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Geochemistry: Supplemental Notes 1
Essential Maths for Geochemistry
1.0 Elementary Problem Solving
When faced with the kinds of problems you find in geochemistry (or anything else for that matter..), the first thing you should do is figure out what is actually happening. It often helps to make a sketch of the problem or write out the relevant chemical reactions.
The second thing you should do is work out the relevant equations based on the physical principles you know which are important for the problem. For example, you might write out an equation that expresses conservation of mass or charge. You might also write out the equilibrium expressions for the relevant reactions.
Once you’ve written out the fundamental equations, do your work with algebraic equations as long as possible; don’t plug in numbers until the last minute.
Most importantly, when you start plugging in numbers to do the calculations, it is vital that you include the unit associated with each number and treat the units like they are algebraic quantities. This approach is formalized in the unit-factor method and is an extremely important technique you need to master and use.
Finally, think about your answer. Does it make sense?
2.0 The “Unit-Factor” method
This is something I am totally fanatical about. It is really the most valuable skill I can teach you. Sadly, it is not found in either the British GCSE or A-Level syllabi for chemistry. If you learn this, you will have a huge competitive advantage over your peers and can apply it to whatever work you do after university. It is the way scientists do calculations and it is also a useful way to do calculations in everyday problems you might encounter as an event coordinator, construction estimator or wherever your geology degree takes you. In North America, the unit-factor method is usually called “dimensional analysis”. If you look on the web under dimensional analysis you’ll find some useful tutorials. However, here in the UK, dimensional analysis usually refers to something similar in spirit but not to the actual technique we will be using. The term “dimensional analysis” has also been hijacked by business management consultants to give a cool-sounding name to one of their trivial theories. Hence, we’ll call it the unit-factor method.
In the unit-factor method, we (1) always write out the units of every quantity and (2) treat the words denoting the units just like they are algebraic variables. For example, let’s calculate how many centimetres there are in 1 foot given that there are 2.54 cm per inch and 12 inches per foot. We set up the calculation like this:
( 1 foot) 12 inches foot
- 54 cm inch
=^?
Think of what we are doing here: each one of the quotients (the unit factors) is just another way of saying “one” since the numerator and denominator of each quotient is the same thing. Hence,
Geochemistry: Supplemental Notes 2
we are just multiplying “1 foot” by 1. If we treat the unit words (“foot”, “inches” etc.) like they are algebraic quantities (they are..), then “foot” divided by “foot” is 1 as is “inch” divided by “inch”. This leaves cm as the only surviving unit; hence, we get
( 1 foot)
12 inches foot
- 54 cm inch
=^30 .48^ cm
Now, supposing we screwed-up and instead of multiplying by 2.54 cm/inch we multiplied by 1 inch/2.54 cm? Our answer would come out to be 4.72 inch^2 cm. Since inch^2 cm is a silly unit, we must have done something wrong. That is the beauty of this method: if you got the units right, you must have set up the calculation correctly.
Example 1 : Petrol in the US costs $3.00 per gallon. What is that in £ per litre? There are 3. litres per gallon and $1.54 per £1. The way to start this problem is to decide which statement is unique to the problem. In this case, it’s $3.00 per gallon. “Per” means division. Then we multiply that by a series of quotients to gradually transform the units into something we want:
gallon
1 gallon 3.74 litres
=^ £0.52^ /^ litre
Example 2 : I need to make 500 ml of a 0.1 molar solution of NaCl. Given that there is 58.5 g NaCl per mole of NaCl, how much NaCl should I weigh out? Solution:
(^500 ml)
0.1mole NaCl 1 Litre
1 Litre 1000 ml
- 5 g NaCl mole NaCl
=^2.^93 g^ NaCl
Example 3 : Here is one you could easily mess up without dimensional analysis: I need to make 300 ml of a 0.1 m solution of Na. How many grams of Na 2 CO 3 should I weigh out given that there are 106 g of Na 2 CO 3 per mole of Na 2 CO 3?
(^ 300ml )
0.1 mol Na 1 litre
1 Litre 1000 ml
106 g Na 2 CO 3 mole Na 2 CO 3
1mole Na 2 CO 3 2 moles Na
=^ 6.36^ g^ Na 2 CO 3
Example 4 : Environmental regulatory agencies like to express concentrations in mg/kg (i.e., ppm or mg per kg of solution). However, when we do thermodynamic calculations, we need to know concentration in moles/kg. What is the molarity of As in a solution that has 0.001 mg As/kg?
