1) A solution is 5 mM in each of the following ions: Indicate which of the metal

Question

1) 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 = 7.8 (correct)

2) 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-20

Explanation / Answer

1) A solution is 5 mM in each of the following ions:
number    ion    Ksp of M(OH)2; [H+]   Qsp
1    Mg2+    1.8e-11
2    Cd2+    2.5e-14
3    Co2+    1.6e-15
4    Zn2+    4.4e-17
5    Cu2+    2.2e-20

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.

A solution is 5 mM in each of the following ions
[M+2] = 5E-3 M

(i)
pH = 6.00;
[OH-] = 10^-(14-pH) = 1.0E-8 M;
M(OH)2(s) = M(+2)(aq)+ 2OH-(aq)

Ionic product of M(OH)2, Qsp =[M+2]*[OH-]^2
[M+2] = 5E-3 M
[OH-] = 1.0E-8 M;
Qsp = (5E-3)*(1E-8)^2 = 5.0E-19

Qsp(1,2,3,4) > Ksp hence they will precipitate.
pH = 6.00: 1,2,3,4

(ii)
pH = 10.00:
[OH-] = 10^-(14-pH) = 1.0E-4 M;
Qsp = (5E-3)*(1E-4)^2 = 5.0E-11

Qsp(1) > Ksp hence only 1 will precipitate.
pH = 10.00: 1

2)
Cu2+ + 4NH3 = [Cu(NH3)4]2+ (Kf = 4.8e+12)

([Cu(NH3)4]2+)/([Cu2+]*[NH3]^4) = Kf
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.
We assume all the Cu2+ has formed the complex and solve the dissociation equilibria of the complex, [Cu(NH3)4]2+
Dissociation constant, Kd = [Cu2+]*[NH3]^4/([Cu(NH3)4]2+) =1/Kf = 1/4.8E12 = 0.21E-12

Here Cu+2 is the limiting reagent and will be fully consumed to produce complex [Cu(NH3)4]2+

Total volume = 17/1000 L + 1.1 L = 1.117 L

Initial concentration:
[Cu(NH3)4]2+ = 17/1000*1.3/1.117 = 0.02M (assumed to complete formation of complex from Cu+2 and solve dissociation equilibria)
[Cu2+] = 0.02 - 0.02 = 0M
[NH3] = 1.1*0.46/1.117 - 4*0.02 = 0.373 M

Change due to dissociation of complex:
[Cu(NH3)4]2+ = -x M
[Cu2+] = x M
[NH3] = 4*x M

Equilibrium concentration:
[Cu(NH3)4]2+ = 0.02-x M
[Cu2+] = x M
[NH3] = (0.373+4*x) M

Kd = [Cu2+]*[NH3]^4/([Cu(NH3)4]2+) = x*(0.373+4*x)^4/(0.02-x) = 0.21E-12

x will be very small as Kd = 0.21E-12.
So, 0.373+4*x = 0.373;
0.02-x = 0.02

x*(4*x)^4/0.02 = 0.21E-12
x = 4.4E-4 M

a) What is the concentration of NH3 in the resulting solution?
[NH3] =   (0.373+4*x) = 0.375 M

b) What is the concentration of Cu(NH3)42+ in the resulting solution?
[Cu(NH3)42+] = 0.02-x = 0.0196 M

c) What is the concentration of Cu2+ in the resulting solution?
[Cu2+] = x   = 4.4E-4 M

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