Download Geochemistry, Practical - Chemistry and more Study notes Chemistry in PDF only on Docsity!
Geochemistry Problem Set 1
Remember in all these problems to use the “unit-factor” method!
Force and Pressure
Force ( F ) is an influence on a mass m to cause it to accelerate ( a ). This is expressed
as Newton’s 2
nd
Law:
F = m a
The unit of force is the Newton (N) and we have that 1 N = kg m/s
2
. We need this in
geochemistry because the force (F) on an object due to having a mass m of overlying
rock or water is F=mg where g is the acceleration due to gravity (9.8 m/s
2
Pressure (P) is force per unit area (A); hence,
P = mg/A = ρgz
Where ρ is density of the overlying rock or water and z is the depth. The unit of
pressure is the Pascal (1 Pa =1 N/m
2
) although geologists like to express pressure in
bars (1 bar is about 1 atmosphere and 1 bar = 10
5
Pa ). The pressure due to overlying
rock is called lithostatic while that due to overlying water (connected to the surface)
is hydrostatic pressure. As we will learn, pressure causes transformations between
minerals.
- The mineral forsterite (Mg 2
SiO 4
) has a unit cell volume of 289.7 Å
3
. There are 4
formula units per unit cell. Adding up the atomic masses, we find that forsterite has a
mass of 140.7 g/mole. What is the density (in g/cm
3
) of forsterite?
Solution:
1 unitcell
289 .7 Å
3
4 fu
unitcell
- 7 g
mole
1 mole
6.023 x 10
23
fu
24
Å
3
cm
3
- Assuming that density of rock is constant at 2.7 g/cm
3
, what is the lithostatic
pressure (in bars) at a depth of 10 km?
Solution:
10 km
- 7 g
cm
3
1 kg
1000 g
9.8 m
s
2
1000 m
km
3
cm
3
m
3
1 Ns
2
kg m
1 Pa m
2
N
Work:
Work is the transfer of energy to a system via a force. In one dimension, the work
done on a mass by moving it with a constant force F over a distance Δ x is simply
F Δ x. At constant external pressure ( P ), the work done on a system by compression is
- P Δ V where Δ V is the change in volume of the system (note the – sign). What if the
pressure is not constant during the expansion/contraction? Then we need to use
calculus to evaluate the work. The infinitesimal amount of work dW done when we
compress the system by an infinitesimal volume change dV is:
dW = - P dV
The total work W is just the sum of all the infinitesimal - PdV terms. This is
evaluated using a definite integral:
W = − PdV
V 1
V 2
We can evaluate this integral if we know how P changes with V. If we are reversibly
compressing against an ideal gas (i.e., the pressure exerted by the gas is the same as
the pressure applied), then we can write
P =
nRT
V
where n is the number of moles of gas in the container, R is the gas constant (8.
J/mol/K, T is the temperature in Kelvin and V is the volume. This is the ideal gas
law; it is a fundamental reference for the description of all matter. Our integral
becomes:
W = − PdV
V 1
V 2
nRT
V
V 1
V 2
dV = − nRT
dV
V
V 1
V 2
The last integral is very easy and is one of the standard integrals you need to have
memorized (see the notes I gave you).
Problems:
3 ) How much reversible work (in joules) is done on a container with 1 mole of an
ideal gas when compressing it from 10 0 to 10 cm
3 at a constant pressure of 10 bar?
Solution:
Since the pressure is constant, the integral is trivial and we get W = - PΔV =
5
Pa/bar x(- 90 cm
3
) x 1m
3
3
cm
3
x 1Nm
2
/Pa x 1 J/Nm
W = − PdV
V 1
V 2
Don’t do this any more. From now on, use the unit-factor method instead:
2.3 g NaCl
0.2 litres
1.0 moles NaCl
58.4 g NaCl
0.197 moles NaCl
litre
You might have been taught to use dm
3
(cubic decimetres) in calculations for volume.
Don’t do this; scientists use litres (= 1 dm
3
) or ml (= 1 cm
3
) for liquid volumes.
