Transcript Solubility Equilibria - Mesa Community College

Solubility Equilibria
Solubility Product Constant
Ksp
for saturated solutions at equilibrium
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Comparing
values.
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Solubility Product (Ksp) = [products]x/[reactants]y but.....
reactants are in solid form, so Ksp=[products]x
i.e. A2B3(s)  2A3+ + 3B2–
Given:
Ksp=[A3+]2 [B2–]3
AgBr(s)  Ag+ + Br–
In a saturated solution @25oC, the [Ag+] = [Br– ]= 5.7 x 10–7 M.
Determine the Ksp value.
 Br   5.7 x 10   3.3 x 10
K sp  Ag


-7 2
-13
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Problem: A saturated solution of silver chromate was to found
contain 0.022 g/L of Ag2CrO4. Find Ksp
Eq. Expression:
Ag2CrO4 (s)  2Ag+ + CrO42–
Ksp = [Ag+]2[CrO42–]
So we must find the
concentrations of each
ion and then solve
for Ksp.
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Problem: A saturated solution of silver chromate was to
contain 0.022 g/L of Ag2CrO4. Find Ksp
Eq. Expression:
Ag2CrO4 (s)  2Ag+ + CrO42–
Ksp = [Ag+]2[CrO42–]

2
Ag
m ol
Ag+: 0.022 g Ag2CrO4
L
332g 1 Ag2C rO4
CrO4
–2:
1 C rO420.022g Ag2CrO4 m ol
L
332g 1 AgC rO4

K sp  1.33 x 10

mol
Ag
 1.33 x 10–4
L
 6.63 x
10–5
mol C rO42L
 6.63 x 10   1.16 x 10
-4 2
-5
-12
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Problems working from Ksp values.
Given: Ksp for MgF2 is 6.4 x 10–9 @ 25 oC
Find: solubility in mol/L and in g/L
Ksp = [Mg2+][F–]2
MgF2(s)  Mg2+ + 2F–
I.
C.
E.
N/A
N/A
N/A
0
+x
+x
0
+2x
+2x
Ksp= [x][2x]2 = 4x3
6.4 x 10–9 = 4x3


-9
6
.
4
x
10
x3
 1.2 x 10-3  Mg2  MgF2 
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now for g/L:
1.2 x 10-3 mol MgF2
L
62.3 g
mol
 7.3
x 10–2
g MgF 2
L
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The common ion effect “Le Chatelier”
overhead fig 17.16
What is the effect of
adding NaF?
CaF2(s)  Ca2+ + 2F-
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Solubility and pH
CaF2(s)  Ca2+ + 2F–
Add H+ (i.e. HCl)
2F– + H+  HF
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Solubility and pH
Mg(OH)2(s)  Mg2+ + 2OH–
Adding NaOH?
Adding HCl?
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The common ion effect “Le Chatelier”
Why is AgCl less soluble in sea water than in fresh water?
AgCl(s)  Ag+ + Cl–
Seawater contains
NaCl
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Problem: The solubility of AgCl in pure water is 1.3 x 10–5 M.
What is its solubility in seawater where the [Cl–] = 0.55 M?
(Ksp of AgCl = 1.8 x 10–10)
I.
C.
E.
AgCl(s)  Ag+ + Cl–
N/A
0
0.55
N/A
+x
+x
N/A
+x 0.55 + x
Ksp= [x][0.55 + x]
Ksp= [Ag+][Cl–]
try dropping this x
Ksp = 0.55x
1.8 x 10–10 = 0.55x
x = 3.3 x 10–10 = [Ag+]=[AgCl]
“AgCl is much less soluble in seawater”
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more Common ion effect:
a. What is the solubility of CaF2 in 0.010 M Ca(NO3)2?
Ksp(CaF2) = 3.9 x 10–11
CaF2(s)  Ca2+ + 2F–
[Ca2+]
[F–]
I. 0.010
0
C.
+x
+2x
2x
E. 0.010 + x
Ksp=[Ca2+][F-]2
Ksp= [0.010 + x][2x]2  [0.010][2x]2 = 0.010(4x2)
3.9 x 10–11 = 0.010(4x2)
x = 3.1 x 10–5 M Ca2+ from CaF2 so = M of CaF2
Now YOU determine the solubility of CaF2 in 0.010 M NaF.
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Answer:
3.9 x 10–7 M Ca2+
CaF2(s)  Ca2+ + 2F–
0
0.010
+x
2x
x
0.010 + 2x
Ksp = [x][0.010 + 2x]2  x(0.010)2
3.9 x 10-11 =x(0.010)2
x = 3.9 x 10-7
What does x tell us
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Reaction Quotient (Q): will a ppt. occur?
Use Q (also called ion product) and compare to Ksp
Q < Ksp
reaction goes
Q = Ksp
Equilibrium
Q > Ksp
reaction goes
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Problem: A solution is 1.5 x 10–6 M in Ni2+. Na2CO3 is added to
make the solution 6.0 x 10–4 M in CO32–.
Ksp(NiCO3) = 6.6 x 10–9.
Will NiCO3 ppt?
We must compare Q to Ksp.
NiCO3  Ni2+ + CO32–
Ksp = [Ni2+][CO32–]
Q = [Ni2+][CO32–]
Q = [1.5 x 10–6][6.0 x 10–4] = 9.0 x 10–10
Q < Ksp
no ppt.
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Problem: 0.50 L of 1.0 x 10–5 M Pb(OAc)2 is combined with 0.50 L
of 1.0 x 10–3 M K2CrO4.
a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–14
Pb(OAc)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KOAc(aq)
then:
PbCrO4(s)  Pb2+ + CrO42–
-5
[Pb2+]: 0.50 L 1.0 x 10 mol Pb(O Ac)2
L
Ksp= [Pb2+][CrO42–]
2
1 Pb2
mol
Pb
 5.0 x 10–6
L
1 Pb(O Ac)2 1 L
[CrO42-]: 0.50 L 1.0 x 10-3 m ol K C rO 1 C rO422
4
1 K 2C rO4
L
1L

