EOC exercise 17.66

[Zellmer, Robert]

Someone asked about EOC exercise 17.66.  I've been asked about this
question by other students in the past.

For this problem, you're trying to completely dissolve 0.020 mol of NiC2O4
by adding NH3 to form a complex with the Ni^2+.  This is a little different
than asking what's the solubility of a solid in a particular soln (where some
solid remains).  There's a couple of ways to do this.

Also, it's asking for the what the conc. of NH3 will have to be in solution
(i.e. at equilibrium) in order to have all of the solute dissolved.  Think of it
as what does the conc. of NH3 have to be in the solution in order to get that
conc. of the solute (or the conc. of Ni^2+) in solution.  It would be more
difficult to determine the actual amount of NH3 which would have to be
added to get this conc. of NH3 because then you would also have to
account for the NH3 that would react with water, although almost all of
the NH3 added would wind up in the complex due to the size of the Kf.


1)  In the solns manual they look at the C2O4^2- since it doesn't form
a complex.  For 0.020 mol of the NiC2O4 to dissolve you will need to get
0.020 mol of C2O4^2- in soln.  Since there's 1 L of soln you need a conc
of 0.020 M C2O4^2-.  You can use the Ksp to determine the conc. of Ni^2+
which would remain when the conc. of C2O4^2- is 0.020 M.  This will give
a conc of free Ni^2+ of 2 x 10^-8 M.  To get this the NH3 had to react
with the Ni^2+ which was in the original NiC2O4.  Thus, there needs to be
essentially 0.020 M of the Ni(NH3)6^2+ in soln.  You need to use the Kf
to calculate the [NH3] in soln at equilibrium to get 0.020 M Ni(NH3)6^2+
and 2 x 10^-8 M free Ni^2+.

      [Ni(NH3)6^2+]              (0.020)
K = ------------------- = -------------------- = 1.2 x 10^9
     [Ni^2+] [NH3]^6      (2 x 10^-8) [NH3]^6

That was the last step they did to get [NH3] = 0.307 M.


2) You could also do this by coupling the Ksp and Kf equations to get
an overall reaction eqn.

    NiC2O4   <=>   Ni^2+   +   C2O4^2-            Ksp
    Ni^2+   +   6 NH3  <=>  Ni(NH3)6^2+           Kf
------------------------------------------------------
    NiC2O4  +  6 NH3  <=>  Ni(NH3)6^2+  +  C2O4^2- K = Ksp*Kf

The value of K is (4 x 10^-10)(1.2 x 10^9) = 0.48

      [Ni(NH3)6^2+] [C2O4^2-]
K = --------------------------- = 0.48
              [NH3]^6

      (0.020) (0.020)
K = ------------------ = 0.48
          [NH3]^6

This gives [NH3] = 0.307 M (same as in #1).


Dr. Zellmer
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