0.001mg As kg
1 g 1000 mg
1 mole As 7 4. 92 g A s
=^ 1.3^ x^10 − (^8) moles As/kg
Geochemistry: Supplemental Notes 4
j) The work done by compressing something at constant pressure is PΔV where ΔV is the change in volume. What is the work (in Joules) in compressing a piston by 1 cm^3 at a constant pressure of 1 bar (10^5 Pa).
k) A wall is 5 m^2 in area. I wish to tile it with tiles that are each 10 cm by 10 cm. Each box contains 25 tiles. How many boxes do I need?
l) While tiling the above bathroom wall, I intend to apply a 3 mm thick layer of adhesive backing. How many gallons of adhesive do I need? There are 3.740 litres/gallon and 1000 cm^3 /litre.
m) You plan to drive from Calais to Monaco (900 km). How much will this cost in £? Your car gets 30 miles to the gallon. Petrol in France is 1.41 euros/litre and there are 4.54 litres/imperial gallon. There are 1.25 euros to the pound and 1.61 km/mile.
n) Light travels at 186,000 miles/sec. The sun is 93 x 10^6 miles from Earth. How long, in minutes, does it take for light to travel from the sun to the Earth.
o) Power is the rate at which energy is generated or consumed; it is measured in Watts (1 W= 1J/s). How many kJ are in 1 kWh (kilo-watt-hour)?
p) The UK receives 1000 kWh/m^2 /yr from the sun. It we could tap this with 10% efficiency, what fraction of Britain must be covered with a giant solar cell to meet the energy needs of the UK given that each person uses 58 x10^6 J/day and there are 60 x 10^6 people in the UK. The area of Britain is 209331 km^2.
q) How many atoms of Fe are in 3.2 g Fe given that 1 mole of Fe is 55.85 g and there are 6.023 x 1023 atoms/mole.
r) What is the approximate mass of the earth’s core if its radius is 3400 km and its average density is 11.0 g/cm^3. Recall that the volume of a sphere is 4/3πr^3.
s) Stars form in cold molecular clouds of interstellar gas which have a typical density of 10^6 atoms/cm^3. If the sun has a mass of 2 x 10^30 kg, and is 75% H (mass of 1 g/mole), what was the volume (in cubic light years) of the of the interstellar cloud needed to make the sun (for simplicity, ignore mass loss due to nucleosynthesis etc.).
3.0 Numbers and Calculations
3.1 Numbers and Significant Figures
Every measured quantity has an uncertainty. The precision of a measured quantity is indicated by the number of significant figures. The rules for significant figures are that
•Leading zeros are never significant. Hence 0.002 has only one sig fig.
•Trailing zeros only significant if the decimal point is indicated. Hence, 0.020 has two sig figs but 0.02 has only one.
Geochemistry: Supplemental Notes 5
To apply these rules, it is always best to use scientific notation. Here are some examples to clarify:
Measurement Number of Sig. Figs
Scientific Notation
4.72 (^3) 4.72 x 10^0
(^42000 2) 4.2 x 10^4
0.00032 (^2) 3.2 x 10-^4
0.01004 (^4) 1.004 x 10-^2
0.010040 (^5) 1.0040 x 10-^2
After doing your calculations, don’t give answers like “3.2384792333457” just because your calculator showed that many figures on the screen. Except for isotope ratios, there is no quantity in geochemistry that is known to more than 4 significant figures. You should only give the significant figures. When doing calculations, note the following rules for significant figures:
Addition : the last digit retained is determined by the first doubtful digit.
1.3752 + 0.033 = 1.
Multiplication/Division : round the answer to the number of sig. fig.’s of the quantity with the least precision.
(3.2)(2.3684) = 7.
Logarithms (described later..): the number of digits past the decimal is the number of sig. figs.
log(100.3) = 2.
4.0 Basic Algebra:
Many of you are still shaky with basic algebraic manipulations. To make matters worse, now you will be applying maths to another subject (geochemistry) and you may be easily intimidated by the somewhat complicated notation we need to use. In the maths units you’ve had, you worked with variables such as “ x ” and “ y ”. Here, our variables will often be things like [HCO 3 - ] or XCa. In these problems, it is important not to confuse chemical subscripts and superscripts with exponents! Even when we use simple variables like P and V , we will often need to use subscripts and superscripts to denote different states or time conditions. Usually, a subscript of
Geochemistry: Supplemental Notes 7
x^2 x + y
≠ x + x^2 y
However division does obey the distributive property over multiplication.
c) Rules for exponents:
- To multiply two powers of the same number, we add the exponents.
x^2 x^4 = x^6
- To divide two powers of the same number, we subtract the exponents.
x^4 /x^2 = x^2
- To raise a power to a new power, we multiply the exponents.