5.0 x
10-4
molC rO42L
Q = [5.0 x 10-6][5.0 x 10-4] = 2.5 x 10-9
compare to Ksp:
Q > Ksp
so a ppt. will occur
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b. find the Eq. conc. of Pb2+ remaining in solution after the
PbCrO4 precipitates. Ksp(PbCrO4) = 1.8 x 10–14
Since [Pb2+] = 5.0 x 10-6 and [CrO42-] = 5.0 x 10-4 and there is a
1:1 stoichiometry, Pb2+ is the limiting reactant.
PbCrO4(s)  Pb2+ + CrO42–
0
5.0 x 10-4 - 5.0 x 10-6 = 5.0 x 10-4
I. (after ppt.)
+x
+x
C.
x
5.0 x 10–4 + x
E.
Ksp = [x][5.0 x 10–4 + x]
Try dropping the “+ x” term.
1.8 x 10–14 = x(5.0 x 10-4)
x = [Pb2+] = 3.6 x 10–11
This is the concentration of Pb2+ remaining in solution.
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Complex ion formation:
AgCl(s)  Ag+ + Cl– Ksp= 1.8 x 10–10
Ag+(aq) + NH3(aq)  Ag(NH3)+(aq)
Ag(NH3)+(aq) + NH3(aq)  Ag(NH3)2+(aq)
Ag+(aq) + 2NH3(aq)  {Ag(NH3)2}+(aq)
formation or stability constant:
Kf

Ag ( NH ) 

 1.7 x 10
Ag NH 

3 2

2
3
For Cu2+:
Cu2+ + 4NH3  [Cu(NH3)4]2+(aq)
K1 x K2 x K3 x K4 = Kf = 6.8 x 1012
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7
Solubility and complex ions:
Problem: How many moles of NH3 must be added to dissolve 0.050 mol of
AgCl in 1.0 L of H2O? (KspAgCl = 1.8 x 10–10 ; Kf[Ag(NH3)2]+ = 1.6 x 107)
AgCl(s)  Ag+ + Cl–
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
AgCl(s) + 2NH3  Ag(NH3)2+(aq) + Cl–
sum of RXNS:


Ag ( NH )  Cl 

 K

K overall

3 2
NH3 
2
–3
=
2.9
x
10
x
K
sp
f
Now use Koverall to solve the problem:
Koverall= 2.9 x
10–3
Ag(NH )  Cl   0.0500.050

=
3 2
NH3 
2

NH3  2
[NH3]eq = 0.93 but ..... How much NH3 must we add?
[NH3]total= 0.93 + (2 x 0.050) = 1.03 M
2 ammonia’s for each Ag+
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Fractional Precipitation: “ppting” one ion at a time. Compounds must have
different Ksp values (i.e. different solubilities)
Example: Ksp CdS = 3.9 x 10–29 and KspNiS = 3.0 x 10–21
? Which will ppt. first?
least soluble
Problem: A solution is 0.020 M in both Cd2+ and Ni2+. Just before NiS begins
to ppt., what conc. of Cd2+ will be left in solution?
Approach: Find conc. of S2– ion when Ni2+ just begins to ppt. since Cd2+
will already be ppting. Then use this S2– conc. to find Cd2+.
NiS(s)  Ni2+ + S2–
Ksp= 3.0 x 10–21 = [0.020][S2–]
[S2–] = 1.5 x 10–19 M when Ni2+ just begins to ppt.
So:
CdS(s)  Cd2+ + S2– Ksp= 3.9 x 10–29 = [Cd2+][1.5 x 10–19]
[Cd2+] = 2.6 x 10–10 M when NiS starts to ppt.
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