1. Jul 19, 2006

### 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 seperate 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)

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

thanks

2. Jul 19, 2006

### PPonte

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

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

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

Last edited by a moderator: Jul 19, 2006
3. Jul 19, 2006

### yellowduck

Is it possible that I have equation #3 wrong.

4. Jul 19, 2006

### PPonte

You have #3 correct, I think.

5. Jul 20, 2006

### 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:

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

6. Aug 6, 2006

### 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)?