1) A solution is 0.020 M in each Ca 2+ and Cd 2+ . Adjusting the pH of the solut
ID: 911108 • Letter: 1
Question
1) A solution is 0.020 M in each Ca2+ and Cd2+. Adjusting the pH of the solution to which of the following values would achieve the best separation by precipitation of the hydroxides?
Best pH = 12 (this is correct)
What would be the resulting concentrations of the ions if the solution were adjusted to this pH?
[Ca2+] = ___ M
[Cd2+] = ____ M
2) A solution is 5 mM in each of the following ions:
Indicate which of the metal ions would precipitate (or start to precipitate) at each of the following pH values. Indicate your answer with the number of the ion. Use 0 to indicate no precipitate. If more than one precipitate is expected, list the numbers in increasing order and separate them with commas. For example, 3,4,5 is ok but 5,4,3 is not.
pH = 6.00: _____________
pH = 10.00: _____________
What is the pH to the nearest 0.1 pH unit at which Co(OH)2 begins to precipitate?
pH = ____________
3) Consider the following equilibrium:
A solution is made by mixing 17.0 mL of 1.30 M CuSO4 and 1.10 L of 0.460 M NH3. Assume additive volumes to answer the following questions. Give all answers to three sig figs.
a) What is the concentration of NH3 in the resulting solution?
[NH3] = _________ M
b) What is the concentration of Cu(NH3)42+ in the resulting solution?
[Cu(NH3)42+] = _________ M
c) What is the concentration of Cu2+ in the resulting solution?
[Cu2+] = _________ M
number ion Ksp of M(OH)2 1 Mg2+ 1.8e-11 2 Cd2+ 2.5e-14 3 Co2+ 1.6e-15 4 Zn2+ 4.4e-17 5 Cu2+ 2.2e-20Explanation / Answer
1) pH=12
pOH=14-12=2
pOH=-log[OH]
or, [OH-]=10^-pOH=10^-2=0.01 M
[OH-]=0.01 M
As Ca(OH)2Ca2+ + 2OH-
Cd(OH)2Cd2+ + 2OH-
So [Ca2+]=[Cd2+]=1/2 [OH]=1/2*0.01M=0.005
[Ca2+]=[Cd2+]=0.005M
2)As all the salts ionizes in the same way giving equal number of cations and anions in solution, we can simply compare the Ksp to find out which is more soluble.
Mg(OH)2Mg2+ +2OH-
Cd(OH)2Cd2+ +2OH-
Co(OH)2Co2+ +2OH-
Zn(OH)2Zn2+ + 2OH-
Cu(OH)2Cu2+ + 2OH-
The order of Ksp is 1>2>3>4>5 so is their solubility order, Mg(OH)2 being the most soluble and Cu(OH)2 being the least at both ph=6.00 and pH=10.00
At pH=6.00
pOH=14-6.00=8
[OH-]=10^-pOH (see part A)
[OH-]=10^-8=10^-8M
1)Ksp=[Mg2+][OH-]^2
Let [Mg2+]=S ,[OH-]=2S
In the presence of [OH-]=10^-8M
[Mg2+]=S’ ,[OH-]=2S’+10^-8M
1.8*10^-11=(S’) (2S’+10^-8M)^2
Or, 1.8*10^-11=(S’) (10^-8M)^2 [2S’<<10^-8]
Or,S’=(1.8*10^-11)/ (10^-8M)^2=1.8*10^-3M=[Mg2+]
1.8*10^5M=[Mg2+]
Similarly ,calculating solubility of other metal hydroxides,
2)Ksp=[Cd2+][OH-]^2
2.5*10^-14=(S’) (10^-8M)^2
S’=[Cd2+]=2.5*10^2M
3) Ksp=[Co2+][OH-]^2
1.6*10^-15=(S’) (10^-8M)^2
S’=[Cd2+]=2.5*10M
4) Ksp=[Zn2+][OH-]^2
4.4*10^-17=(S’) (10^-8M)^2
S’=[Zn2+]=4.4*10^-1M
5) Ksp=[Cu2+][OH-]^2
2.2*10^-20=(S’) (10^-8M)^2
S’=[Cu2+]=2.2*10^--4M
So [Cu2+]<[Zn2+]<[Co2+]<[Cd2+]<[Mg2+] order of solubility
So they precipitate in reverse order, 1<2<3<4<5
At pH=10.00,
Also similar calculations would give same results,
3)
Moles of Cu2+=17.0 ml*1.30M=0.017 L*1.30 mol/L=0.0221 moles
Moles of NH3=1.10L*0.460 mol/L=0.506 mol
Prepare ICE table,
[Cu2+]
[NH3]
[Cu(NH3)4]2-
initial
0.0221
0.506
0
After complete rxn
0
0.506-4*0.0221=0.418
0.0221
change
+x
+4x
-x
equilibrium
x
0.418+4x
0.0221-x
Kf=[Cu(NH3)4]2-/[Cu2+][NH3]^4
Or, Kf=(0.0221-x)/(0.418+4x)^4 (x)
Or,kf=(0.0221)/(0.418)^4 (x)
4.8*10^12=(0.0221)/(0.418)^4 (x)
X=(0.0221)/(0.418)^4 (4.8*10^12)=0.15*10^-12 moles
X=[Cu2+]=0.15*10^-12 moles
a)[NH3]= 0.418+4x=0.418moles (approx) [4x<<<0.418]
[NH3]=0.418moles/(0.017L+1.10L)=0.418 moles/(1.117L)=0.020M
b) [Cu(NH3)4]2-=0.0221-x=0.0221 moles
[Cu(NH3)4]2-=0.0221 moles/(0.017L+1.10L)= 0.0221 moles/(1.117L)=0.020M
c)[Cu2+]=0.15*10^-12 moles/total volume=0.15*10^-12 moles/(0.017L+1.10L)
= 0.15*10^-12 moles/1.117L=0.13*10^-12M
[Cu2+]
[NH3]
[Cu(NH3)4]2-
initial
0.0221
0.506
0
After complete rxn
0
0.506-4*0.0221=0.418
0.0221
change
+x
+4x
-x
equilibrium
x
0.418+4x
0.0221-x
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