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How can you use ICE tables to solve multiple coupled equilibria?


How to calculate the concentration of all relevant species in a buffer of a given pH?Determining equilibrium concentrations from initial conditions and equilibrium constantDetermining solubility of silver sulfate in its saturated solutionIs LiOH a weaker base than NaOH?What's the Kp value?Equilibrium Bond Dissociationice tables equilibirium qConfusion about the ICE tableHow do I find the theoretical pH of a buffer solution after HCl and NaOH were added, separately?Q. 36 2018 molar solubility of CaF2 at pH3 given molar solubility at pH7 and pKa of HF













2












$begingroup$


If I have a problem involving multiple coupled equilibrium reactions, such as




Calcium fluoride, $ceCaF2$, has a molar solubility of $pu2.1e−4 mol L−1$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_mathrma$ of $ceHF$ is 3.17.




The relevant reactions are:



$$ceCaF2(s) <=> Ca^2+(aq) + 2 F-(aq)$$ and
$$ceHF(aq) <=> H+(aq) + F-(aq)$$



They are coupled because fluoride occurs in both of them.



Is there a way to use ICE tables to organize the information (stoichiometry, initial concentrations, mass balance) as a way to solve the problem?



For example, I could try to set up one ICE table for each reaction (the column for $ceH+$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):



$$
beginarray
hline
&[ceCa^2+] & [ceF-] \
hline
I & pu2.1e−4 & pu4.2e−4 \
hline
C & +x & +2x \
hline
E & pu2.1e−4+x & pu4.2e−4+2x \
hline
endarray
$$



and



$$
beginarray
hline
&[ceHF] & [ceH+] & [ceF-] \
hline
I & 0 & textN/A & pu4.2e−4 \
hline
C & +x &textN/A & -x \
hline
E & +x & 10^-3.00 & pu4.2e−4 - x\
hline
endarray
$$



However, how do the fluoride concentrations "talk to each other" in the two tables? Is the $x$ in one table the same as the $x$ in the other table?










share|improve this question











$endgroup$
















    2












    $begingroup$


    If I have a problem involving multiple coupled equilibrium reactions, such as




    Calcium fluoride, $ceCaF2$, has a molar solubility of $pu2.1e−4 mol L−1$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_mathrma$ of $ceHF$ is 3.17.




    The relevant reactions are:



    $$ceCaF2(s) <=> Ca^2+(aq) + 2 F-(aq)$$ and
    $$ceHF(aq) <=> H+(aq) + F-(aq)$$



    They are coupled because fluoride occurs in both of them.



    Is there a way to use ICE tables to organize the information (stoichiometry, initial concentrations, mass balance) as a way to solve the problem?



    For example, I could try to set up one ICE table for each reaction (the column for $ceH+$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):



    $$
    beginarray
    hline
    &[ceCa^2+] & [ceF-] \
    hline
    I & pu2.1e−4 & pu4.2e−4 \
    hline
    C & +x & +2x \
    hline
    E & pu2.1e−4+x & pu4.2e−4+2x \
    hline
    endarray
    $$



    and



    $$
    beginarray
    hline
    &[ceHF] & [ceH+] & [ceF-] \
    hline
    I & 0 & textN/A & pu4.2e−4 \
    hline
    C & +x &textN/A & -x \
    hline
    E & +x & 10^-3.00 & pu4.2e−4 - x\
    hline
    endarray
    $$



    However, how do the fluoride concentrations "talk to each other" in the two tables? Is the $x$ in one table the same as the $x$ in the other table?










    share|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      If I have a problem involving multiple coupled equilibrium reactions, such as




      Calcium fluoride, $ceCaF2$, has a molar solubility of $pu2.1e−4 mol L−1$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_mathrma$ of $ceHF$ is 3.17.




      The relevant reactions are:



      $$ceCaF2(s) <=> Ca^2+(aq) + 2 F-(aq)$$ and
      $$ceHF(aq) <=> H+(aq) + F-(aq)$$



      They are coupled because fluoride occurs in both of them.



      Is there a way to use ICE tables to organize the information (stoichiometry, initial concentrations, mass balance) as a way to solve the problem?



      For example, I could try to set up one ICE table for each reaction (the column for $ceH+$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):



      $$
      beginarray
      hline
      &[ceCa^2+] & [ceF-] \
      hline
      I & pu2.1e−4 & pu4.2e−4 \
      hline
      C & +x & +2x \
      hline
      E & pu2.1e−4+x & pu4.2e−4+2x \
      hline
      endarray
      $$



      and



      $$
      beginarray
      hline
      &[ceHF] & [ceH+] & [ceF-] \
      hline
      I & 0 & textN/A & pu4.2e−4 \
      hline
      C & +x &textN/A & -x \
      hline
      E & +x & 10^-3.00 & pu4.2e−4 - x\
      hline
      endarray
      $$



      However, how do the fluoride concentrations "talk to each other" in the two tables? Is the $x$ in one table the same as the $x$ in the other table?










      share|improve this question











      $endgroup$




      If I have a problem involving multiple coupled equilibrium reactions, such as




      Calcium fluoride, $ceCaF2$, has a molar solubility of $pu2.1e−4 mol L−1$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_mathrma$ of $ceHF$ is 3.17.




