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Unformatted text preview: 1 + Ag PRECIPITATION REACTIONS
Chapter 19 Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2 AgCl(s)
(s) Analysis of
Silver Group Ag+(aq) + Cl(aq)
Ag When solution is SATURATED , expt. shows
When
SATURATED
that [Ag+] = 1.67 x 105 M.
This is equivalent to the SOLUBILITY of AgCl.
This
SOLUBILITY of
What is [Cl ]?
This is also equivalent to the AgCl solubility. 2+ Pb Hg2 Analysis of
Silver Group 2+ AgCl PbCl2 Hg2Cl2 4 Analysis of
Silver Group AgCl PbCl2 Hg2Cl2 AgCl(s)
(s) + Cl(aq) Saturated solution has
Saturated
[Ag+] = [Cl] = 1.67 x 10 5 M
Use this to calculate Kc
Kc = [Ag+] [Cl] Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2 All salts formed in
this experiment are
said to be
INSOLUBLE and
form when mixing
moderately
concentrated
solutions of the
metal ion with
chloride ions. Ag+ Pb2+ Hg22+ Ag+(aq)
Ag 2 Analysis of
Silver Group Although all salts formed in this experiment
are said to be insoluble, they do dissolve
to some SLIGHT extent.
AgCl(s)
Ag+(aq) + Cl(aq)
(s)
Ag
When equilibrium has been established, no
more AgCl dissolves and the solution is
SATURATED . 5 Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2 AgCl(s)
(s)
Kc = [Ag+] Analysis of
Silver Group Ag+(aq) + Cl(aq)
[Cl] = 2.79 x 1010 Because this is the product of “ solubilities ”,
we call it = (1.67 x 10 5)(1.67 x 10 5) Ksp = solubility product constant = 2.79 x 10 10 See Table 19.2 and Appendix J Page 1 3 6 7 Lead(II) Chloride Solubility of Lead(II) Iodide 8 Consider PbI 2 dissolving in water
PbI2(s) √ Pb2+(aq) + 2 I (aq)
(s)
Calculate Ksp
if solubility = 0.00130 M
if Solubility of Lead(II) Iodide Solution Consider PbI 2 dissolving in water
PbI2(s) √ Pb2+(aq) + 2 I (aq)
(s)
Calculate Ksp
if solubility = 0.00130 M
if
Solution 1. Solubility = [Pb 2+] = 1.30 x 10 3 M PbCl2(s)
Pb2+(aq ) + 2 Cl(aq)
(s)
Ksp = 1.9 x 10 5 9 2. Ksp = [Pb2+] [I]2
= [Pb2+] {2 • [Pb2+]}2
[Pb [I] = ?
[I Ksp = 4 [Pb2+]3 = 4 (solubility) 33
= 4 (solubility) [I] = 2 x [Pb2+] = 2.60 x 10 3 M Ksp = 4 ((1.30 x 10 3))33 = 8.8 x 10 9
Ksp = 4 1.30 x 10 3 = 8.8 x 10 9 10 Precipitating an Insoluble Salt
Hg2Cl2(s)
(s) Hg22+(aq) + 2 Cl(aq) 11 Precipitating an Insoluble Salt
Hg2Cl2(s)
(s) Hg22+(aq) + 2 Cl(aq) 12 Precipitating an Insoluble Salt
Hg2Cl2(s)
(s) Hg22+(aq) + 2 Cl(aq) Ksp = 1.1 x 10 18 = [Hg2 2+] [Cl]2 Ksp = 1.1 x 10 18 = [Hg2 2+] [Cl]2 Ksp = 1.1 x 10 18 = [Hg2 2+] [Cl]2 If [Hg 22+] = 0.010 M, what [ Cl] is req’d to just Recognize that Solution Ksp = product of [Cl] that can exist when [Hg 22+] = 0.010 M, begin the precipitation of Hg 2Cl2 ?
