MTRL 358 - Major Assign #1 - SOLUTIONS - Winter 2023 T1
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Course
358
Subject
Material Science
Date
Jan 9, 2024
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Assignment 1, 2023 - Chemistry review and Hydrometallurgical Chemistry
1. In any process chemical industry, including hydrometallurgy, it is important to gain an understanding
of the periodic table. This question begins to address that. Match the compound or element on the right
to the definition or term on the left. There will be one unique compound/element per term or definition.
By process of elimination determine the most suitable match for each phrase.
Strong acid
H
2
SO
4
The most electronegative metal
Au
Main group metal
Pb
Strong oxidant
Cl
2
Most important industrial base
CaO
Strong base
NaOH
Strong reducing agent
Na
Transition metal
Ti
Strong electrolyte
NaCl
Strongly sulphophilic metal
Fe
Amphoteric ion
HSO
3
-
Weak acid
H
2
S
Semi-metal
As
Incorrect chemical formula
NaCO
3
2. Fill in the blanks:
(a) In an electron transfer reaction when an oxidizing agent reacts it gets
reduced
.
(b) Reduction is
gain
of electrons.
(c) A reducing agent
loses
electrons.
(d) What is the term we use for -log
10
[H
+
]?
pH
(e) A salt is a compound composed of
ions
.
3. Select the most comprehensive and correct definition for a base in aqueous solution. A base is a
compound that
(a) has an ionizable OH
-
group and raises the pH.
(b) has an ionizable OH
-
group and lowers the pH.
Faculty of Applied Sciences
Department of Materials Engineering
MTRL 358: Hydrometallurgy I
Assignment #1 – SOLUTIONS
(c) accepts a proton and raises the pH
.
(d) accepts a proton and lowers the pH.
4. Which of the following compounds
(a) will act as an acid in water:
CH
4
,
CO
2
, HCO
3
-
, HNO
3
, NH
3
(b) will acts as a base in water:
HCO
3
-
, SO
4
2-
, NaCH
3
CO
2
, N
2
, C
6
H
5
OH (phenol)
5. Answer the following questions:
(a) Why is the chemical formula Mg
2
PO
4
incorrect?
PO
4
3-
has a 3- charge; Mg
+2
has a +2 charge.
To form a balanced compound the simplest formula must be
Mg
3
(PO
4
)
2
.
(b) What is the formula for the highest possible oxide of titanium?
Ti is a group IVa element. It can form up to 4 electron-pair bonds. Oxygen is a group VIb element. It can
form 2 electron pair bonds by gaining two electrons. Hence the highest oxide of Ti is TiO2. The 2 O atoms
each gain 2 electrons from Ti (formally) and the Ti shares (loses, formally) 4 electrons.
(c) What is the value of n in the formula: H
3
AsO
n
where As forms the maximum number of electron pair
bonds possible? (The H atoms are bound to oxygens.) Briefly explain your answer.
As is a group Vb element. It will form 5 electron pair bonds. Each O forms two electron pair bonds. Since
there are three H atoms are bound to O atoms, the groups are –OH where one electron pair bond still
can form between O and As. This gives As-(OH)
3
. But this would leave As as only trivalent. Hence one
more O atom forms two electron pair bonds with As, i.e. (HO)
3
As=O. Each O is divalent; the As is
pentavalent. The formula is H
3
AsO
4
.
6. Indicate which of the following are soluble salts in water:
HCl, Ag
2
S,
Na
2
S
, C
12
H
22
O
11
(sugar), S
8
, NH
3
, CuO, PbCl
2
, CaO, CO
2
, H
2
, Te
Only Na
2
S. Note that HCl, C
12
H
22
O
11
and NH
3
are all soluble, but, not salts. S
8
, H
2
, CO
2
and Te are not salts
and not soluble. Ag
2
S, CuO, PbCl
2
and CaO are insoluble salts.
7. Perform the required calculations for the following aqueous solutions:
(a) Calculate the pH of a solution made up of 5.00 mL of 1.2 M HCl diluted to precisely 250 mL.
5 x 1.2 M H
+
/250 = 0.024 M H
+
(HCl is a strong acid; fully dissociates.)
pH = -log
10
(0.024) =
1.62
(b) A solution contains a total ammonia concentration of 0.8 M (the sum of ammonia and ammonium
concentrations). The volume is 1.5 L and the pH is 9.6. What volume (mL) of aqueous ammonia (density =
0.90 g/mL and 28% by mass, i.e. 280 g NH
3
/kg of solution) and mass of ammonium sulfate (in g) were
added to make up the solution? Use K
b
for ammonia = 1.8 x 10
-5
.
[NH
3
] + [NH
4
+
] = 0.8
pH = pK
a
+ log([NH
3
]/[NH
4
+
]) = 9.6
K
a
= 1 x 10
-14
/1.8 x 10
-5
= 5.556 x 10
-10
pK
a
= 9.2553
9.6 = 9.2553 + log([NH
3
]/[NH
4
+
])
10
0.3447
= 2.212 = [NH
3
]/[NH
4
+
]
[NH
3
] = 2.212[NH
4
+
]
2.212[NH
4
+
] + [NH
4
+
] = 0.8 = 3.212[NH
4+
]
[NH
4
+
] = 0.2491 M
[NH
3
] = 0.5509 M
moles NH
4
+
= 1.5 L x 0.2491 mol/L =
0.37365 mol
There are two moles of NH
4
+
per mole (NH
4
)
2
SO
4
, so we need 0.37365/2 mol
(NH
4
)
2
SO
4
, or 132.136 g/mol x 0.37365/2 =
24.69 g
.
1L of aqueous ammonia weighs 900 g. 28% of this is NH
3
, i.e. 252 g NH
3
.
252 g NH
3
/17.031 g NH
3
/mol NH
3
= 14.797 mol/L
The volume needed is 1.5 L solution x 0.5509 mol NH
3
/L/14.797 mol NH
3
/L
= 0.0558 L =
55.8 mL
(c) An acetic acid solution has a pH of 2.78. What concentration of the acetic acid (in molar) was added to
make up the solution? Take K
a
for acetic acid to be 1.75 x 10
-5
. Acetic acid has the formula CH
3
CO
2
H.
HA
=
H
+
+ A
-
K
a
Co – x
x
x M
(Let x = [H
+
] = [A
-
] at equilibrium)
x
2
/(Co - x) = K
a
x
2
/K
a
+ x = Co
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