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.

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

3. Graph 1
 PbS Analysis Data
Weight PbS used 0.1003g
Digested PbS solution volume 100ml
Volumetric flask 250ml
Dilution factor 50
Pb AA result 17.266 mg/L
[Pb] 0.8633 mg/L
Weight% of Pb 86.1
Theoretical weight% Pb 86.6
Table 4
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0 5 10 15 20 25 30
Absorbance
[Pb]
Absorbancevs [Pb]

4.  Leaching Samples Analytical Data
Sample ID
[exp. #-time min)]
Diluction
factor
AA result [Pb]
(mg/L)
Undiluted [Pb]
(mg/L)
Corrected [Pb]
in samples (g/L)
1-0
25
0.033 0.825 0.000
1-5 3.637 90.925 0.090
1-15 5.486 137.15 0.136
1-25 8.399 209.975 0.209
1-35
50
6.54 327 0.326
1-45 8.477 423.85 0.423
1-60 10.717 535.85 0.535
1-80 13.278 663.9 0.663
Table 5
Sample ID
[exp. #-time min)]
Diluction
factor
AA result [Pb]
(mg/L)
Undiluted [Pb]
(mg/L)
Corrected [Pb]
in samples (g/L)
2-0
25
0.399 9.975 0.000
2-5 8.559 213.975 0.204
2-11 13.257 331.425 0.321
2-20
50
11.97 598.5 0.589
2-30 13.836 691.8 0.682
2-45 16.208 810.4 0.800
2-60 17.351 867.55 0.858
2-80 17.535 876.75 0.867
Table 6
Sample ID
[exp. #-time min)]
Diluction
factor
AA result [Pb]
(mg/L)
Undiluted [Pb]
(mg/L)
Corrected [Pb]
in samples (g/L)
3-0
25
0.805 20.125 0.000
3-3 24.269 606.725 0.587
3-6.5 19.84 496 0.476
3-10
50
15.413 770.65 0.751
3-15 17.266 863.3 0.843
3-25 17.721 886.05 0.866
3-35 17.817 890.85 0.871
3-45 17.758 887.9 0.868
Table 7

5. Sample ID
[exp. #-time min)]
Diluction
factor
AA result [Pb]
(mg/L)
Undiluted [Pb]
(mg/L)
Corrected [Pb]
in samples (g/L)
4-0
25
0 0 0.000
4-5.5 4.535 113.375 0.113
4-15 8.142 203.55 0.204
4-25 11.596 289.9 0.290
4-35
50
7.506 375.3 0.375
4-45 9.2 460 0.460
4-60 11.193 559.65 0.560
4-80 13.063 653.15 0.653
Table 8
 Mass of Pb Leached and α Values
Experiment 1
Time
(min)
Sample vol. (mL)
Undiluted [Pb]
(g/L)
Corrected
[Pb] (g/L)
Cumulative mass
Pb in samples (g)
0
7
3.30E-05 0.000 0.000000
5 3.64E-03 0.090 0.000631
15 5.49E-03 0.136 0.001585
25 8.40E-03 0.209 0.002418
35 6.54E-03 0.326 0.003747
45 8.48E-03 0.423 0.005244
60 1.07E-02 0.535 0.006706
80 1.33E-02 0.663 0.008387
Table 9
Experiment 2
Time
(min)
Sample vol. (mL)
Undiluted [Pb]
(g/L)
Corrected
[Pb] (g/L)
Cumulative mass
Pb in samples (g)
0
7
3.99E-04 0.000 0.000000
5 8.56E-03 0.204 0.001428
11 1.33E-02 0.321 0.003678
20 1.20E-02 0.589 0.006370
30 1.38E-02 0.682 0.008892
45 1.62E-02 0.800 0.010376
60 1.74E-02 0.858 0.011606
80 1.75E-02 0.867 0.012070
Table 10

6. Experiment 3
Time
(min)
Sample vol. (mL)
Undiluted [Pb]
(g/L)
Corrected [Pb]
(g/L)
Cumulative
mass Pb in
samples (g)
0
7
8.05E-04 0.000 0.000000
3 2.43E-02 0.587 0.004106
6.5 1.98E-02 0.476 0.007437
10 1.54E-02 0.751 0.008585
15 1.73E-02 0.843 0.011156
25 1.77E-02 0.866 0.011964
35 1.78E-02 0.871 0.012157
45 1.78E-02 0.868 0.012170
Table 11
Experiment 4
Time
(min)
Sample vol. (mL)
Undiluted [Pb]
(g/L)
Corrected [Pb]
(g/L)
Cumulative
mass Pb in
samples (g)
0
7
0.00E+00 0.000 0.000000
5.5 4.54E-03 0.113 0.000794
15 8.14E-03 0.204 0.002218
25 1.16E-02 0.290 0.003454
35 7.51E-03 0.375 0.004656
45 9.20E-03 0.460 0.005847
60 1.12E-02 0.560 0.007138
80 1.31E-02 0.653 0.008490
Table 12

