Chemistry-Repost

posted by anonymous

Write formula equations and net ionic equations for the hydrolysis of sodium carbonate in water.

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Chemistry - DrBob222, Friday, April 24, 2009 at 10:50pm
The hydrolysis of Na2CO3 ends us as the hydrolysis of the carbonate ion.
The molecular formula is
Na2CO3 + HOH==NaHCO3 + NaOH
NaHCO3 + HOH ==> H2CO3 + NaOH
The net ionic equations are
CO3^= + HOH ==> HCO3^- + OH^-
HCO3^- + HOH ==> H2CO3 + OH^-
but your teacher PROBABLY want you to show how you get from steps 1 and 2 to steps 3 and 4.

But I am still confused with this. Can you explain it further to me please? Thanks in advance.

  1. DrBob222

    Here is the first equation.

    The molecular formula is
    Na2CO3 + HOH==>NaHCO3 + NaOH

    Change to the ionic equation.
    1. Strong electrolytes are written as the ions.
    2. Slightly ionized materials are written as the molecule.

    If you don't know weak electrolytes, you can look in a set of Ka or Kb tables. If it is listed it is a weak electrolyte. Of course, you know H2O is a weak electrolyte.
    3. Solids (insoluble materials/precipitates) are written as the molecule but there are no ppts in this problem.
    4. Gases are written as the molecule but there are no gases in this equation.

    You may find it easier to remember just #1; i.e., write as ions those materials that are strong electrolytes. Everything else is written as the molecule.
    2Na^+ + CO3^= + HOH ==>Na^+ + HCO3^- + Na^+ + OH^-
    The last one is the complete ionized equation BEFORE you cancel the common ions.
    Now cancel the ions common to both sides. Cancel 2Na^+ on the left with Na^+ and Na^+ on the right. YOu are left with
    CO3^= + HOH ==> HCO3^- + OH^-
    The points you need to remember are these.
    1. Hydrolysis means react with water.
    2. Write the hydrolysis as a double decomposition (double replacement).
    3. Change the double displacement molecular equation into an ionic equation using the four statements above as a guide to know which to write as ions and which to write as the molecule.
    4. Cancel ions common to both sides and you end up with the net ionic equation.
    You try it on the second hydrolysis part. [Note: Your prof may prefer you to write these TWO equations as one. If so just use make all of it in one reactions as in
    Na2CO3 + 2HOH ==> H2CO3 + 2NaOH]
    Check my work.

  2. anonymous

    From your previous post, did you mean the second hydrolysis part is NaHCO3 + HOH ==> H2CO3 + NaOH?
    But I don't understand where does NaHCO3 come from....

    NaHCO3 + HOH ==> H2CO3 + NaOH
    If this is the formula equation for the second hydrolysis part, would it be...
    Na^+ +HCO3^- +HOH ==>H2CO3 +Na^+ +OH^-
    The Na^- cancels out, and it leaves..
    HCO3^- +HOH ==>H2CO3 +OH^-
    So this is the net ionic equations for the second hydrolysis part?

    Thanks.

  3. anonymous

    If we use this formula equation,
    Na2CO3 + 2HOH ==> H2CO3 + 2NaOH
    2Na^+ +CO3^2- +2HOH==> H2CO3 +2Na^+ +2OH^-
    CO3^2- +2HOH==> H2CO3 +2OH^-
    Is the net ionic equation right?

  4. DrBob222

    From your previous post, did you mean the second hydrolysis part is NaHCO3 + HOH ==> H2CO3 + NaOH?
    But I don't understand where does NaHCO3 come from....
    Yes, that's what I meant. I hope that's what I wrote originally. Where did the NaHCO3 come from? From the first hydrolysis of the original Na2CO3.
    Na2CO3 + HOH ==> NaHCO3 + NaOH


    NaHCO3 + HOH ==> H2CO3 + NaOH
    If this is the formula equation for the second hydrolysis part, would it be...
    Na^+ +HCO3^- +HOH ==>H2CO3 +Na^+ +OH^-
    The Na^- cancels out, and it leaves..
    HCO3^- +HOH ==>H2CO3 +OH^-
    So this is the net ionic equations for the second hydrolysis part?
    Yes.

    From your next post,
    If we use this formula equation,
    Na2CO3 + 2HOH ==> H2CO3 + 2NaOH
    2Na^+ +CO3^2- +2HOH==> H2CO3 +2Na^+ +2OH^-
    CO3^2- +2HOH==> H2CO3 +2OH^-
    Is the net ionic equation right?
    Yes, that's right. MOST of the time we deal only with the first hydrolyis [CO3^= + HOH ==> HCO3^- + OH^-] because of the values. The constant for this one is approximately 1 x 10^-3 and for the second one it is 1 x 10^-7 so you see the first one occurs about 10,000 times more than the second. Therefore, most of the time we take the first one into account and ignore the second one.

  5. anonymous

    oh, I finally got it!
    Thanks alot! :)

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