Chemist’s Daily
Chapter 16 - Ksp
1. Write out the Ksp and equilibrium expression for:
PbCO3 ⇔ Pb
PbCO3
Ksp = [Pb
2+
+
2+
Ag2S
Ca(OH)2
Ca(OH)2
+
Ksp = [Ag ]
+
Ksp = [Ca
2-
] [CO3
⇔ 2 Ag
Ag2S
2+
2
[S
2-
CO3
2-
]
+ S
2-
]
⇔
Ca
] [OH ]
-
2+
2
+
2 OH
-
2. Calculate the molar solubility of copper(I) iodide.
Looking in your appendix, Ksp (CuI) = 1 x 10
CuI
⇔ Cu
+
+
Ksp = [Cu ] [I ]
+
1x 10
-
I
= (x) (x)
-12
x = 1 x 10
-6
= x
2
So the molar solubility is 1 x 10
3.
-12
-
-6
The molar solubility of silver sulfate is 1.5 x 10
⇔ 2 Ag
Ag2SO4
1.5 x 10
+
-2
+
-2
2-
] = 1.5 x 10
Ksp =
M.
[Ag ]
+
2
-2
[SO4
M
2-
What is the Ksp for this salt?
2-
mols of silver sulfate can dissolve per liter.
[Ag ] = 2 (1.5 x 10 )
[SO4
+ SO4
-2
= 3.0 x 10
-2
] = (3.0 x 10
This leads to:
M
-2
M)
2
(1.5
x 10
-2
M) = 1.4 x 10
-5
Chemist’s Daily
Chapter 16 - Ksp
1.
Exactly 200 mL of a 0.0040 BaCl2 solution are added to exactly 600 mL of 0.0080 M
K2SO4.
Will a precipitate form?
Solubility rules tell us the only possible precipitate is barium sulfate, so we need only
concern ourselves with the initial concentrations of Ba
2+
and SO4
2-
.
both solutions, so we must calculate the new initial concentrations:
200 mL (0.0040 M Ba
[Ba
2+
2+
] = 0.0010 M
600 mL (0.0080 M SO4
[SO4
2-
) = 800 mL (? M Ba
2-
] = 0.0060 M
The ion product is [Ba
2+
)
2+
2-
]
2-
)
= (0.0010) (0.0060) = 6.0 x 10 .
the Ksp for barium sulfate, 1.1 x 10
-6
-10
precipitate will form.
2.
)
= 800 mL (? M SO 4
] [SO4
-3
Express your answer in g/L.
⇔
Initial
End
+
+
-3
6.5 x 10
Ksp = [Ag ] [Cl ]
+
1.8 x 10
Ag
6.5 x 10
-10
solubility =
Cl
M
M
M silver nitrate solution?
-
x M
-
=
x = 2.8 x 10
-3
-8
(6.5 x 10 ) (x)
(2.8 x 10
-3
-8
mol L ) (143.323 g mol ) = 4.0 x 10
-1
Comparing this to
, we find it to be larger than the Ksp so a
What is the solubility of silver chloride in a 6.5 x 10
AgCl
We are diluting
-1
-6
g/L