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












4












$begingroup$


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




Calcium fluoride, $ce{CaF2}$, has a molar solubility of $pu{2.1e−4 mol L−1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_{mathrm{a}}$ of $ce{HF}$ is 3.17.




The relevant reactions are:



$$ce{CaF2(s) <=> Ca^2+(aq) + 2 F-(aq)}$$ and
$$ce{HF(aq) <=> H+(aq) + F-(aq)}$$



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 $ce{H+}$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):



$$
begin{array}{|c|c|c|}
hline
&[ce{Ca^2+}] & [ce{F-}] \
hline
I & pu{2.1e−4} & pu{4.2e−4} \
hline
C & +x & +2x \
hline
E & pu{2.1e−4}+x & pu{4.2e−4}+2x \
hline
end{array}
$$



and



$$
begin{array}{|c|c|c|}
hline
&[ce{HF}] & [ce{H+}] & [ce{F-}] \
hline
I & 0 & text{N/A} & pu{4.2e−4} \
hline
C & +x &text{N/A} & -x \
hline
E & +x & 10^{-3.00} & pu{4.2e−4} - x\
hline
end{array}
$$



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











$endgroup$












  • $begingroup$
    Your "initial state" seems to be pH 7 so you do know the [H+] at the start.
    $endgroup$
    – Tan Yong Boon
    7 mins ago










  • $begingroup$
    Yes, but the pH is adjusted externally, so we can't take the change in [H+] to deduce how much [HF] and [F-] are changing. For that reason, I left that box as N/A. I could have left the entire column out instead.
    $endgroup$
    – Karsten Theis
    42 secs ago
















4












$begingroup$


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




Calcium fluoride, $ce{CaF2}$, has a molar solubility of $pu{2.1e−4 mol L−1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_{mathrm{a}}$ of $ce{HF}$ is 3.17.




The relevant reactions are:



$$ce{CaF2(s) <=> Ca^2+(aq) + 2 F-(aq)}$$ and
$$ce{HF(aq) <=> H+(aq) + F-(aq)}$$



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 $ce{H+}$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):



$$
begin{array}{|c|c|c|}
hline
&[ce{Ca^2+}] & [ce{F-}] \
hline
I & pu{2.1e−4} & pu{4.2e−4} \
hline
C & +x & +2x \
hline
E & pu{2.1e−4}+x & pu{4.2e−4}+2x \
hline
end{array}
$$



and



$$
begin{array}{|c|c|c|}
hline
&[ce{HF}] & [ce{H+}] & [ce{F-}] \
hline
I & 0 & text{N/A} & pu{4.2e−4} \
hline
C & +x &text{N/A} & -x \
hline
E & +x & 10^{-3.00} & pu{4.2e−4} - x\
hline
end{array}
$$



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











$endgroup$












  • $begingroup$
    Your "initial state" seems to be pH 7 so you do know the [H+] at the start.
    $endgroup$
    – Tan Yong Boon
    7 mins ago










  • $begingroup$
    Yes, but the pH is adjusted externally, so we can't take the change in [H+] to deduce how much [HF] and [F-] are changing. For that reason, I left that box as N/A. I could have left the entire column out instead.
    $endgroup$
    – Karsten Theis
    42 secs ago














4












4








4





$begingroup$


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




Calcium fluoride, $ce{CaF2}$, has a molar solubility of $pu{2.1e−4 mol L−1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_{mathrm{a}}$ of $ce{HF}$ is 3.17.




