The document summarizes the results of an experiment on leaching lead sulfide (PbS) particles using ferric chloride (FeCl3) as the lixiviant. The experiment involved leaching PbS at different times and temperatures, collecting leachate samples, and analyzing them to determine the amount of lead dissolved using atomic absorption spectroscopy. Graphs and tables show the results of lead concentration, mass of lead leached, and fractional lead dissolution over time for the different experimental conditions.
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Lab 5 Report harith edit 8.32pm
1. THE UNIVERSITY OF BRITISH COLUMBIA
DEPARTMENT OF MATERIALS ENGINEERING
MTRL 359
LABORATORY 5: LEACHING
MUHAMMAD HARITH MOHD FAUZI
18204115
2. Introduction
Leaching is one of the core processes in hydrometallurgy. The purpose is to dissolve
minerals of interest from an ore or concentrate in a suitable “lixiviant”. Once a solution of the
metal(s) of interest is obtained it may be purified and treated to recover pure metal.
Results and Data
PbS screen size: 200 -230micron mesh size
Particle size: 63-75 micrometers
Experimental Conditions
Experiment
Weight of
FeCl3.6H2O (g)
Leach solution
volume (ml)
[Fe+3]
(mg/L)
Temperature
PbS (oC)
Weight PbS
used (g)
1 21.9 1000 0.187 21 1.01
2 21.9 1000 0.187 30 0.98
3 21.9 1000 0.187 40 1.05
4 24.5 1000 0.130 21 1.02
Table 1
AA Calibration Data
[Pb] ppm Vol. flask mL
Vol. 100 ppm
Pb standard mL
Required HCl
g/L
Vol. 200 g/L
HCl mL
Blank 100.0 0 10 5
5 100.0 5 10 5
10 100.0 10 10 5
20 100.0 20 10 5
25 100.0 25 10 5
Table 2
Standard (mg/L) Absorbance
0 0.0004
5 0.0849
10 0.1687
20 0.3203
25 0.3941
Table 3
10. Discussion
1. (i) Plot the following
Leach solution [Pb] vs Time
Graph 2 Test 1, 2, and 3
0
200
400
600
800
1000
1200
0 10 20 30 40 50 60 70 80 90
[Pb](mg/L)
Time (min)
Leach solution [Pb] versus Time
Test 1
Test 2
Test 3
11. Graph 3 Test 1 and 4
o 1-(1-α)1/3 versus Time
0
100
200
300
400
500
600
700
800
0 10 20 30 40 50 60 70 80 90
[Pb](mg/L)
Time (min)
Leach solution [Pb] versus Time
Test 1
Test 4
12. Graph 4
1-(2/3)α – (1-α)2/3 versus Time
Graph 5 Test
y = 0.0048x - 0.0073
y = 0.0077x
y = 0.0193x
y = 0.0044x + 0.0157
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 10 20 30 40 50 60 70 80 90
1-(1-α)1/3
Time (min)
1-(1-α)1/3 versus time
Test 1
Test 2
Test 3
Test 4
Linear (Test 1)
Linear (Test 2)
Linear (Test 3)
Linear (Test 4)
y = 9E-09x4 - 2E-06x3 + 0.0002x2 - 0.0009x +
0.0019
0
0.05
0.1
0.15
0.2
0 20 40 60 80 100
1-(2/3)α-(1-α)2/3
Time (min)
1-(2/3)α - (1-α)2/3 versus
Time
Series1
Poly. (Series1)
13. Graph 6 Test 2
Graph 7 Test 3
Graph 8 Test 4
y = 5E-08x4 - 7E-06x3 + 0.0002x2 + 0.0038x +
0.0121
0
0.05
0.1
0.15
0.2
0.25
0.3
0 20 40 60 80 100
1-(2/3)α-(1-α)2/3
Time (min)
1-(2/3)α - (1-α)2/3 versus Time
Series1
Poly. (Series1)
y = 3E-08x4 - 3E-06x3 + 7E-05x2 + 0.001x -
0.0008
-0.05
0
0.05
0.1
0.15
0 10 20 30 40 50
1-(2/3)α-(1-α)2/3
Time (min)
1-(2/3)α - (1-α)2/3 versus Time
Series1
Poly. (Series1)
y = 1E-05x2 + 0.0005x - 0.0013
-0.02
0
0.02
0.04
0.06
0.08
0.1
0.12
0 20 40 60 80 100
1-(2/3)α-(1-α)2/3
Time (min)
1-(2/3)α - (1-α)2/3 versus Time
Series1
Poly. (Series1)
14. The alpha function should not be constrained to the origin.
(ii) Leaching completion
From the Pb analysis data, we can predict that the leaching of PbS did go to completion in
experiment 2 and 3. This is because the last 2-3 samples have the roughly the same [Pb] which
indicates that all the PbS has reacted before the end of the test.
However, the condition of monosized particles is quite false and the simple leaching model thus
fails. This affects the curve of the concentration versus time plot. This can lead to the wrong
extraction of information from the graph. A good model should produce a smooth curve that
close to the line of best fit for any types of equation of line.
