by yellowduck
 P: 16 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) 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
 P: n/a 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. $$NaOH \text{(s)} + H^+ \text{(aq)} \longrightarrow Na^+ \text{(aq)} + H_2O \text{(l)}$$
 P: 16 Is it possible that I have equation #3 wrong. Should the net ionic equation look like your final answer above?
P: n/a

 P: 92 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: $$NaOH(s) + H^{+} \rightarrow Na^{+} + H_{2}O$$