How to Write Net Ionic Equations for Dissolution Reactions

[yellowduck]

This should be easy but I feel there is something I am mission

I need to write the net ionic equations and then add #1 & #2 and compare to #3.

Equation #1 - write net ionic equation for dissolution of solid NaOH in water
Equation: NaOH (s) + H2O --> Na+ (aq) + OH- (aq) + H2O (l)
Net: NaOH (s) --> Na+ (aq) + OH- (aq)
(it says in the book to separate strong bases such as NaOH but if I do all ions cancel out?)

Equation #2 - write net ionic equation for aqueous soloutions of NaOH & HCl
Equation: NaOH (aq) + HCl (aq) ---> NaCl (aq) + H2O (l)
Net: OH- (aq) + H+ (aq) --> H2O (l)

Equation 3: Solid NaOH and aqueous HCl
Equation: NaOH (s) + HCl (aq) ---> NaCl (aq) + H2O
OH- (s) + H+ (aq) --> H2O (l)

Add #1 & #2
This is where I get lost. I think it should be net equation #2. The difference is in the solid/aqueous state.

thanks

 

[PPonte]

1. Add #1 to #2.

NaOH \text{(s)} + OH^- \text{(aq)} + H^+ \text{(aq)} \longrightarrow Na^+ \text{(aq)} + OH^- \text{(aq)} + H_2O \text{(l)}

2. Delete the common compounds on both members of the equation.

NaOH \text{(s)} + H^+ \text{(aq)} \longrightarrow Na^+ \text{(aq)} + H_2O \text{(l)}

 

[yellowduck]

Is it possible that I have equation #3 wrong.
Should the net ionic equation look like your final answer above?

 

[PPonte]

Is it possible that I have equation #3 wrong.
Should the net ionic equation look like your final answer above?

You have #3 correct, I think.

 

[sdekivit]

Solid NaOH in HCl-solution. The HCl-solution contains H^{+} and Cl^{-} but the solid is just NAOH. So the reaction at 3 should be:

NaOH(s) + H^{+} \rightarrow Na^{+} + H_{2}O

 

[dissolver]

So if it states solid, then you must write out the entire formula even in the net ionic equation (because it is infered that the solid is not dissolved in the solution)?