CHEM 11111_2013_05_20 - University of Kelaniya

CHEM 11111
Calculations in Chemistry
Dr. S. Sri Skandaraja
Department of Chemistry
University of Kelaniya

Logarithms
10 n Number = base exponent
When base is 10 then ‘n’ is the logarithm of
‘N’.
n = log 10 N
log 10 10 = 1
log 10 1000 = 3
log 10 0.001 = -3

The concept of pH
pH = -log[H + ]
The H + concentration of a acid
solution is 0.053 M.
Calculate the pH of the solution.
pH = -log [H+] = - log (0.053)
= - (-1.27572413…..)
Accurate pH value
= 1.28

The pH of a solution is 10.5
Calculate the OH - concentration of
the solution.
pH = -log [H + ] = 10.5
Hence pOH = 3.5 = - log [OH - ]
[OH - ] = antilog 10 (-3.5)
= 3.1622776…..x 10 -4
Accurate [OH - ] value
= 3 x 10 -4 M

log MN = log M + log N
log M/N = log M - log N
eg. k a = [H + ][A - ] / [HA]
hence
pH
pk
a
[ A ]
log
[ HA ]
log M p = p log M
log [H + ] 4 = 4 log [H + ]
= -4pH
Henderson-Hasselbalch equation
K b ??

Natural logarithms (ln)
e n
Number = base exponent
n = log e N = ln N
2.718
ln MN = ln M + ln N
ln M/N = ln M - ln N
ln M p = p ln M
ln (1) = 0
‘e’ has a value of

Eg. in chemistry ( thermodynamics)
DG o = -RT ln K or DG = DG o + RT ln K
Eg.
If the equilibrium constant of a reaction is 4.6 x 10 3 at a
temperature of 298 K, calculate the change in standard
Gibbs free energy for the reaction at this temperature
( R= 8.314 J K -1 mol -1 ).
Answer
= 20895.374…
significant figures ?
= 2.1 x 10 4
Units?

Nernst equation
Ox +ne
Red
E
E
0
RT
nF
ln
[Re d]
[ Ox]
R= gas constant = 8.314 J mol -1 K -1
T= temperature = 298.1 K (25 0 C)
n = number of electrons transferred in the half cell reaction
F = Faraday’s constant = 96490 C
E = reduction potential
Fe 3+ + e Fe 2+ E 0 Fe3+/Fe2+=+0.68 V

If log 10 P = q
P= 10 q
Hence
ln P = ln 10 q
= q ln 10
But ln 10 = 2.303
Hence ln P = 2.303 log 10 P
E
E
0
RT
nF
ln
[Re d]
[ Ox]
• At 298.1K (25 0 C): 2.303 RT/F= 0.0592
E cell
E
0
2.303RT
nF
log
[Re d]
[ Ox]
E cell
E
0
0.0592
n
log
[Re d]
[ Ox]

Eg. in chemistry ( Kinetics)
rate = k [reactants] n
log(rate) = log(k) + n log [reactants]
log (rate)
log [reactants]

Eg. in chemistry ( Kinetics)
First order rate law
[A] = [A 0 ] e -kt
t = time
[A] = concentration of A at time ‘t’
[A 0 ] = initial concentration of A
k = Rate constant
ln[A] = ln[A 0 ] – kt

Eg. in chemistry ( Kinetics)
Arrhenius Equation
k = A e -Ea/RT
T= temperature
A = Arrhenius constant or pre-exponential factor
E a = activation energy
R = universal gas constant
k = rate constant
ln k = ln A - E a /RT
y = c + mx

Another definition.
e x
1
x
x
2
2!
x
3
3!
x
4
4!
.....
If x is very small.
e x 1
x

Using Calculators
To get
• log
•Anti-log
•ln
•Anti-ln (e)
•EXP key
?

Trigonometric functions
b
a
tana
sina
cosa
tana
a
b
a
c
b
c
a
b
sin
cos
a
a

2
2
2
c
b
a
c
a
sina
2
2
2
)
(sin
a
c
a
c
b
cosa
2
2
2
)
(cos
b
c
a
2
2
2
2
2
)
(cos
)
(sin
c
c
c
a
a
1
)
(cos
)
(sin
2
2
a
a

Values of trigonometric function
‣Cosecant(cosec)
hypotenuse/opposite side
‣cosec = c/a
‣Secant(sec)
hypotenuse/adjacent side
‣sec = c/b
‣Cotangent(cot)
adjacent side/opposite side
‣cot = b/a
17

Some important relationships
‣ sin 2 A + cos 2 A = 1
‣ 1 + tan 2 A = sec 2 A
‣ 1 + cot 2 A = cosec 2 A
‣ sin(A+B) = sinAcosB + cosAsin B
‣ cos(A+B) = cosAcosB – sinAsinB
‣ tan(A+B) = (tanA+tanB)/(1 – tanAtan B)
‣ sin(A-B) = sinAcosB – cosAsinB
‣ cos(A-B) = cosAcosB+sinAsinB
‣ tan(A-B) = (tanA-tanB)(1+tanAtanB)
‣ sin2A = 2sinAcosA
‣ cos2A = cos 2 A - sin 2 A
‣ tan2A = 2tanA/(1-tan 2 A)
‣ sin(-A) = sin (A)
‣ Cos(-A) = Cos(A)
‣ Tan(-A) = Tan(A)

Values of Trigonometric function
0 30 45 60 90
Sine 0 0.5 1/2 3/2 1
Cosine 1 3/2 1/2 0.5 0
Tangent 0 1/ 3 1 3 Not defined
Cosecant
Not
defined
2 2 2/ 3 1
Secant 1 2/ 3 2 2 Not defined
Cotangent
Not
defined
3 1 1/ 3 0

X-Ray Diffraction: Bragg’s Law
2d sin = nl
Extra distance
traveled by wave 2
= 2d sin