7.CHEM 1622002 HOURLY EXAM III + ANSWERSEQUILIBRIASOLUBILITY PRODUCTWhen three moles of CaCl2are added to one mole of Na2SO4and two moles of NaF, all contained in one liter, of aqueous solution, which of the following is true? (CaSO4Ksp = 6.1 x 105and CaF2Ksp = 5.3 x 109)
A. Precipitation of CaSO4is not completeB.CaF2and CaSO4will precipitate
Chapter 15B practice problems
50
C
.
Precipitation of CaSO
4
is complete
D.
Only CaSO
4
will precipitate
E.
Only CaF
2
will precipitate
CaSO
4
Ca
2+
+
SO
4
2
Ksp = 6.1 x 10
5
X
X
X x X = 6.1 x 10
5
X = 0.78 x 10
2
= [Ca
2+
] to initiate CaSO
4
precipitation
CaF
2
Ca
2+
+
2F

Ksp = 5.3 x 10
9
X
2X
X x (2X)
2
= 4X
3
= 5.3 x 10
9
X = 1.1 x 10
3
= [Ca
2+
] to initiate CaF
2
precipitation
Since it takes less Calcium ion to initiate the CaF
2
precipitation, then calcium fluoride will precipitate
first.
Calcium fluoride will
precipitate since it requires only 1.1 x 10
3
M Calcium ion, and there is 3
molar calcium ion present.
If the stoichiometric amount of calcium
completely
reacts with 1 mole of
fluoride to precipitate CaF
2
, then this would leave two moles of calcium to react with the one mole of
sulfate.
Since it requires only 0.078 x 10
2
M Ca
2+
to initiate precipitation of sulfate, then calcium sulfate
will also precipitate.
We can stop here and choose “B” as the answer.
However, if we want to investigate
whether “C” might also be an answer, then:
To determine whether the CaSO
4
precipitation is complete, assume that the CaSO4 precipitation reaction
goes to completion, and then work backwards to determine how much sulfate is in solution.
Ca
2+
+
SO
4
2
CaSO
4
K = 1/Ksp = 1/6.1 x 10
5
= 1.64 x 10
4
Initial
2
1
0
Change
1
1
+1
Completion
+1
0
+1
Change
+X
+X
X
Equilibrium
1+X
+X
(1+X) x X = 6.1 x 10
5
Assume that X is very small; simplify.
1 x X = 6.1 x 10
5
X = 6.1 x 10
5
= SO
4
2
concentration remaining in solution after precipitation of CaSO
4
(6.1 x 10
5
/1) x 100 = 0.0061% SO
4
2
left in solution after precipitation of CaSO
4
.
Therefore, the
precipitation of CaSO
4
is complete making “C” also a correct answer.
Chapter 15B practice problems
51
Chapter 15B practice problems
52
25.
CHEM 1622002 HOURLY EXAM III + ANSWERS
EQUILIBRIA
SOLUBILITY PRODUCT
A solution is made by mixing 200 mL 0.10 M CaCl
2
and 200 mL 1.0 M NaF.
If the K
sp
for CaF
2
is 5.3 x 10
9
, the concentration of Ca
2+
and F

in solution will be
0.02 moles of CaCl
2
& 0.2 moles of NaF in 0.400 L
Therefore, 0.05 M CaCl
2
and 0.50 M NaF
Considering the excess of F

and the small K
sp
, assume that the precipitation reaction between the Ca
2+
and
the F

goes to completion, i.e., the entire 0.05 M Ca
2+
is used up to form 0.05 M CaF
2
precipitate, and then
work backwards to find out how much Ca
2+
and F

are remaining in solution.
Therefore, remaining is 0.40
M F

.
An equilibrium equation may be written from left to right or right to left; it makes no difference.
CaF
2
Ca
2+
+
2F

K
sp
for CaF
2
is 5.3 x 10
9
Initial
0
0.05
0.50
Change
+0.05
0.05
0.10
Completion
+0.05
0
+0.40
Change
+0.05  X
X
0.40 + 2X
X
0.40 + 2X
X x (0.40 + 2X)
2
= 5.3 x 10
9
Simplifying:
X x (0.40)
2
= 5.3 x 10
9
X = 3.31 x 10
8
M Ca
2+
+ 0.40 M F

A
.
Ca
2+
= 3.3 x 10
8
, and F

= 0.40 M
B.
Ca
2+
= 0.10 M, and F

= 6.0 x 10
6
M
C.
Ca
2+
= 0.050 M, and F

= 0.050 M
D.
Ca
2+
= 0.30 M, and F

= 0.040 M
E.
Ca
2+
= 1.1 x 10
3
M, and F

= 1.1 x 10
3
M
1.
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 Solubility, Solubility equilibrium