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Net ionic compounds

  1. Oct 1, 2005 #1
    1. Calcium + Water yields calcuim hydroxide + hydrogen
    Need balanced and net ionic equation, I got
    Ca + 3/2H20 =Ca(OH)2 +H
    I can't figure out how to balance it further and I don't know how to get the net equation cause calcuim hydroxide is soluble which means it wouldn't be in the net equation right?

    2. Silver(I)nitrate + potassium iodide yields silver(I)iodide + potassium nitrate
    I got
    AgNO2- + KI = AgI + KNO2-
    Ag+ + I- = AgI
    Potassium nitrate is soluble so it's not in the net equation but silver iodide is not soluble so it is. Is it balanced ok? Am I right? Heeelllpp!
  2. jcsd
  3. Oct 1, 2005 #2
    [tex] \text{Ca}}\left( {{\text{OH}}} \right)_2 [/tex] is soluble in water.
    Therefore, you will observe hydrolysis, as:
    [tex] {\text{H}}_{\text{2}} {\text{O}} \rightleftharpoons {\text{H}}^ + \left( {aq} \right) + {\text{OH}}^ - \left( {aq} \right) [/tex]

    *Or you balance it as:
    [tex] 2{\text{H}}_{\text{2}} {\text{O}} \rightleftharpoons {\text{H}}_{\text{3}} {\text{O}}^{\text{ + }} \left( {aq} \right) + {\text{OH}}^ - \left( {aq} \right) [/tex]

    *Actually, silver(I) nitrate is--> [tex] \text{AgNO} _ 3 [/tex] :wink:

    You are right, [tex] \text{KNO} _ 3 [/tex] and [tex] \text{AgNO} _ 3 [/tex] and [tex] {\text{KI}} [/tex] are all obviously soluble. Looking at your chemicals, you will observe [tex] {\text{AgI}} [/tex] precipitate. Therefore, the net ionic reaction equation will omit the potassium cations and the nitrate anions:

    [tex] {\text{Ag}}^ + \left( {aq} \right) + {\text{I}}^ - \left( {aq} \right) \rightleftharpoons {\text{AgI}}\left( s \right) [/tex]

    Remember those common solubility rules. They can help greatly :smile:
    Last edited: Oct 1, 2005
  4. Mar 4, 2007 #3
    ok, I understand how to work out soluble and nonsoluble compounds but I do not know how to write Net Ionic Equations. So if I had:

    BaCl2(aq) + 2AgNO3(aq) -->Ba(NO3)2(aq)+2AgCl(s)

    Would the Net Ionic equation be:
    a) 2Ag+(aq) + 2Cl-(aq) -->2Ag+Cl-(s)

    b) Ag+(aq) + Cl-(aq) --> Ag+Cl-(s)

    c) 2Ag+(aq) + Cl2-(aq) --> 2Ag+Cl-(s)

    My current guess would be (c)
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