Solubility Review Questions Worksheet With Answers Page 2
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14SOLUBILITY REVIEW QUESTIONS
Solubility Problem Set 2
2+
2-
-10
1.
In a saturated solution of FeS, the [Fe
] and the [S
] are both 6.08 x 10
M.
Calculate the value of K
.
sp
2+
2-
2.
Find [Ca
] and [SO
] in a saturated solution of CaSO
.
4
4
2+
-
-11
3.
Find [Ca
] and [F
] in a saturated solution of CaF
. K
for CaF
is 4.9 x 10
.
2
sp
2
-49
4.
Find the solubility of Ag
S in water in g/L. K
for Ag
S is 1.6 x 10
.
2
sp
2
5.
Will a precipitate form if 10.0 mL of 0.010 M AgNO
is mixed with 25.0 mL
3
of 0.10 M Na
CO
?
2
3
6.
Will a precipitate form if 5.0 mL of 0.0040 M AgNO
is added to 15.0 mL of a solution
3
-
containing 1.5 mg of bromide ions? (How does the mass of Br compare to Br
?)
7.
Calculate the K
for each of the salts whose solubility is:
sp
-3
a) CaSO
: 5.0 x 10
M
b) SrF
: 12.2 mg/100 mL.
4
2
8.
Can the temperature of 2.5 L of a saturated AgCl solution containing 38 mg of dissociated
o
AgCl be 25
C? If the dissociation of the solid phase to the aqueous phase is endothermic,
o
how does the temperature of this saturated solution compare to one at 25
C?
9.
Compare the molar solubility of PbI
in pure water and in 0.10 M Pb(NO
)
2
3
2.
+
2+
10.
A 1.0 L solution contains 0.010 M Ag
ions and 0.010 M Sr
ions.
a)
Which ion precipitates first when K
CO
is added to the mixture? Assume no
2
3
change in volume.
b)
What is the concentration of the ion that is precipitated first when the second ion
begins to precipitate?
11.
Barium nitrate reacts with potassium sulphate solution and forms insoluble barium
sulphate. What volume of 0.40 M Ba(NO
)
solution is required to precipitate the sulphate
3
2
ions in 25.00 mL of 0.80 M K
SO
?
2
4
12.
A 225 mL sample of tap water containing the chloride ion requires 36.42 mL
of 0.100 M AgNO
to titrate. What is the chloride concentration in the tap water?
3
Answers to Solubility Problem Set 1 (for questions on the other side):
-3
1. 8.4 x 10
M, 1.1 g/L, 0.11 g/100 mL
-5
+
-4
2-
-5
2. s = 6.5 x 10
, [Ag
] = 1.3 x 10
M, [CrO4
] = 6.5 x 10
M
-11
-6
20
2+
20
-
3. 1.9 x 10
M
4. 1.8 x 10
M
5. 1.0 x 10
Pb
ions, 2.0 x 10
IO
ions
3
-8
6. K
= 1.3 x 10
> K
, therefore a ppt will form
TIP
sp
-5
-4
7. K
(Mg(OH)
) = 6.4 x 10
, K
(SrSO
) = 6.6 x 10
, two ppt’s form
TIP
2
TIP
4
8. 0.51 g
+
-
9.
a)
1. add NaCl: Ag
+ Cl
-----> AgCl
(filter)
(aq)
(aq)
(s)
2+
2-
2. add Na
SO
: Sr
+ SO
-----> SrSO
(filter)
2
4
(aq)
4
(aq)
4(s)
2+
3-
3. add Na
PO
: 3 Cu
+ 2 PO
-----> Cu
(PO
)
(filter)
3
4
(aq)
4
(aq)
3
4
2(s)
2+
2-
b)
1. add Mg(NO
)
: Mg
+ SO
-----> MgSO
(filter)
3
2
(aq)
3
(aq)
3(s)
2+
2-
2. add Ba(NO
)
: Ba
+ SO
-----> BaSO
(filter)
3
2
(aq)
4
(aq)
4(s)
2+
-
3. add Pb(NO
)
: Pb
+ 2 Br
-----> PbBr
(filter)
3
2
(aq)
(aq)
2(s)
10. rate(dissolution) = increase, rate(crystallization) = increase, s = no change, K
= no change
sp
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