Logarithms and PH scale
Goal is to find an equilibrium between water , hydronium and hydroxide
2H2O  H3O+ + OH[H3O+] is the concentration of hydronium (or H+) in moles/liter
[OH-] is the concentration of hydroxide in moles/liter
We expect the process to be balanced when the same number of H2O molecules are
created by joining as are lost through separation. For the separation the amount of water
is so large
1 liter = 1000 cm3  1000gm/liter= 55.6 moles/liter of water
that we can assume that the disassociation rate is some constant value that isn’t changing
(disassociation fraction is too small to change the amount of water) and that equilibrium
occurs when there is enough [H3O+] and [OH-] to cause the same number of H2O to
form. We made the hand waving argument that the product of concentrations
Product = [H3O+] [OH-] (collision model: What is the likelihood of these
two objects colliding in a solution)
will be the key factor and that we need to adjust this product until it reaches
1x10-14 = 1x10-7 x 1x10-7 (pure water)
Hopefully, this is a plausible result.
Now we can change the concentrations by adding either H3O+or OH- directly into the
water. This can be done by finding molecules that dissolve in water and release the H+ or
OH-. Two examples of molecules that dissolve and add H+ or OH- are
2H2O+ HCl  H2O + H+ + CL2H2O+ NaOH  H2O + Na+ + OH-
increase [H3O+]
increase [OH-]
The question becomes what happens to the equilibrium condition if there is not a
matching number of H+ and OH-. The answer is nothing. Equilibrium i.e. the balance
between disassociation and association is achieved when
Product = [H3O+] [OH-]=1x10-14
Here are some examples of concentrations that are possible.
Concentration of
Concentration
PRODUCT
[H3O+]
(Moles/Liter)
of [OH-]
(Moles/Liter)
(Moles/Liter) (Moles/Liter)
1x10-7
1x10-3
1x10-5
1x10-12
1x10-8
1x10-7
1x10-11
1x10-9
1x10-2
1x10-6
1x10-14
1x10-14
1x10-14
1x10-14
1x10-14
Let’s remind ourselves what these numbers mean.
Complete the table
power or 10
102
100
10
0.1=
1
10
1
10000
1,000,000
1,000,000,000,000
0.0001=
Multiply the following then write the problem using powers of 10.
10 x 10=
1
x100=
10000
1
1
x
=
100 100
1000x1000=
10 x 10=100
101x101=102
The powers of 10 can be expressed simply by stating the exponent
million
ten thousand
one billionth
one tenth
one hundredth
106
6
Can any number be represented as 10 raised to some power?
water
acid
acid
base
base
3=10x
.05=10x
14,632=10x
1
=10x
3
where x=
where x=
where x=
where x=
Logarithms
A  10a There is a number a such that 10 raised to the power a is equal to A .
a is then the logarithm of A .
For any number we can find a number that serves as the exponent for 10 to give the
number back.
B  10b
log B  b
From the work done so far you should realize that when two numbers are multiplied their
logarithms add as shown below
AB  C
A  10 a , B  10b , C  10 c
C  10 a  10b  10 a b  10 c
so  a  b  c
log( AB )  log A  log B
We need to use this feature of logarithms to express the PH scale. This is simply a
method or preferred way to express the concentration of [H3O+] and [OH-] in a given
solution.
PH= -log( [H+] ) the negative logarithm of the concentration of H+
POH= -log( [OH-] ) the negative logarithm of the concentration of OHUsing these facts
calculate the PH of a 0.1 M solution of HCl (all HCl ionizes in solution)
calculate the PH of a 5 M solution of NaOH (all NaOH ionizes in solution)
.
1. What is the pH and pOH of the following solutions?
pH
pOH
(a) 0.01 M HCl
(b) Pure water
(c) 0.001 M HNO3
(d) 50 mL of 0.5 M KOH
(e) 200 mL of 4 mM NaOH
(f) 50 mL of acid with 7e-6 moles of H3O+ in it
(g) 100 mL of base with 3e-7 moles of OH- in it
2. (a) If you were to mix (f) and (g) above together, would it be an acid or a base?
(b) How many moles of excess hydronium or hydroxide would it have?
(d) What would the pH and pOH of this mixture be?