That is, what is the maximum [ Cl] that can be in solution with 0.010 M Hg 22+ without
forming Hg2Cl 2? maximum ion concs.
maximum
Precip . begins when product of
begins
ion concs. EXCEEDS the K sp.
ion Page 2 [Cl ] = K sp
0.010 = 1 .1 x 108 M If this conc. of Cl is just exceeded, Hg 2Cl2
begins to precipitate. Precipitating an Insoluble Salt
Hg2Cl2(s)
(s)
Ksp = 1.1 x 13 14 Separating Metal Ions Hg22+(aq) + 2 Cl(aq) A solution contains 0.020 M Ag+ and Pb2+. Add
CrO42 to precipitate red Ag 2CrO4 and yellow
PbCrO4. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10 12
Ksp for PbCrO4 = 1.8 x 10 14
for
1.8
Solution 2+ Ag +
2+
Cu2+, Ag+, Pb2+ 10 18 Now raise [ Cl] to 1.0 M. What is the value of
[Hg22+] at this point? Ksp Values
Ksp Values
AgCl
1.8 x 1010
AgCl
1.8 x 1010
PbCl22
1.7 x 105
PbCl
1.7 x 105
PbCrO44 1.8 x 1014
1.8 x 1014
PbCrO
1.8 Solution
[Hg22+] = Ksp / [Cl]2
= Ksp / (1.0)2 = 1.1 x 10 18 M The substance whose Ksp is first
exceeded precipitates first.
The ion requiring the lesser amount
of CrO 42 ppts. first. The concentration of Hg 2 2+ has been reduced
by 1016 ! 16 15 Separating Salts by Differences in K sp
sp 17 18 Separating Salts by Differences in K sp
sp Separating Salts by Differences in K sp
sp Separating Salts by Differences in K sp
sp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42 to precipitate
red Ag2CrO4 and yellow PbCrO4. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 1012
Ksp for PbCrO4 = 1.8 x 1014
for
1.8 A solution contains 0.020 M Ag+ and Pb2+. Add
CrO42 to precipitate red Ag 2CrO4 and yellow
PbCrO4. PbCrO 4 ppts. first.
Ksp (Ag2CrO4)= 9.0 x 1012
Ksp (PbCrO4) = 1.8 x 1014
(PbCrO
How much Pb 2+ remains in solution when Ag+
begins to precipitate? A solution contains 0.020 M Ag+ and Pb2+. Add CrO42 to precipitate
red Ag2CrO4 and yellow PbCrO4.
Ksp (Ag2CrO4)= 9.0 x 1012
Ksp (PbCrO4) = 1.8 x 1014
(PbCrO
How much Pb2+ remains in solution when Ag+ begins to precipitate? Solution
Calculate [CrO 42] required by each ion.
[CrO42] to ppt. PbCrO 4 = Ksp / [Pb2+]
[Pb
= 1.8 x 10 14 / 0.020 = 9.0 x 10 13 M
1.8
[CrO42] to ppt. Ag2CrO4 = Ksp / [Ag+]2
= 9.0 x 10 12 / (0.020) 2 = 2.3 x 10 8 M
9.0 PbCrO4 precipitates first Page 3 Solution
We know that [CrO 42] = 2.3 x 10 8 M to begin to
ppt. Ag2CrO4.
What is the Pb 2+ conc . at this point?
