Acids and Bases II
Dr. Ron Rusay
Summer 2004
© Copyright 2004 R.J. Rusay
Water as an Acid and a Base
 Amphoteric
substances can act as
either an acid or a base
• Water acting as an acid:
NH3 + H2O  NH4+1 + OH-1
• Water acting as a base:
HCl + H2O  H3O+1 + Cl-1
• Water reacting with itself as both:
H2O + H2O  H3O +1 + OH-1
Water as an Acid and a Base
Water is amphoteric. It can behave
either as an acid or a base.
H2O(l) + H2O(l)  H3O+(aq) + OH(aq)

acid 1 base 1
conj
acid 2
conj
base 2
The equilibrium expression for pure
water is:
Kw = [H3O+(aq)] [OH(aq) ]

Water: Self-ionization
Autoionization of Water
 Water
is an extremely weak electrolyte
therefore are only a few ions present:
Kw = [H3O+1] [OH-1] = 1 x 10-14 @ 25°C
• NOTE: the concentration of H3O+1 and OH-1 are
equal
• [H3O+1] = [OH-1] = 10-7 M @ 25°C
• Kw is called the ion product constant for
water: as [H3O+1] increases, [OH-] decreases
and vice versa.
Acidic and Basic Solutions
Acidic solutions have:
a larger [H+1] than [OH-1]
 Basic solutions have:
a larger [OH-1] than [H+1]
 Neutral solutions have
[H+1] = [OH-1] = 1 x 10-7 M

[H+1]
-14
1
x
10
=
[OH-1]
[OH-1]
-14
1
x
10
=
[H+1]
The pH Scale
 pH

 log [H+]  log [H3O+]
1 pH unit corresponds to a factor of 10
 pH
in water ranges from 0 to 14.
Kw = 1.00  1014 = [H+] [OH]
-log Kw = -log [H3O+1] -log [OH-1]
pKw = pH + pOH = 14.00
 As pH rises, pOH falls (Sum = 14.00).
pH & pOH

pH = -log[H3O+1]
pOH = -log[OH-1]
• pHwater = -log[10-7] = 7 = pOHwater

[H+1] = 10-pH [OH-1] = 10-pOH
pH < 7 is acidic; pH > 7 is basic, pH = 7 is
neutral
 The lower the pH, the more acidic the
solution; The higher the pH, the more basic
the solution.
 1 pH unit corresponds to a factor of 10
 pOH = 14 - pH

There are no theoretical limits on the values of
pH or pOH. (e.g. pH of 2.0 M HCl is -0.301, the
pH at Iron Mountain, California is ~ -2 to -3)
What’s in these household products?
Acids or bases?
Strong or weak?
Should you be concerned about safety?
The pH of Some Familiar
Aqueous Solutions
[H3O+]
[OH-]
=
KW
[H3O+]
[OH-]
[H3O+]>
[OH-]
[H3O+]<
[OH-]
acidic
solution
What’s your diet?
Your urine will tell!
neutral
solution
basic
solution
[H3O+] =
[OH-]
The pH Scale
What is the pH of 6M hydrochloric acid?
Example #1
Determine the [H+1] and [OH-1] in a
10.0 M H+1 solution

Determine the given information and
the information you need to find
Given [H+1] = 10.0 M

Find [OH-1]
Solve the Equation for the Unknown
Amount
1
Kw  [H ] x [OH ]
Kw
-1
[OH ]  1
[H ]
-1
Example #1 (continued)
Determine the [H+1] and [OH-1] in a
10.0 M H+1 solution

Convert all the information to Scientific
Notation and Plug the given
information into the equation.
Given [H+1] = 10.0 M
= 1.00 x 101 M
Kw = 1.0 x 10-14
Kw
[OH ]  1
[H ]
-1
-14
1.0
x
10
-15
[OH -1 ] 

1.0
x
10
M
1
1.00 x 10
Example #2
Calculate the pH of a solution with a
[OH-1] = 1.0 x 10-6 M
 Find
the concentration of [H+1]
Kw
[H ] 
1
[OH ]
1
-14
1.0 x 10
-8
[H ] 

1.0
x
10
M
-6
1.0 x 10
1
Example #2 continued
Calculate the pH of a solution with a
[OH-1] = 1.0 x 10-6 M
the [H+1] concentration into your
calculator and press the log key
 Enter
log(1.0 x 10-8) = -8.0
 Change
the sign to get the pH
pH = -(-8.0) = 8.0
Example #3
Calculate the pH and pOH of a
solution with a [OH-1] = 1.0 x 10-3 M
the [H+1] or [OH-1]concentration
into your calculator and press the log
key
 Enter
log(1.0 x 10-3) = -3.0
 Change
the sign to get the pH or pH
pOH = -(-3) = 3.0
 Subtract
the calculated pH or pOH from
14.00 to get the other value
pH = 14.00 – 3.0 = 11.0
Example #4
Calculate the [OH-1] of a solution with a pH of 7.41

If you want to calculate [OH-1] use pOH, if you
want [H+1] use pH. It may be necessary to
convert one to the other using 14 = [H+1] +
[OH-1]
pOH = 14.00 – 7.41 = 6.59
Enter the pH or pOH concentration into your
calculator
 Change the sign of the pH or pOH

-pOH = -(6.59)

Press the button(s) on you calculator to take
the inverse log or 10x
[OH-1] = 10-6.59 = 2.6 x 10-7
Calculating the pH of a Strong,
Monoprotic Acid
A
strong acid will dissociate 100%
HA  H+1 + A-1
 Therefore the molarity of H+1 ions will be
the same as the molarity of the acid
 Once the H+1 molarity is determined, the
pH can be determined
pH = -log[H+1]
Example #5
Calculate the pH of a 0.10 M HNO3 solution
the [H+1] from the acid
concentration
 Determine
HNO3  H+1 + NO3-1
0.10 M HNO3 = 0.10 M H+1
the [H+1] concentration into your
calculator and press the log key
 Enter
log(0.10) = -1.00
 Change
the sign to get the pH
pH = -(-1.00) = 1.00
Buffered Solutions
Buffered Solutions resist change in pH when
an acid or base is added to it.
 Used when need to maintain a certain pH in the
system, eg. Blood.
 A buffer solution contains a weak acid and its
conjugate base
 Buffers work by reacting with added H+1 or OH-1
ions so they do not accumulate and change the
pH.
 Buffers will only work as long as there is
sufficient weak acid and conjugate base
molecules present.

Buffers
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