Lesson 5.

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Solubility
Lesson 5
Trial Ion Product
We have learned that when two ionic solutions are mixed and if one
product has low solubility, then there is a reaction where a
precipitate will form.
Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)
low solubility
The solubility chart on page 4 predicts this reaction, but only if the
solution is 0.10 M or greater. If the molarity is less than 0.10 M, then
the reaction may or may not happen.
A trial ion product must be calculated to predict the reaction of all
solutions less than 0.10 M.
The capacity of a solution to dissolve a solid is described by the
Ksp.
Pb(NO3)2
NaCl
Pb2+
PbCl2(s)
⇌
2ClPb2+
+
2Cl-
The Ksp represents the limit of the solution to dissolve PbCl2.
You can add Pb2+ and Cl- until the ion concentrations are equal to the
Ksp.
The solution is saturated and addition PbCl2 must sit on the bottom
not dissolved.
1.
200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M
NaCl, will a precipitate occur?
1.
200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M
NaCl, will a precipitate occur?
PbCl2(s)
⇌
Pb2+
+
2Cl-
Write a dissociation equation for the compound with low solubility.
1.
200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M
NaCl, will a precipitate occur?
PbCl2(s)
⇌
Pb2+
+
2Cl0.10 M
0.20 M
Write a dissociation equation for the compound with low solubility.
List the initial molarities of each ion.
1.
200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M
NaCl, will a precipitate occur?
PbCl2(s)
⇌
Pb2+
+
2Cl200
500
0.10 M
0.040 M
300
500
0.20 M
0.12 M
Write a dissociation equation for the compound with low solubility.
List the initial molarities of each ion.
Reduce each molarity by the dilution factor: V1/V2.
1.
200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M
NaCl, will a precipitate occur?
PbCl2(s)
⇌
Pb2+
+
2Cl200
500
0.10 M
0.040 M
TIP
TIP
=
=
=
300
500
0.20 M
0.12 M
[Pb2+][Cl-]2
[0.040][0.12] 2
5.8 x 10-4
Write a dissociation equation for the compound with low solubility.
List the initial molarities of each ion.
Reduce each molarity by the dilution factor: V1/V2.
Write the Ksp or TIP (trial ion product) and solve.
1.
200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M
NaCl, will a precipitate occur?
PbCl2(s)
⇌
Pb2+
+
2Cl200
500
0.10 M
0.040 M
TIP
TIP
Ksp
=
=
=
=
300
500
0.20 M
0.12 M
[Pb2+][Cl-]2
[0.040][0.12] 2
5.8 x 10-4
1.2 x 10-5 TIP > Ksp ppt forms
Write a dissociation equation for the compound with low solubility.
List the initial molarities of each ion.
Reduce each molarity by the dilution factor: V1/V2.
Write the Ksp or TIP (trial ion product) and solve.
Compare to real Ksp
2.
Will a precipitate form if 20.0 mL of 0.010M CaCl2 is
mixed with 60.0 mL of 0.0080 M Na2SO4?
CaSO4(s)
⇌
Ca2+
20
80
0.010 M
+
0.0025 M
Ksp
SO42-
60
0.0080 M
80
0.0060 M
[Ca2+][SO42-]
[0.0025][0.0060]
1.5 x 10-5
TIP
TIP
=
=
=
=
7.1 x 10-5 TIP < Ksp no ppt forms
3.
Will a precipitate form when equal volumes of 0.020 M CaCl2
and 0.040 M AgNO3 are mixed.
The Cl- is doubled
AgCl(s)
⇌
Ag+
+
Cl-
1
0.040 M
2
0.020 M
TIP
TIP
=
=
=
[Ag+][Cl-]
[0.020][0.020]
4.0 x 10-4
Ksp
=
1.8 x 10-10
TIP > Ksp
ppt forms
1
0.040 M
2
0.020 M
4.
Consider the two saturated solutions AgCl and Ag2CrO4.
Which has the greater Ag+ concentration?
AgCl ⇌
s
Ksp
Ag+ + Cls
=
s2
s
Ag2CrO4 ⇌ 2Ag+
s
+
2s
Ksp
=
CrO42s
4s3
1.8 x 10 -10 = s2
1.1 x 10-12 =4s3
s = 1.3 x 10-5 M
s = 6.5 x 10-5 M
[Ag+] = 1.3 x 10-5 M
[Ag+] = 2s = 1.3 x 10-4 M
Ag2CrO4 has the greater Ag+ concentration
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