      The relevant reactions are:



      $$ceCaF2(s) <=> Ca^2+(aq) + 2 F-(aq)$$ and
      $$ceHF(aq) <=> H+(aq) + F-(aq)$$



      They are coupled because fluoride occurs in both of them.



      Is there a way to use ICE tables to organize the information (stoichiometry, initial concentrations, mass balance) as a way to solve the problem?



      For example, I could try to set up one ICE table for each reaction (the column for $ceH+$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):



      $$
      beginarray
      hline
      &[ceCa^2+] & [ceF-] \
      hline
      I & pu2.1e−4 & pu4.2e−4 \
      hline
      C & +x & +2x \
      hline
      E & pu2.1e−4+x & pu4.2e−4+2x \
      hline
      endarray
      $$



      and



      $$
      beginarray
      hline
      &[ceHF] & [ceH+] & [ceF-] \
      hline
      I & 0 & textN/A & pu4.2e−4 \
      hline
      C & +x &textN/A & -x \
      hline
      E & +x & 10^-3.00 & pu4.2e−4 - x\
      hline
      endarray
      $$



      However, how do the fluoride concentrations "talk to each other" in the two tables? Is the $x$ in one table the same as the $x$ in the other table?







      equilibrium






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago









      Mathew Mahindaratne

      3,870318




      3,870318










      asked 3 hours ago









      Karsten TheisKarsten Theis

      2,685434




      2,685434




















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.



          Here is the combined ICE table:



          $$
          beginarray
          hline
          &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
          hline
          I & pu2.1e−4 & pu4.2e−4 & textN/A & 0 \
          hline
          C & +x & +2x-y & textN/A & +y \
          hline
          E & pu2.1e−4+x & pu4.2e−4+2x-y & 10^-3.00 & +y \
          hline
          endarray
          $$



          Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.



          What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.



          Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):



          $$
          beginarray
          hline
          &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
          hline
          I & 0 & 0 & textN/A & 0 \
          hline
          C & +p & +2p-q & textN/A & +q \
          hline
          E & +p & +2p-q & 10^-3.00 & +q \
          hline
          endarray
          $$






          share|improve this answer











          $endgroup$












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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.



            Here is the combined ICE table:



            $$
            beginarray
            hline
            &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
            hline
            I & pu2.1e−4 & pu4.2e−4 & textN/A & 0 \
            hline
            C & +x & +2x-y & textN/A & +y \
            hline
            E & pu2.1e−4+x & pu4.2e−4+2x-y & 10^-3.00 & +y \
            hline
            endarray
            $$



            Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.



            What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.



            Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):



            $$
            beginarray
            hline
            &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
            hline
            I & 0 & 0 & textN/A & 0 \
            hline
            C & +p & +2p-q & textN/A & +q \
            hline
            E & +p & +2p-q & 10^-3.00 & +q \
            hline
            endarray
            $$






            share|improve this answer











            $endgroup$

















              4












              $begingroup$

              The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.



              Here is the combined ICE table:



              $$
              beginarray
              hline
              &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
              hline
              I & pu2.1e−4 & pu4.2e−4 & textN/A & 0 \
              hline
              C & +x & +2x-y & textN/A & +y \
              hline
              E & pu2.1e−4+x & pu4.2e−4+2x-y & 10^-3.00 & +y \
              hline
              endarray
              $$



              Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.



              What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.



              Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):



              $$
              beginarray
              hline
              &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
              hline
              I & 0 & 0 & textN/A & 0 \
              hline
              C & +p & +2p-q & textN/A & +q \
              hline
              E & +p & +2p-q & 10^-3.00 & +q \
              hline
              endarray
              $$






              share|improve this answer











              $endgroup$















                4












                4








                4





                $begingroup$

                The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.



                Here is the combined ICE table:



                $$
                beginarray
                hline
                &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
                hline
                I & pu2.1e−4 & pu4.2e−4 & textN/A & 0 \
                hline
                C & +x & +2x-y & textN/A & +y \
                hline
                E & pu2.1e−4+x & pu4.2e−4+2x-y & 10^-3.00 & +y \
                hline
                endarray
                $$



                Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.



                What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.



                Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):



                $$
                beginarray
                hline
                &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
                hline
                I & 0 & 0 & textN/A & 0 \
                hline
                C & +p & +2p-q & textN/A & +q \
                hline
                E & +p & +2p-q & 10^-3.00 & +q \
                hline
                endarray
                $$






                share|improve this answer











                $endgroup$



                The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.



                Here is the combined ICE table:



                $$
                beginarray
                hline
                &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
                hline
                I & pu2.1e−4 & pu4.2e−4 & textN/A & 0 \
                hline
                C & +x & +2x-y & textN/A & +y \
                hline
                E & pu2.1e−4+x & pu4.2e−4+2x-y & 10^-3.00 & +y \
                hline
                endarray
                $$



                Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.



                What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.



                Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):



                $$
                beginarray
                hline
                &[ceCa^2+] & [ceF-] & [ceH+]&[ceHF] \
                hline
                I & 0 & 0 & textN/A & 0 \
                hline
                C & +p & +2p-q & textN/A & +q \
                hline
                E & +p & +2p-q & 10^-3.00 & +q \
                hline
                endarray
                $$







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 1 hour ago









                Mathew Mahindaratne

                3,870318




                3,870318










                answered 3 hours ago









                Karsten TheisKarsten Theis

                2,685434




                2,685434



























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