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Adding Net Ionic Equations

  1. Jul 19, 2006 #1
    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.

  2. jcsd
  3. Jul 19, 2006 #2
    1. Add #1 to #2.

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

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

    [tex]NaOH \text{(s)} + H^+ \text{(aq)} \longrightarrow Na^+ \text{(aq)} + H_2O \text{(l)}[/tex]
    Last edited by a moderator: Jul 19, 2006
  4. Jul 19, 2006 #3
    Is it possible that I have equation #3 wrong.
    Should the net ionic equation look like your final answer above?
  5. Jul 19, 2006 #4
    You have #3 correct, I think.

    :approve: :wink:
  6. Jul 20, 2006 #5
    Solid NaOH in HCl-solution. The HCl-solution contains [tex]H^{+}[/tex] and [tex]Cl^{-}[/tex] but the solid is just NAOH. So the reaction at 3 should be:

    [tex]NaOH(s) + H^{+} \rightarrow Na^{+} + H_{2}O[/tex]
  7. Aug 6, 2006 #6
    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)?
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