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Acid/base balancing question (given products)

  1. Oct 10, 2006 #1
    I have a question and I dont even know where to begin. Perhaps someone can guide me.

    For these products:
    RbBr(aq)
    NaCH3CO2(aq)
    Fe(ClO4)3 (aq)

    I need to write the balanced equation and the net ionic euqation for each of them. nd indicate their states (aq, s, l)... (s) is the precipitation one.

    Please?!!!
     
  2. jcsd
  3. Oct 10, 2006 #2
    well, you are given products sooo...

    you know a reaction consists of reactants and products right?

    reactants ----> products

    for example:
    2Rb(s) + Br(2) (aq) ---- 2RbBr (aq)

    (i hope you know what a diatomic molecule is? H, O, F, Br, I, Cl ... these elements cannot be free as a single atom)

    but from that written equation we can breakdown the ionic equation...

    2Rb+(s) + 2Br-(aq) ------ 2Rb+(aq) + 2Br-(aq) (RbBr breaks down because it is soluble...search
    Solubility on wikipedia.org if you're not familiar with the rules)

    Thus the net ionic equation (where spectator ions cancel out):

    2Rb+(s) ----- 2Rb+(aq) (2Br cancel because they're spectator ions) (Rb dont cancel b/c they're in different states)

    ^^^Here's your answer for part 1...You only need to see it once for these kinds of problems, but if you need more help PM or post here
     
    Last edited: Oct 10, 2006
  4. Oct 11, 2006 #3
    Example: How can Lao(SOo), be produced? Ans. From the reaction of La(OH), (base) with HrSOo
    (acid)
    Molecularz 2La(OH), (s) + 3 HrSO4 (aq) -+ Lq(SOa), (aq) + 6 HrO (l)
    Net ionic: La(OH), (s) + 3 H. (aq) -+ Lf* (aq) + 3 H2O (l) (Note lowest whole #'s in equation)

    I was given that example... So I actually have to use the acid and base in the reaction. That is why I am confused.
     
  5. Oct 11, 2006 #4
    ^^^hey by the way, i just remembered an arguement a few days ago about somethin on the question i helpled u with. I'm not sure the 2Br cancels out...you might wanna check for that!
     
  6. Oct 11, 2006 #5
    hmmmm that's weird...it might have been a mistake by the teacher/prof. I can see that H2SO4 breaks down (because it is a strong acid) and the LaOH is insoluble because of the solubility rule. Other than that i cannot understand the balancing part of this question--for some reason! i hope someone comes in here to save you
     
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