look like this:

OH- + NH4ClO2 => NH4OH- + ClO2-

The reason why you dont break up the NH4ClO2 is because it isnt strong acid, so you cant break it up. It'll make more sense when you get into equilibrium.