( x^2 y −^1 )^3 = x^6 y^3
- To raise a product to a power, we raise each factor to the power. To raise a quotient to a power, we raise both numerator and denominator to that power.
x^2 y z
1 / 2 = x
y z
1 / 2
4.1 Exercises.
Rearrange the following equations to obtain an expression for x:
a) x + 5 = 3x + 9
b) x/y = x^2 y
c) 4x/y = y^2 /x
d) x(3-y) = 4x
5.0 Quadratic equations
If an equation contains both x^2 and x , then it is a quadratic equation. There will be two roots (values of x) of the equation and they can be found using the quadratic formula. We’ll need
Geochemistry: Supplemental Notes 8
to use the quadratic formula when we do certain equilibrium calculations. A good example is ion-exchange equilibria. The equilibrium constant expression for the reaction
Na 2 - Smectite + Ca+2^ = Ca-Smectite + 2 Na+
is
K =
( NCa )[ Na +]^2 ( NNa )^2 [ Ca +^2 ]
where NNa is something called the “equivalent fraction” of Na (the amount of interlayer charge due to Na+) and [Na+] is the aqueous concentration of Na+. Since NNa + NCa = 1 , we get
K =
( 1 − NNa )[ Na +]^2 ( NNa )^2 [ Ca +^2 ]
which can be rearranged to give
K Ca +^2 ^2 + [ Na +]^2 NNa − [ Na +]^2 = 0
We now have a quadratic equation for NNa. When we get an equation into the form
Ax^2 + Bx + C = 0
where A , B and C are constants, we can find the two values of x using the quadratic formula:
x =
− B ± B^2 − 4 AC
2 A
Memorize this. Note that, although there will be two possible values for x , only one of them will be physically reasonable (e.g., if x is a concentration, only one of the values will be positive).
5.1 Exercises: Solve the following quadratic equations (round answers to 3 s.f.):
a) 4x^2 = 2x + 5 (Ans. x = 1.39 or x = - 0. 896
b) x^2 + 3x = 2x^2 (Ans: x = 0 or x = 3)
c) 5 = (6 + x)^2 (Ans: x= - 3.76)
d) x^2 /(0.2-x) = 1.3 x 10-^5 (Ans: x = 1.61 x 10-^3 and - 1.62 x 10-^3 )
e) x(0.10+x)/(0.10 - x) = 1.0 x 10-^2 (Ans: x = 0.001 or - 0.991)
Geochemistry: Supplemental Notes 10
sense as we go through the lectures. For now, just remember that it is against the law to take the log of a quantity with units.
It is important that you develop a good intuition with log quantities: you should instantly recognize, for example that a solution of pH=2 is 100 times more acidic than one that is pH=4. If a solution is initially at pH=6 and I double the acidity, the pH will become 5.7 (not 3 or 12 or whatever..) since - log (2 x 10-^6 ) = - log (2) - log(10-^6 ) = - 0.3 + 6.
In the lectures and practicals, I’m going to teach you how to use logs to make equilibrium concentration calculations much simpler. It is important that you master the rules for taking logs of quotients and products given above. For example, if we have an equilibrium
H 2 CO 3 -^ = 2H+^ + CO 3 -^2
With equilibrium constant expression
K =
[^ H ]
2 CO 3
^ −^2
^
HCO 3
^ −
^
we will take the p-operator (= - log) of both sides to get
pK = 2 pH + pCO 3 − pHCO 3
(we drop the brackets to simplify the notation..).