[Pb2+] = K sp / [CrO42] = 1.8 x 10 14 / 2.3 x 10 8 M
= 7.8 x 107 M
7.8
Lead ion has dropped from 0.020 M to < 10 6 M 19 20 Simultaneous Equilibria Simultaneous Equilibria Section 19.8 • PbCl2(s) + CrO42 • Salt Salt
PbCl
PbCl2
PbCrO4 PbCrO4 + 2 Cl PbCl 2(s)
(s) Ksp PbCl 2
PbCrO 4 1.8 x 1014 Complex Ions
Add NH3 to light blue [Cu(H 2O)4]2+ >
light blue Cu(OH)2 and then deep blue [Cu(NH3)4]2+ Ksp
1.7 x 105
1.8 x 1014 Pb 2+ + CrO42 1.7 x 105 The combination of metal ions
with small molecules such
as H2O and NH3 > PbCrO4 + 2 Cl Pb2+ + 2 Cl COMPLEX IONS K1 = Ksp PbCrO4 K2 = 1/Ksp
[Cu(NH3)4]2+ • Knet = K1 • K2 = 9.4 x 108
• Net reaction is productfavored 22 Dissolving Precipitates
by forming Complex Ions
Formation of complex ions explains why you
can dissolve a ppt. by forming a complex
ion.
AgCl(s) ¸ Ag+ + Cl (s)
Ag
Ag+ + 2 NH3 > Ag(NH3)2 Ksp = 1.8 x 10 10
+ Kform = 1.6 x 10 7 AgCl(s) + 2 NH3 ¸ Ag(NH3)2+ + ClAg
Knet = Ksp • Kform = 2.9 x 10 3 Page 4 21 Section 19.10 Section 19.8 • Add CrO42 to solid PbCl 2. The less
soluble salt, PbCrO 4, precipitates
• PbCl2(s) + CrO42 Simultaneous Equilibria and
Complex Ions All metal ions form
All metal ions form
complex ions with water
complex ions with water
—and are of the type
—and are of the type
n+
[M(H22O)x]n+where x = 4
[M(H O)x] where x = 4
and 6.
and 6. 23 24 Common Ion Effect Adding an Ion “Common” to an
Adding an Ion “Common” to an
Equilibrium
Equilibrium 25 The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and
in
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
1.1
BaSO4(s) √ Ba2+(aq) + SO42(aq)
(s) Solution
Ksp = [Ba2+] [SO42] = (0.010 + y) (y)
Because y < 1.1 x 10 5 M (= x, the solubility in
pure water), this means 0.010 + y is about
equal to 0.010. Therefore,
Ksp = 1.1 x 10 10 = (0.010)(y)
y = 1.1 x 10 8 M = solubility in presence of
added Ba2+ ion. 27 The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and
in
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
1.1 Calculate the solubility of BaSO4 in (a) pure water and
in
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
1.1 BaSO4(s) √ Ba2+(aq) + SO42(aq)
(s) Calculate the solubility of BaSO 4 in (a) pure
in
water and (b) in 0.010 M Ba(NO3)2.
Ksp for BaSO 4 = 1.1 x 10 10
1.1
BaSO4(s)
Ba2+(aq) + SO42(aq)
(s)
Solution
Solubility in pure water = [Ba 2+] = [SO42] = x
Ksp = [Ba2+] [SO42] = x 2
x = (Ksp)1/2 = 1.1 x 105 M
Solubility in pure water = 1.1 x 10 5 mol/L The Common Ion Effect 26 The Common Ion Effect BaSO4(s) √ Ba2+(aq) + SO42(aq)
(s) Solution Solution
Solubility in pure water = 1.1 x 10 5 mol/L.
Now dissolve BaSO 4 in water already
containing 0.010 M Ba 2+.
Which way will the “common ion” shift the
equilibrium? ___ Will solubility of BaSO 4 be
less than or greater than in pure water?___ 28 initial
change
equilib.
equilib 29 The Common Ion Effect
Calculate the solubility of BaSO4 in (a) pure water and
in
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
1.1
BaSO4(s) √ Ba2+(aq) + SO42(aq)
(s) SUMMARY
Solubility in pure water = x = 1.1 x 10 5 M
Solubility in presence of added Ba 2+
= 1.1 x 10 8 M
1.1
Le Chatelier’s Principle is followed! Page 5 [Ba2+]
[Ba
0.010
+y
0.010 + y [SO42]
0
+y
y ...
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 Fall '01
 Reynolds
 Organic chemistry, Reaction, Solubility

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