Practice Exercise 17.18 error

[Zellmer, Robert]

There's in error in the answer in the back of the book for Practice
Ex. 17.18 that follows Sample Exercise 17.18.    I sent out something
about this on Wed but I got a couple of questions about it so here's
more of an explanation

The problem deals with precipitating Mg(OH)2 and Cu(OH)2 from
a soln of 0.050 M Mg^2+ and 0.050 M Cu^2+ by adding OH-.
It wants to know which ppt first and the conc. necessary to ppt
each.

You use the Ksp values for Mg(OH)2 and Cu(OH)2 and the conc.
of the Mg^2+ and Cu^2+ to determine the [OH-] required to ppt
both, technically the equilibrium conc of OH- in a saturated soln.

Doing this you get the following:

For Mg(OH)2:

For a saturated soln [OH-] = 1.9 x 10^-5 M.  Thus, to get
Mg(OH)2 to ppt [OH-] > 1.9 x 10^-5 M.

For Cu(OH)2:

For a saturated soln [OH-] = 9.8 x 10^-10 M.  Thus, to get
Cu(OH)2 to ppt [OH-] > 9.8 x 10^-10 M.

This means Cu(OH)2 will ppt first.

This is the answer given for which ppt first.  However, they
have the wrong conc. for the OH- required to begin to ppt
the Cu(OH)2.  The incorrect answer in the book is 1.5 x 10^-9 M).

To go a step further, what conc. of OH- will give maximum
separation of the Mg^2+ and Cu^2+?  You already have the
answer above.  As you add OH- the Cu(OH)2 starts to ppt.
It will continue to ppt w/o any Mg(OH)2 precipitating until the
conc. of OH- exceeds 1.9 x 10^-5 M.  That's the conc. of
OH- to give a saturated soln of Mg(OH)2 but no ppt.  That's
what you got above when calculating the conc of OH- that
will cause the Mg(OH)2 to ppt.

The amount of Cu^2+ which remains in soln when the
Mg(OH)2 starts to ppt can be found by using the Ksp for
Cu(OH)2 and the conc. of OH- which gives a saturated
solution of Mg(OH)2 (before it ppt) to calc the conc. of
Cu^2+ with this conc of OH-.

Ksp = [Cu^2+] * [OH-]^2

[Cu+2] = Ksp/[OH-]^2 = (4.8 x 10^-20)/(1.9 x 10^-5)^2

[Cu^2+] = 1.33 x 10^-10 M

Only 2.7 x 10^-7 % of the Cu^2+ remains in soln, a very
effective separation.

Dr. Zellmer
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