The relevant reactions are:



$$ce{CaF2(s) <=> Ca^2+(aq) + 2 F-(aq)}$$ and
$$ce{HF(aq) <=> H+(aq) + F-(aq)}$$



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 $ce{H+}$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):



$$
begin{array}{|c|c|c|}
hline
&[ce{Ca^2+}] & [ce{F-}] \
hline
I & pu{2.1e−4} & pu{4.2e−4} \
hline
C & +x & +2x \
hline
E & pu{2.1e−4}+x & pu{4.2e−4}+2x \
hline
end{array}
$$



and



$$
begin{array}{|c|c|c|}
hline
&[ce{HF}] & [ce{H+}] & [ce{F-}] \
hline
I & 0 & text{N/A} & pu{4.2e−4} \
hline
C & +x &text{N/A} & -x \
hline
E & +x & 10^{-3.00} & pu{4.2e−4} - x\
hline
end{array}
$$



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, $ce{CaF2}$, has a molar solubility of $pu{2.1e−4 mol L−1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_{mathrm{a}}$ of $ce{HF}$ is 3.17.




The relevant reactions are:



$$ce{CaF2(s) <=> Ca^2+(aq) + 2 F-(aq)}$$ and
$$ce{HF(aq) <=> H+(aq) + F-(aq)}$$



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 $ce{H+}$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):



$$
begin{array}{|c|c|c|}
hline
&[ce{Ca^2+}] & [ce{F-}] \
hline
I & pu{2.1e−4} & pu{4.2e−4} \
hline
C & +x & +2x \
hline
E & pu{2.1e−4}+x & pu{4.2e−4}+2x \
hline
end{array}
$$



and



$$
begin{array}{|c|c|c|}
hline
&[ce{HF}] & [ce{H+}] & [ce{F-}] \
hline
I & 0 & text{N/A} & pu{4.2e−4} \
hline
C & +x &text{N/A} & -x \
hline
E & +x & 10^{-3.00} & pu{4.2e−4} - x\
hline
end{array}
$$



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 3 hours ago









Mathew Mahindaratne

3,870318




3,870318










asked 6 hours ago









Karsten TheisKarsten Theis

2,705434




2,705434












  • $begingroup$
    Your "initial state" seems to be pH 7 so you do know the [H+] at the start.
    $endgroup$
    – Tan Yong Boon
    7 mins ago










  • $begingroup$
    Yes, but the pH is adjusted externally, so we can't take the change in [H+] to deduce how much [HF] and [F-] are changing. For that reason, I left that box as N/A. I could have left the entire column out instead.
    $endgroup$
    – Karsten Theis
    42 secs ago


















  • $begingroup$
    Your "initial state" seems to be pH 7 so you do know the [H+] at the start.
    $endgroup$
    – Tan Yong Boon
    7 mins ago










  • $begingroup$
    Yes, but the pH is adjusted externally, so we can't take the change in [H+] to deduce how much [HF] and [F-] are changing. For that reason, I left that box as N/A. I could have left the entire column out instead.
    $endgroup$
    – Karsten Theis
    42 secs ago
















$begingroup$
Your "initial state" seems to be pH 7 so you do know the [H+] at the start.
$endgroup$
– Tan Yong Boon
7 mins ago




$begingroup$
Your "initial state" seems to be pH 7 so you do know the [H+] at the start.
$endgroup$
– Tan Yong Boon
7 mins ago












$begingroup$
Yes, but the pH is adjusted externally, so we can't take the change in [H+] to deduce how much [HF] and [F-] are changing. For that reason, I left that box as N/A. I could have left the entire column out instead.
$endgroup$
– Karsten Theis
42 secs ago




$begingroup$
Yes, but the pH is adjusted externally, so we can't take the change in [H+] to deduce how much [HF] and [F-] are changing. For that reason, I left that box as N/A. I could have left the entire column out instead.
$endgroup$
– Karsten Theis
42 secs ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

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:



$$
begin{array}{|c|c|c|c|c|}
hline
&[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
hline
I & pu{2.1e−4} & pu{4.2e−4} & text{N/A} & 0 \
hline
C & +x & +2x-y & text{N/A} & +y \
hline
E & pu{2.1e−4}+x & pu{4.2e−4}+2x-y & 10^{-3.00} & +y \
hline
end{array}
$$



