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Overview: Significance of Chemical Equilibrium

  • Page ID
    301974
  • Suppose you are a chemist involved in developing a new product for a small manufacturing company.  Part of the process leads to the formation of the compound lead phosphate.  The lead phosphate will end up in the wastewater from the process.  Since you are a small facility, instead of having your own wastewater treatment plant, you will discharge the wastewater to the local municipal wastewater treatment plant.  The municipal wastewater treatment plant faces strict requirements on the amount of lead that is permitted in their end products.  A wastewater treatment plant ends up with "clean" water and a solid sludge.  Most lead ends up in the sludge, and the Environmental Protection Agency has set a limit on how much lead is permitted in the sludge.  Most municipalities will require you to enter into a pre-treatment agreement, under which you will need to remove the lead before discharging to the plant.  For example, the City of Lewiston, Maine will require you to discharge a material that contains no more than 0.50 mg of total lead per liter.

    Lead phosphate is a sparingly soluble material so most of it will actually be a solid in your waste, thereby allowing you to filter it out before discharge to the treatment plant.

    What is the concentration of total dissolved lead in the discharge from your facility?

    What we need to consider here is the reaction that describes the solubility of lead phosphate.  Lead phosphate has the formula Pb3(PO4)2, and the accepted practice for writing the solubility reaction of a sparingly soluble compound that will dissociate into a cation and anion is shown.  The solid is always shown on the left, or reactant, side.  The dissolved ions are always shown on the product side.

    \[\ce{Pb3(PO4)2(s)  \leftrightarrow   3Pb^2+(aq)  +  2PO_4^3- (aq)}\nonumber\]

    Next, we can write the equilibrium constant expression for this reaction, which is as follows:

    \[\ce{K_{sp}} = \ce{[Pb^2+]^3[PO4^3- ]^2}\nonumber\]

    This general equilibrium constant expression for a sparingly soluble, ionic compound is known as the solubility product, or Ksp.  Note that there is no term for the solid lead phosphate in the expression.  One way to view this is that a solid really cannot have variable concentrations (moles/liter) and is therefore not important to the expression. Ksp values have been measured for many substances and tables of these numbers are available.  The Ksp for lead phosphate is known and is 8.1×10-47.  What this means is that any solution that is in contact with solid lead phosphate will have a solubility product ([Pb2+]3[\(\ce{PO_4^3-}\)]2) that exactly equals its Ksp (8.1×10-47).

    There is a complication to this process though.  It turns out that the phosphate ion is a species that appears in the dissociation reactions for a substance known as phosphoric acid (H3PO4).  Acids and their corresponding conjugate bases are very important in chemistry and the properties of many acids and bases have been studied.  What can happen in this case is that the phosphate ion can undergo a set of stepwise protonations, as shown below.

    \[\begin{array}{c}
    \ce{&Pb3(PO4)2 &↔ &3Pb^2+ &+ &2PO_4^3- \\
    &&&&&⇕\\
    &&&&&HPO_4^2- \\
    &&&&&⇕\\
    &&&&&H_2PO_4^- \\
    &&&&&⇕\\
    &&&&&H_3PO_4}
    \end{array}\nonumber\]

    If we wanted to calculate the solubility of lead phosphate in water, we would need to consider the effect of protonation of the phosphate on the solubility.  Remember, the Ksp expression only includes terms for Pb2+ and \(\ce{PO_4^3-}\), and it is the product of these two that must always equal Ksp if some solid lead phosphate is in the mixture.  Protonation of the phosphate will reduce the concentration of \(\ce{PO_4^3-}\).  If the concentration of \(\ce{PO_4^3-}\) is reduced, more of the lead phosphate must dissolve to maintain Ksp.

    We can look up relevant equilibrium constants for the dissociation of phosphoric acid.  There is an accepted practice in chemistry for the way in which these reactions are written, and the series for phosphoric acid is shown below.  This describes the chemistry of an acid and the equilibrium constant expressions are known as Ka values, or acid dissociation constants.

    \[\begin{array}{c}
    \ce{&H3PO4  &+  &H2O  &↔  &H_2 PO_4^- &+  &H3O+} &\hspace{40px}\mathrm{K_{a1}}\\
    &\ce{H_2PO_4^- &+  &H2O  &↔  &HPO_4^2- &+  &H3O+}   &\hspace{40px}\mathrm{K_{a2}}\\
    &\ce{HPO_4^2- &+  &H2O  &↔  &PO_4^3- &+  &H3O+}     &\hspace{40px}\mathrm{K_{a3}}
    \end{array}\nonumber\]

     

    \[\mathrm{K_{a1}=\dfrac{[H_2PO_4^-][H_3O^+]}{[H_3PO_4]}\qquad K_{a2}=\dfrac{[HPO_4^{2-}][H_3O^+]}{[H_2PO_4^-]}}\nonumber\]

    \[\mathrm{K_{a3}=\dfrac{[PO_4^{3-}][H_3O^+]}{[HPO_4^{2-}]}}\nonumber\]

    But before we can proceed, there is still one other complication to this process.  It turns out that the lead cation has the possibility of forming complexes with other anions in solution.  One such anion that is always present in water is hydroxide (\(\ce{OH-}\)).  The hydroxide complex could be another insoluble one with lead.  More important, though, is whether lead can form water-soluble complexes with the hydroxide ion.  A species that complexes with a metal ion is known as a ligand.  It turns out that hydroxide can form water-soluble complexes with lead ions, and that there are three of them that form in a stepwise manner.  The equations to represent this are always written with the metal ion and ligand on the reactant side and the complex on the product side, as shown below.

