Chem104 Quiz 5 ANSWERS
Self Check Due by 5 PM Tuesday.
1. The reaction:
[Cu(H
2
O)
6
] + + Cl-  [CuCl(H
2
O)
5
] + + H
2
O has the equilibrium constant K = 0.28. 100mL of this reaction system is at equilibrium with these concentrations
[Cu(H
2
O)
62+
] = 0.40M, [Cl-] = 7.1 M, [CuCl(H
2
O)
5+
] = 0.80 M. and then 2.0 g LiCl is added to the 100mL of solution.
(a). What will happen to the distribution of the species in this reaction? Prove your prediction by calculating and interpreting Q.
Addition of more Cl- to the equilibrium will push the reactants to make more products; the equilibrium shifts right.
Calculation of new [Cl]:
(2.0 g / 42.3 g mol -1 ) / 0.100 L = 0.473 M Cl- added.
New [Cl-] before equilibrium is re-established 7.1 M + 0.473 M = 7.573 M
NOTE: at this point it is best to keep all the sig figs until the end. Experience will show you this. Truncating the sig figs too soon in a Keq calc can mean losing the answer which is often as small as the new concentrations (the “x”)
Q= [Cu-Cl[ / [Cu2+][Cl-] = 0.80 / 0.40 x 7.573 = 0.264 < K eq
= 0.28.
When Q < K, the reaction shift right to make more prodcts.
(b) What are the new equilibrium concentrations established after the lithium chloride addition?
Create the ICEbox:
Initial
[Cu2+]
0.40 M
Change
Equil
-x
0.40 –x
[Cl]
7.573 M
-x
7.573 –x
[Cu-Cl]
0.80 M
+x
0.80 + x

K eq
= 0.28 = [Cu-Cl[ / [Cu2+][Cl-] = ( 0.80 + x ) / ( 0.40 –x ) ( 7.573 –x )
0.28 ( 0.40 –x ) ( 7.573 –x ) = ( 0.80 + x )
0.848 – 2.23x - 0.28 x 2 = 0.8 + x
0 = -0.048 + 3.23x + 0.28 x 2 x = 0.0154
So final equilibrium concentrations are:
[Cu2+] = 0.40 - 0.0154 = 0.385 M
[Cl-] = 7.573 - 0.0154 = 7.56 M
[Cu-Cl] = 0.80 + 0.0154 = 0.815 M
(c) What is the equilibrium constant for this reaction?
[CuCl(H
2
O)
5
] + + H
2
O  [Cu(H
2
O)
6
] + + Cl-
This reaction the reverse of the one in (a), so its K will be 1/ Keq in (a) or 1/ 0.28.
The Keq for the reaction in (c) is 3.57.