7. Experiment 1
Time
(min)
Sample vol. (ml)
Cumulative mass
Pb in samples (g)
Leach vol.
remaining (ml)
Mass Pb in the
leach solution (g)
0
7
0.000000 993 0.0000
5 0.000631 986 0.0895
15 0.001585 979 0.1350
25 0.002418 972 0.2057
35 0.003747 965 0.3185
45 0.005244 958 0.4105
60 0.006706 951 0.5155
80 0.008387 944 0.6343
Table 13
Experiment 2
Time
(min)
Sample vol. (mL)
Cumulative mass
Pb in samples (g)
Leach vol.
remaining (ml)
Mass Pb in the
leach solution (g)
0
7
0.000000 993 0.0000
5 0.001428 986 0.2026
11 0.003678 979 0.3184
20 0.006370 972 0.5784
30 0.008892 965 0.6669
45 0.010376 958 0.7772
60 0.011606 951 0.8272
80 0.012070 944 0.8303
Table 14
Experiment 3
Time
(min)
Sample vol. (mL)
Cumulative mass
Pb in samples (g)
Leach vol.
remaining (ml)
Mass Pb in the
leach solution (g)
0
7
0.000000 993 0.0000
3 0.004106 986 0.5825
6.5 0.007437 979 0.4733
10 0.008585 972 0.7381
15 0.011156 965 0.8248
25 0.011964 958 0.8415
35 0.012157 951 0.8402
45 0.012170 944 0.8313
Table 15

8. Experiment 4
Time
(min)
Sample vol. (mL)
Cumulative mass
Pb in samples (g)
Leach vol.
remaining (ml)
Mass Pb in the
leach solution (g)
0
7
0.000000 993 0.0000
5.5 0.000794 986 0.1126
15 0.002218 979 0.2015
25 0.003454 972 0.2852
35 0.004656 965 0.3668
45 0.005847 958 0.4465
60 0.007138 951 0.5394
80 0.008490 944 0.6251
Table16
Experiment 1
Time
(min)
Total mass Pb
leached (g)
α 1-(1-α)1/3 1-2/3 α – (1-α)2/3
0 0.0000 0.0000 0 0.0000
5 0.0895 0.1092 0.037826113 0.0054
15 0.2245 0.1649 0.058295516 0.0144
25 0.4302 0.2512 0.091912365 0.0573
35 0.7487 0.3889 0.151392776 0.0828
45 1.1592 0.5012 0.206942471 0.1281
60 1.6748 0.6294 0.281727687 0.1565
80 2.3091 0.7745 0.39133154 0.1585
Mass PbS used (g) 1.01
Mass Pb added (g) 0.81902
Table 17
Experiment 2
Time
(min)
Total mass Pb
leached (g)
α 1-(1-α)1/3 1-2/3 α – (1-α)2/3
0 0.0000 0.0000 0 0.0000
5 0.2026 0.2104 0.07571907 0.0726
11 0.5209 0.3307 0.125256438 0.0433
20 1.0994 0.6007 0.263650047 0.1427
30 1.7662 0.6926 0.325102985 0.2135
45 2.5434 0.8072 0.422281921 0.2337
60 3.3706 0.8591 0.479625371 0.2320
80 4.2009 0.8624 0.483679262 0.2210
Mass PbS used (g) 0.98
Mass Pb added (g) 0.96283

9. Table 18
Experiment 3
Time
(min)
Total mass Pb
leached (g)
α 1-(1-α)1/3 1-2/3 α – (1-α)2/3
0 0.0000 0.0000 0 0.0000
3 0.5825 0.6592 0.301514377 0.0021
6.5 1.0558 0.5357 0.225638787 0.0070
10 1.7939 0.8353 0.451872826 0.0148
15 2.6187 0.9335 0.594791191 0.0259
25 3.4602 0.9524 0.637504283 0.0410
35 4.3005 0.9509 0.633799175 0.0653
45 5.1318 0.9409 0.610384899 0.0968
Mass PbS used (g) 1.05
Mass Pb added (g) 0.88361
Table 19
Experiment 4
Time
(min)
Total mass Pb
leached (g)
α 1-(1-α)1/3 1-2/3 α – (1-α)2/3
0 0.0000 0.0000 0 0
5.5 0.1126 0.1321 0.046117414 0.001741873
15 0.3141 0.2364 0.085970339 0.006278368
25 0.5993 0.3346 0.126978907 0.016484711
35 0.9661 0.4303 0.171020643 0.034844932
45 1.4127 0.5238 0.219111021 0.053493991
60 1.9520 0.6327 0.283865835 0.086688913
80 2.5771 0.7333 0.35628571 0.130044604
Mass PbS used (g) 1.02
Mass Pb added (g) 0.883176
Table 20

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

19. 𝑘 𝑠2 =
0.0070522∗ 0.00002
0.0000345
= 0.00408
Based on equation:
𝑡 =
[1 − (1 − 𝛼)1 3⁄
]
𝑘 𝑠2
For α = 0.95,
𝑡 =
[1 − (1 − 0.95)1 3⁄
]
0.00408
= 154.8 𝑚𝑖𝑛
For α = 0.99,
𝑡 =
[1 − (1 − 0.99)1 3⁄
]
0.00408
= 192.29 𝑚𝑖𝑛