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 $y$):



$$
begin{array}{|c|c|c|c|c|}
hline
&[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
hline
I & 0 & 0 & text{N/A} & 0 \
hline
C & +p & +2p-q & text{N/A} & +q \
hline
E & +p & +2p-q & 10^{-3.00} & +q \
hline
end{array}
$$






share|improve this answer











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    1 Answer
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    1 Answer
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    active

    oldest

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    active

    oldest

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    5












    $begingroup$

    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:



    $$
    begin{array}{|c|c|c|c|c|}
    hline
    &[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
    hline
    I & pu{2.1e−4} & pu{4.2e−4} & text{N/A} & 0 \
    hline
    C & +x & +2x-y & text{N/A} & +y \
    hline
    E & pu{2.1e−4}+x & pu{4.2e−4}+2x-y & 10^{-3.00} & +y \
    hline
    end{array}
    $$



    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 $y$):



    $$
    begin{array}{|c|c|c|c|c|}
    hline
    &[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
    hline
    I & 0 & 0 & text{N/A} & 0 \
    hline
    C & +p & +2p-q & text{N/A} & +q \
    hline
    E & +p & +2p-q & 10^{-3.00} & +q \
    hline
    end{array}
    $$






    share|improve this answer











    $endgroup$


















      5












      $begingroup$

      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:



      $$
      begin{array}{|c|c|c|c|c|}
      hline
      &[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
      hline
      I & pu{2.1e−4} & pu{4.2e−4} & text{N/A} & 0 \
      hline
      C & +x & +2x-y & text{N/A} & +y \
      hline
      E & pu{2.1e−4}+x & pu{4.2e−4}+2x-y & 10^{-3.00} & +y \
      hline
      end{array}
      $$



      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 $y$):



      $$
      begin{array}{|c|c|c|c|c|}
      hline
      &[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
      hline
      I & 0 & 0 & text{N/A} & 0 \
      hline
      C & +p & +2p-q & text{N/A} & +q \
      hline
      E & +p & +2p-q & 10^{-3.00} & +q \
      hline
      end{array}
      $$






      share|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        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:



        $$
        begin{array}{|c|c|c|c|c|}
        hline
        &[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
        hline
        I & pu{2.1e−4} & pu{4.2e−4} & text{N/A} & 0 \
        hline
        C & +x & +2x-y & text{N/A} & +y \
        hline
        E & pu{2.1e−4}+x & pu{4.2e−4}+2x-y & 10^{-3.00} & +y \
        hline
        end{array}
        $$



        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 $y$):



        $$
        begin{array}{|c|c|c|c|c|}
        hline
        &[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
        hline
        I & 0 & 0 & text{N/A} & 0 \
        hline
        C & +p & +2p-q & text{N/A} & +q \
        hline
        E & +p & +2p-q & 10^{-3.00} & +q \
        hline
        end{array}
        $$






        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:



        $$
        begin{array}{|c|c|c|c|c|}
        hline
        &[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
        hline
        I & pu{2.1e−4} & pu{4.2e−4} & text{N/A} & 0 \
        hline
        C & +x & +2x-y & text{N/A} & +y \
        hline
        E & pu{2.1e−4}+x & pu{4.2e−4}+2x-y & 10^{-3.00} & +y \
        hline
        end{array}
        $$



        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 $y$):



        $$
        begin{array}{|c|c|c|c|c|}
        hline
        &[ce{Ca^2+}] & [ce{F-}] & [ce{H+}]&[ce{HF}] \
        hline
        I & 0 & 0 & text{N/A} & 0 \
        hline
        C & +p & +2p-q & text{N/A} & +q \
        hline
        E & +p & +2p-q & 10^{-3.00} & +q \
        hline
        end{array}
        $$







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 3 hours ago









        Mathew Mahindaratne

        3,870318




        3,870318










        answered 6 hours ago









        Karsten TheisKarsten Theis

        2,705434




        2,705434






























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