Examples of Strong Acids and
Bases
Weak Acids, Ka
=[H+] of acetic
acid, pH, and % ionization
We discussed the balance
between H+ and
Given [H+], [
Measuring pH
H
MO = H+ +
Red Yellow
At pH < 2 the HMO species predominates in water. At pH>2 the yellow
HX + H2O = H3O+ + X- the equilibrium strongly favors the
products. Strong means almost 100%
ionization.
The following are strong acids
HCl Hydrochloric
Acid Ka = 107
HBr Hydrobromic Acid Ka = 109
HI Hydroiodic Acid Ka
= 1010
HNO3 Nitric
Acid
H2SO4 Sulfuric Acid (1st H+ coming off
is strong, the second is <100%, but not too much less)
HClO4, 3 Perchloric and Chloric Acid
Please note, HF is not a
strong acid.
Since these almost
completely ionize, we can assume [H+] = [HX][1],
so it it easy to calculate the pH.
What is the pH of a .010 M
HCl solution?
Well since this is a strong
acid, the [H+] = 1.0X10-2M and pH=Log(1.0X10-2)=2.00. I underlined the significant figures that
came from
1.0 X 10-2
Strong bases include group I and 2 metal hydroxides (note group 2
hydroxides are not as soluble as group 1).
NaOH, KOH, Ca(OH)2,
Strong means almost 100%
ionization, so for group 1 hydroxides, [
What is the pH of a 0.01 M
NaOH solution?
Well, [
Metal oxides, hydrides, and nitrides
are also strong bases. The group 1 metal
compounds are soluble but other group oxides, hydrides and nitrides have low
solubility.
All of these are stronger
bases than
O-2 + H2O = 2
For weak acids, the
ionization is much less than 100%
HA + H2O = H3O+ +
A-
Kc = [H3O+] [A-]
/ [HA] [H2O]
Since [H2O] is
fairly constant for dilute solutions, we usually write
Ka = [H3O+] [A-]
/ [HA] = [H+] [A-] / [HA]
here
you can think of the reaction as HA = H+ + A-
Where Ka is
called the acid-dissociation constant.
The larger Ka is, the stronger the acid. (For example, the Ka for HCl is ~
107.
See table 16.2 (page 607)
Write the expression for Ka
for the following acids
(CH3)2NH2+, H2PO4-,
HCHO2)
Calculate the pH of a 0.010
M acetic acid solution and the % ionized.
The Ka for acetic acid is 1.8 X 10-5. Remind people that this [HCl] gave a pH of
2.00
HC2H3O2 = H+ + C2H3O2-
|
Initial |
1.0 X 10-2 |
0 * |
0 |
|
Change |
-x |
+x |
+x |
|
Equilibrium |
1.0 X 10-2 – x |
X |
X |
*We are neglecting the [H+]
from the autoionization of water because it is to
small to contribute much.
Ka = x2 / (1.0 x 10-2
–x) ~
x2 / 1.0 X 10-2
(Lets see if we can simplify
by neglecting x compared to 0.010 M.
Solve for x=4.2 X 10-4M (which is indeed very small compared
to
1.0X10-2M.)
pH = -Log (4.2 X 10-2) = 3.37
(compare to same [HCl] giving a pH of 2.00).
The % ionized will be =
amount ionized / initial amount X 100%
= 4.2 X 10-4 / 1.0 X 10-2 X 100% =
4.2 %
Note, as long as [H+]
> 4.5 X 10-7M , [H+] from the autoionization
of water produces less than 5% of the hydronium ions in solution.
Further note, [HA]equilibrium=[HA]initial –x
Acid –Base calculations are
limited to about +/- 5%, so as long as x does not exceed +/-5% of [HA]initial
We assume that x is
negligible and
[HA]equilibrium ~
[HA]initial
For this to work
[HA]initial
> 400 X Ka
This means the stronger the acid, the less
likely you can make the assumption that [HA]equilibrium ~ [HA]initial.
[1] [H+]
= [HX] when we can neglect the contribution of [H+] from water. This is a good assumption as long as [HX]
> 10-6 M. At [HX] < 10-6
M we can no longer neglect the acid species from water.
[2] Similar
to above, [