Samples that should be omitted from the alpha function plots.
Experiment 2
Time (min) Undiluted [Pb] (mg/L)
45 810.4
60 867.55
80 876.75
Table 21
Experiment 3
Time (min) Undiluted [Pb] (mg/L)
25 886.05
35 890.85
45 887.9
Table 22
These are some of the points that should be omitted since the undiluted concentration are roughly
the same the same towards the end reaction. This indicates that all the PbS has fully reacted at
the end of the experiment.
Based on the leaching results, the function 1-(1-α)1/3 best fit the leaching results. This is because
the values from the function produce curves that are closer to linear line of best fit compares to
another function. Thus, the function 1-(1-α)1/3 is more reliable in showing the experimental data
obtained from the experiments.
15. 2. (i) Calculate rate constants.
The rate constant is equal to the slope of the linear best fit of the chosen function.
Graph 9
Summary of rate constant for each experiment.
Experiment Rate constant (slope), k
1 0.0044
2 0.0077
3 0.0193
4 0.0048
Table 23
y = 0.0048x - 0.0073
y = 0.0077x
y = 0.0193x
y = 0.0044x + 0.0157
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 10 20 30 40 50 60 70 80 90
1-(1-α)1/3
Time (min)
1-(1-α)1/3 versus time
Test 1
Test 2
Test 3
Test 4
Linear (Test 1)
Linear (Test 2)
Linear (Test 3)
Linear (Test 4)
16. (ii) Plot the natural logarithm of the rate constants versus 1/T.
Graph 10
Calculation of pre-exponential factor A
ln 𝑘 = ln 𝐴 −
𝐸𝑎
𝑅
(
1
𝑇
)
From this formula, ln A is the y-intercept. Thus to find the value for A, equate it with the
y- intercept from the line equation.
ln 𝐴 = 18.952
𝐴 = 𝑒18.952
𝐴 = 1.701 ∗ 108
min^(−1 )
Calculation of activation energy in kJ/mol.
In this case, the activation energy is equal to the slope of the line equation.
𝐸𝑎
𝑅
= 7195.8
𝐸𝑎 = 7195.8 ∗ 8.314
y = -7195.8x + 18.952
-6
-5
-4
-3
-2
-1
0
0.00315 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345
lnk
1/T
Plot of natural logarithm of K vs (1/T)
ln (k)
Linear (ln (k))
17. = 59825.88
𝐽
𝑚𝑜𝑙
= 59.826
𝑘𝐽
𝑚𝑜𝑙
(iii) The activation energy is consistent with the chosen leaching model. The activation
energy of chemical reaction is between the range of 42 -105 kJ/mole. Our activation
energy falls within this range.
3. (i) Proving.
𝑘 𝑠 =
𝑉𝑚 𝑘′ 𝑓(𝐶 𝐵)
𝑟𝑜
Since we are assuming that Vm and CB are constants, we eliminate these from the equation and
get:
𝑘 𝑠 =
𝑘′
𝑟𝑜
Rate constant for a PbS particle of initial size ro1:
𝑘 𝑠1 =
𝑘′
𝑟𝑜1
𝑘′
= 𝑘 𝑠1 ∗ 𝑟𝑜1
Rate constant for a PbS particle of initial size ro2:
𝑘 𝑠2 =
𝑘′
𝑟𝑜2
Therefore, substitute for k’ and get
𝑘 𝑠2 =
𝑘 𝑠1 ∗ 𝑟𝑜1
𝑟𝑜2
( 𝑠ℎ𝑜𝑤𝑛)
18. (ii) Value of ks at 32 degree Celcius.
(ii) Value of ks at 32 degree Celsius.
Using the equation [27] and our own data values,
𝑘 𝑠 = 𝐴𝑒−𝐸𝑎/𝑅𝑇
𝑘 𝑠 = (1.701 ∗ 108
)𝑒
(
−59825.884
8.314∗(32+273)
)
𝑘 𝑠 = 0.009649
(iii) Time taken
From equation 21:
𝑡 =
[1 − (1 − 𝛼)1 3⁄
] ∗ 𝑟𝑜
𝑉𝑚 𝑘′ 𝑓(𝐶 𝐵
𝑜)
=
[1 − (1 − 𝛼)1 3⁄
]
𝑘 𝑠2
Substituting equation from part i),
𝑘 𝑠2 =
𝑘 𝑠1 ∗ 𝑟𝑜1
𝑟𝑜2
𝑘 𝑠1 = 𝐴𝑒−𝐸𝑎/𝑅𝑇
𝑘 𝑠1 = (1.701 ∗ 108) 𝑒−(59825 .884 )/(8.314∗(28+273))
𝑘 𝑠1 = 0.0070522
To find ro1
𝑟𝑜1 =
0.00004
2
= 0.00002𝑚
And ro2,
𝑟𝑜2 =
0.000063 + 0.000075
4
= 0.0000345𝑚
Coming back to the other equation,
𝑘 𝑠2 =
𝑘 𝑠1 ∗ 𝑟𝑜1
𝑟𝑜2