Sometimes you’ll see volumes in m
3
. Solid volumes are often in cm
3
In chemistry, concentrations in aqueous solutions are usually given in moles/litre of
solution. In geochemistry, however, liquid volumes are often poorly defined since
they change so greatly with pressure and temperature. Moreover, we often deal with
very concentrated solutions (hydrothermal ore-forming fluids can be 10-20 moles
NaCl/l) so that the amount of water in a litre is unclear. To cope with this, we use a
concentration unit called molality which is the number of moles of a solute per 1 kg
of water.
A very common unit in geochemistry is parts per million (ppm). These are
common when talking about the concentration of trace metals in rocks. In aqueous
geochemistry, these units are common when communicating to regulatory agencies.
When discussing trace metals in seawater, we might use parts per billion (ppb) or
even parts per trillion (ppt). These are somewhat awkward units to use until you
recognize that 1 ppm = 1 mg/kg. For example, a mine waste-water contains 1.0 x 10
m As. What is the concentration in ppm (mg/kg) given the atomic mass of As is 74.
g/mole?
1.0 × 10
mole As
kg H 2
O
74.9 g As
mole As
1000 mg
g
0.749 mg As
kg H 2
O
or 0. 749 ppm As
Now try this: A solution contains 23.0 ppm Cr. What is the concentration in moles
Cr/kg water given that there are 52.0 g Cr/mole? Here is the way do to it:
- 0 mg Cr
kg H 2
O
1 g
1000 mg
1 mole Cr
52.0 g Cr
4.42 x 10
moles Cr
kg H 2
O
or 4. 4 x 10
− 4
m Cr
This explicit writing-out of the units might seem to be a needless tedium to you;
however, it will save you much grief down the road and will enable you to solve
problems that you couldn’t do otherwise. Cr and As are carcinogenic. Getting your
calculations wrong may land you in court!
For the following, you need the atomic masses: (Na =22.99, K = 39.10, O=16.00,
N=14.01, C= 12.01, Cd =112.4, Cl =35.45, Pb = 207.2)
8 ) 15 grams of NaCl are added to a beaker. It is dissolved in 200 ml of water and
diluted up to 250 ml. What is the Na
concentration in moles/liter in the final
solution?
15 g NaCl
200 ml
×
200 ml
250 ml
×
1000 ml
litre
×
1 mole NaCl
- 4 g NaCl
×
1 mole Na
mole NaCl
= 1 .03 moles Na/litre
9 ) 3 grams of KNO 3
is dissolved in 50 liters of water. What is the (NO 3
concentration in ppm? 1 ppm is a mg/kg.
3.0 g KNO 3
50 litres
×
1 mol NO 3
101.1 g KNO 3
×
1 litre
kg
×
62 g NO 3
1 mole NO 3
×
1000 mg
1 g
×
1 ppm
mg/kg
= 36.8 ppm
10 ) 5.3 grams of Na 2
CO
3
are dissolved and diluted up to 350 ml. What is the final
concentration of Na in moles/liter?
5.3 g Na 2
CO
3
350 ml
×
1000 ml
litre
×
2 mole NaCl
mole Na 2
CO
3
×
1 mole Na 2
CO
3
106 g Na 2
CO
3
= 0.286 moles Na/litre
11 ) How much Pb(NO 3
2
is needed to make up 450 ml of a 0.1 ppm Pb solution?
450 ml ×
1 kgH 2
O
1000 ml
×
0.1mg Pb
kg
×
1 g
1000 mg
×
- 2 g Pb(NO 3
2
207.2 g Pb
= 7. 19 x 10
− 5
gPb(NO 3
2
- 1 gram of soil was added to a centrifuge tube; to this was added 25 ml of 10%
HNO3 which dissolved all of the lead in the soil. The concentration of Pb in the
acidic solution was 250 ppm. What was the concentration of Pb in the soil?
250 mg Pb
kg soln
×
1kg soln
1000 mlsoln
×
25 ml soln
1.0 g soil
×
1000 gsoil
kg soil
mgPb
kgsoil
or 6250 ppm
