Will Mn(OH)2 precipitate from a 0.01 M solution of MnCl2 at pH = 9? Ksp (Mn(OH)2) = 1,0 x 10-13. Discuss from solubility diagrams and prove by calculation. calculation. Co is a metal that is present in every lithium battery to stabilize its charge and to increase stability. Every cell phone, tablet and electric car relies on the availability of cobalt, 97% of which is extracted from mines in places like the Congo, where essentially slave labor is used. Cobalt ions form complexes with e.g. CN- . You see solubility diagrams and fraction diagrams for CoCO3 with and without the addition of 10 mM CN- . . Explain whythe solubility of cobalt carbonate increases at high pH when cyanide ions are present. Write reaction formulas for the formation of the dominant cobalt cyanide complexes.

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter15: Complex Ion And Precipitation Equilibria
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Problem 70QAP
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Will Mn(OH)2 precipitate from a 0.01 M solution of MnCl2 at pH = 9? Ksp
(Mn(OH)2) = 1,0 x 10-13. Discuss from solubility diagrams and prove by calculation.
calculation.

 Co is a metal that is present in every lithium battery to stabilize its charge and to
increase stability. Every cell phone, tablet and electric car relies on the availability of
cobalt, 97% of which is extracted from mines in places like the Congo, where essentially slave labor is used.
Cobalt ions form complexes with e.g. CN- .
 You see solubility diagrams and fraction diagrams for CoCO3 with and without the addition of 10 mM CN- .
. Explain whythe solubility of cobalt carbonate increases at high pH when cyanide ions are present.
Write reaction formulas for the formation of the dominant cobalt cyanide complexes.

 

t= 25°C
I= varied
[CN TOT = 0.00
[CO32-]TOT = 0.00
COCO3(s)
Log Solubl.
-2
Log Solubl.
+
-6
-8
t= 25°C
= 10.00 mM
[CN TOT
[CO32-]TOT = 0.00
COCO3(s)
0
-2
-4
-6
0
-8
2
0
NI
2
4
4
6
pH
6
PH
Log {COCO3(s)} = 0.00
[NH3] TOT 0.00
CO3²-
8
10
8
Log (COCO3(s)} = 0.00
[NH3]TOT = 0.00
10
12
CO3²-
CN-
12
t=25°C
I= varied
0.00
[CN-]TOT =
[CO32-]TOT = 0.00
Co²+
1.0
Fraction
0.8
Fraction
0.6
0.4
0.2
0.0
t= 25°C
I= varied
[CN TOT
[CO32-]TOT 0.00
1.0
0.8
0.6
0.4
0.2
0
0.0
= 10.00 mM
2
-
Co²+
0
2
4
6
PH
Log (CoCO3(s)} = 0.00
[NH3]TOT = 0.00
6
PH
COOH+
8
Cd(CN) 3
10
8
Co(OH)2 (cr)
Log (CoCO3(s)}= 0.00
[NH3]TOT 0.00
12
10
Ço(OH)2(cr)
Co(CN)5³
12
Transcribed Image Text:t= 25°C I= varied [CN TOT = 0.00 [CO32-]TOT = 0.00 COCO3(s) Log Solubl. -2 Log Solubl. + -6 -8 t= 25°C = 10.00 mM [CN TOT [CO32-]TOT = 0.00 COCO3(s) 0 -2 -4 -6 0 -8 2 0 NI 2 4 4 6 pH 6 PH Log {COCO3(s)} = 0.00 [NH3] TOT 0.00 CO3²- 8 10 8 Log (COCO3(s)} = 0.00 [NH3]TOT = 0.00 10 12 CO3²- CN- 12 t=25°C I= varied 0.00 [CN-]TOT = [CO32-]TOT = 0.00 Co²+ 1.0 Fraction 0.8 Fraction 0.6 0.4 0.2 0.0 t= 25°C I= varied [CN TOT [CO32-]TOT 0.00 1.0 0.8 0.6 0.4 0.2 0 0.0 = 10.00 mM 2 - Co²+ 0 2 4 6 PH Log (CoCO3(s)} = 0.00 [NH3]TOT = 0.00 6 PH COOH+ 8 Cd(CN) 3 10 8 Co(OH)2 (cr) Log (CoCO3(s)}= 0.00 [NH3]TOT 0.00 12 10 Ço(OH)2(cr) Co(CN)5³ 12
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