    \[\begin{array}{lccccc}
    \ce{Pb^2+(aq)  &+  &OH- (aq)  &↔  &Pb(OH)+(aq)}         &\qquad \mathrm{K_{f1}}\\
    \ce{Pb(OH)+(aq)  &+  &OH- (aq)  &↔  &Pb(OH)2(aq)}       &\qquad \mathrm{K_{f2}}\\
    \ce{Pb(OH)2(aq)  &+  &OH- (aq)  &↔  &Pb(OH)_3^- (aq)}   &\qquad \mathrm{K_{f3}}
    \end{array}\nonumber\]

    The equilibrium constant expressions are shown below, and these are known as formation constants (Kf).

    \[\mathrm{K_{f1}=\dfrac{[Pb(OH)^+]}{[Pb^{2+}][OH^-]} \qquad K_{f2}=\dfrac{[Pb(OH)_2]}{[Pb(OH)^+][OH^-]}}\nonumber\]

    \[\mathrm{K_{f3}=\dfrac{[Pb(OH)_3^-]}{[Pb(OH)_2][OH^-]}}\nonumber\]

    The important thing to realize is that any complexation of lead ions by hydroxide will lower the concentration of Pb2+.  Since [Pb2+] is the concentration in the Ksp expression, complexation of lead ions by hydroxide will cause more lead phosphate to dissolve to maintain Ksp.  Since all soluble forms of lead are toxic, this increase in lead concentration is a potential problem.  We can now couple these reactions into our scheme that describes the solubility of lead phosphate in this solution.

    \[\begin{array}{c}
    \ce{Pb3(PO4)2 &↔ &3Pb^2+ &+ &2PO_4^3- \\
    &&⇕&&⇕\\
    &&Pb(OH)^+&&HPO_4^2- \\
    &&⇕&&⇕\\
    &&Pb(OH)_2&&H_2PO_4^- \\
    &&⇕&&⇕\\
    &&Pb(OH)_3^- &&H_3PO_4}
    \end{array}\nonumber\]

    This is now quite a complicated set of simultaneous reactions that take place.  Our goal in the equilibrium unit of this course will be to develop the facility to handle these types of complicated problems.

     

     

    Before we get started into this process, there are a couple of other general things to know about chemical equilibrium.  Consider the general reaction shown below.

    \[\mathrm{aA  +  bB  \leftrightarrow   cC  +  dD}\nonumber\]

    One way of describing equilibrium is to say that the concentrations do not change.  The concentrations of the species in this solution represent a macroscopic parameter of the system, and so at the macroscopic level, this system is static.

    Another way of describing equilibrium is to say that for every forward reaction there is a corresponding reverse reaction.  This means at the microscopic level that As and Bs are constantly converting to Cs and Ds and vice versa, but that the rate of these two processes are equal.  At the microscopic level, a system at equilibrium is dynamic.

    Unless you have taken physical chemistry, I am fairly certain that everything you have learned until this point has taught you that the following expression can be used to describe the equilibrium state of this reaction.

    \[\mathrm{K = \dfrac{[C ]^c [D ]^d}{[A ]^a [B ]^b}}\nonumber\]

    Well it turns out that this expression is not rigorously correct.  Instead of the concentrations of reagents, the actual terms we need in an equilibrium constant expression are the activities of the substances.  The expression shown below is the correct form of the equilibrium constant, in which aA represents the activity of substance A.

    \[\mathrm{K = \dfrac{[a _C  ]^c  [a _D  ]^d }{[a _A  ]^a  [a _B  ]^b  }}\nonumber\]

    If you examine the group of As and Bs below, hopefully you can appreciate that the A shown in boldface is “inactive”.  For that A species to react with a B, another A species must move out of the way.

    A         B         A         A         A         B         B         A

    If the correct form of the equilibrium constant expression uses the activities of the chemicals, why have you always been taught to use concentrations?  It turns out that in most situations we do not have reliable procedures to accurately calculate the activities of substances.  If we did, we would almost certainly use the correct form of the expression.  Since we do not know how to evaluate the activities of substances under most circumstances, we do the next best thing and use concentrations as an approximation.  This means that all equilibrium calculations are at best approximations (some better than others).  In other words, equilibrium calculations usually provide estimations of the situation, but not rigorously correct answers.  Because the entire premise is based on an approximation, this will often allow us to make other approximations when we perform equilibrium calculations.  These approximations will usually involve ignoring the contributions of minor constituents of the solution.

    One last thing we ought to consider is when the approximation of using concentration instead of activity is most valid.  Perhaps a way to see this is to consider a solution that has lots of A (the concentration of A is high) and only a small amount of B (the concentration of B is low).  Inactivity results if a similar species is in the way of the two reactants getting together.  Since the concentration of B is low, there is very little probability that one B would get in the way of another and prevent it from encountering an A.  For A, on the other hand, there are so many that they are likely to get in each other’s way from being able to encounter a B.  Concentration is a better approximation of activity at low concentrations.  The example I have shown with A and B implies there is no solvent, but this trend holds as well if the substances are dissolved in a solvent.  Notice as well that the activity can never be higher than the concentration, but only lower.

    How low a concentration do we need to feel fully comfortable in using the approximation of concentrations for activities?  A general rule of thumb is if the concentrations are less than 0.01 M then the approximation is quite a good one.  Many solutions we will handle this term will have concentrations lower than 0.01 M, but many others will not.  We do not need to dwell excessively on this point, but it is worth keeping in the back of one’s mind that calculations of solutions with relatively high concentrations are always approximations.  We are getting a ballpark figure that lets us know whether a particular process we want to use or study is viable. 

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