6.1 Exercises: Evaluate the log’s of the following expressions:
a)
x^2 y z^3
( Ans: 2log(x)+log(y)-3log(z) )
b)
x^3 zy^1 /^2 a
( Ans: 3log(x)+ log(z)+(1/2)log(y)-log (a) )
c)
x^2 y z^3
x z
3 / 2 ( Ans: (7/2)log(x)+log(y)-( 9 /2)log(z) )
d)
x −^2 y z^3
− 1 ( Ans: 2log(x)+3log(z)-log(y) )
e) log((10-^2 )^3 ) ( Ans: - 6 )
Geochemistry: Supplemental Notes 11
7.0 Simultaneous Equations
When solving problems in aqueous geochemistry, we will often find ourselves with several simultaneous equations. If the number of equations is the same as the number of variables (and all the equations are actually different so that no equation can be derived from the others) then the problem will have a unique solution. If all of the equations are linear (i.e., all the variables are in their first power) then the problem is easy to solve. However, problems involving simultaneous chemical equilibria will generate simultaneous non-linear equations. For example, if we have a soil with anhydrite and anglesite present, then the composition of the soil pore water will be controlled by the two solubility equilibria
CaSO 4 (anhydrite) = Ca+2^ + SO 4 -^2
PbSO 4 (anglesite) = Pb+2^ + SO 4 -^2
The equilibrium constant expressions will give two simultaneous non-linear equations in three variables:
KCaSO 4 = [Ca+^2 ][SO^ − 42 ]
KPbSO 4 = [Pb+^2 ][SO 4 −^2 ]
Coupled with these two equations is a linear equation for the conservation of charge:
[Pb+^2 ] + [Ca+^2 ] = [SO^ − 42 ]
We could solve these three equations exactly by combining them to get a single quadratic equation for one of the variables and then plugging that result into the other two equations to solve for the remaining variables. An easier way, that I will teach you later, is to recognize which variables are much smaller than the others. In this case, it turns out that KPbSO4 is much smaller than KCaSO4 ; consequently, [Pb+2] << [Ca+2]. It follows that, to a very close approximation, [Ca+2] = [SO 4 -^2 ]. The important thing to know at this point is that, as long as the number of unique equations is the same as the number of unknowns, the problem will have a solution. If they are not linear equations, we can either solve the problem by making a few approximations or we can solve the problem numerically using a computer.
8.0 Functions and their derivatives:
One quantity (say, y ) is a function of another quantity (say, x ) if its value depends on the value of that other quantity. If y is a function of x , we might write “ y ” as “ y(x) ” to express the functional relationship (Note: this doesn’t mean we multiply y by x). Often, thermodynamic quantities are functions of more than one variable. For example, the volume ( V ) of a gas is a function of both pressure ( P ) and temperature ( T ). Hence we sometimes might want to write volume as V(P,T) to express this functional relationship. In the case of an ideal gas, we actually know what the function is (via the ideal gas law):
V(P,T) = nRT/P
Geochemistry: Supplemental Notes 13
In the case of a linear function y(x) = mx + b (where m and b are constants), the derivative of y with respect to x is simply the slope ( m ) of the line (the average rate of change over any interval and the instantaneous rate are the same). However, in the case of the function y = x^2 , the derivative is also the slope at a given value of x and the slope keeps changing (it gets steeper and steeper with x). Derivatives are very useful quantities. For a moving object, the derivative of position with respect to time is speed. The derivative of speed with respect to time is acceleration. (Hence, we can also say that acceleration is the second derivative of position with respect to time).
In maths lectures, they tend to indicate “the derivative of y with respect to x” as y’(x). In science and engineering, we like to write the derivative of y with respect to x as
dy dx
The quantities dy and dx are called differentials (note that the d ’s are not separate variables..); the quantity dx means a change in x so small as to not be measureable but still greater than 0. We need these things to do thermodynamics since a reversible process is one in which we go from one state to another via a series of infinitesimally small changes. Scientists and engineers like to treat dy and dx like any other quantity and “divide both sides of the equation by dx …” etc. Some mathematicians hate this (since you cannot divide something by zero and dx is, essentially, zero) but we do it anyway. In fact, dx isn’t zero.
Here are some simple rules to remember which will enable you to bluff your way around all the derivatives you’ll find in your career:
y ( x ) = axn^ :
dy dx
= naxn −^1
y ( x ) = c
dy dx
= 0 (where c is a constant)
Figure 1 : The slope of the line a-a’ is Δy/Δx. This is the average rate of change the function y=x^2 over the interval Δx. The slope of the line b-b’ is the instantaneous rate of change of y when x = x 1. The slope of a-a’ approaches that of b-b’ as Δx goes to zero.
Geochemistry: Supplemental Notes 14
y ( x ) = aebx^ : dy dx
= abebx^ (where e is 2.71828…)
y ( x ) = ln( x ) :
dy dx
x
Some functions are the sum of two or more functions (e.g., as in a polynomial):
y ( x ) = g ( x ) + h ( x ) :
dy dx
dg dx
dh dx
(i.e., the derivative of a sum is just the sum of the derivatives). Hence, the derivative of
y ( x ) = 3 x^2 + 2 x + 3
is just
dy dx
= 6 x + 2
Some functions are the product of two functions:
y ( x ) = g ( x ) h ( x ) : dy dx
= h ( x ) dg dx
Hence, the derivative of
y ( x ) = x^2 e − x
is
dy dx
= 2 xe − x^ − x^2 e − x
(Note: even though you get lots of problems like this in a typical calculus course, you will probably never encounter anything this complicated in your work. However, it is absolutely essential that you understand the concept of derivative and be able to work out derivatives of simple functions.)
8.1 Exercises: Work out the derivatives of the following functions:
a) y = 4x + 7 (Ans: dy/dx = 4)
b) y = 3x^2 + 2 (Ans: dy/dx = 6x)
c) y = 3/x + 2x (Ans: dy/dx = - 3/x^2 +2)
d) y = log (x) (Ans: dy/dx = 1/(2.303x) )
Geochemistry: Supplemental Notes 16
b) V = nRT/P ( n and R are constant .) (Ans: dV = (nR/P)dT - (nRT/P^2 )dP
c ) G= U-TS+PV (Ans: dG = dU – TdS – SdT + PdV + VdP)
9.2 The exact differential (advanced topic) :
Suppose we have a function G(P,T). Then the differential of G
dG =
∂ G
∂ P
T
dP +
∂ G
∂ T
P
dT
is an exact differential if
∂ T
P
∂ G
∂ P
T
∂ P
T
∂ G
∂ T
P
If the differential of G is exact, then G is a state function. It turns out that the whole point of thermodynamics is to express the world in terms of state functions. To understand the significance of state function, we have to learn the idea of path integral.
10.0 Integration and integrals:
There are two ways to view integrals: on the one hand, the integral of a function means the “antiderivative” of the function (i.e., the new function for which our original function is the derivative). Hence, integration is just doing differentiation backwards:
y ( x ) = axn^ : y ( x ) dx =
a n + 1
∫ x
n + (^1) + C
The C is simply an unknown constant. Note that if we take the derivative of the right hand side of the = sign, we’ll get our original function back regardless of C. What if n = - 1? This is a very special integral:
y ( x ) = x −^1 : ∫ y ( x ) dx = ln( x ) + C
The ln function is the log to base e (= 2.71828…)and shows up everywhere in nature.
y ( x ) = ae^ λ x^ : y ( x ) dx = a λ
∫ e
λ x
The other kind of integral is a called a definite integral ; this is a sum. When we perform a definite integration of y(x) over the range from x=a to x=b , we evaluate the sum all of the dy ’s over that range. Suppose y(x) = 3x. Then the evaluation of the definite integral of y(x) from x = 1 to x = 2 looks like this:
Geochemistry: Supplemental Notes 17
3 xdx 1
2
∫ =^
x^2 1
2
( 2 )^2 −
( 1 )^2 =
(I’ve omitted the unknown constant C since it gets cancelled in a definite integral). An important example we’ll be doing is evaluating the pressure-volume integral to predict Δ G of an ideal gas when we go from an initial pressure of P 1 to a final pressure of P 2 :
VdP P 1
P 2
∫ =^
nRT P
dP P 1
P 2
∫ =^ nRT^ (ln^ P 2 −^ ln^ P 1 )^ =^ nRT^ ln^
P 2
P 1
What if our function depends on more than one variable, say P and T? When we integrate our function, we must do so along a path going from P 1 T 1 to P 2 T 2.
11.0 Differential equations:
Often in science, we don’t know the actual (“analytic”) form of a function and our goal is to find that out. Usually, some physical law will give us an equation for the derivative of the function. For example, Newton’s law will give us the second derivative (acceleration) of position with respect to time. The diffusion equations we will learn will give allow us to relate fluxes to concentration gradients. These differential equations can be solved if we have some kind of boundary condition. A very important example of a differential equation with a boundary condition in geology is radioactive decay. We know that the activity A of some radioactive isotope will change with time according to the law:
dA dt
= − λ A
where λ is the decay constant. This means that the rate by which our isotope disappears is a function of the amount of that isotope that is present at a given time. (This is also the same math for compound interest, population growth and rates of simple chemical reactions). We can solve this differential equation for A (t) if we have the boundary condition that at t = 0, A = A 0 (where A 0 is the initial activity). If we multiply both sides by dt and divide both sides by A , we get
dA A
= − λ dt
Now we can integrate both sides over the range t = 0 to t = t. When t = 0, A = A 0 ; when t = t, A = A. This means we integrate both sides like this..
dA A 0^ A
A
∫ =^ −^ λ dt
0
t
These are easy integrals. The left hand side is just:
