**If the Hydrogen ion concentration is 0.1
moles/liter **

**Then the [OH ^{-}] could be
found by the equation above: **

**[OH ^{-}] = 1 X
10^{-14} / 1 X 10^{-1} = 1 X 10^{-13}
**

**The pOH = -log[OH ^{-}] =
-log(1 X 10^{-13}) = -(log 1 + log 10^{-13}) = -(0 +
-13) = -(-13) = 13 **

**For a [H ^{+}] = 0.1 = 1 X
10^{-1} **

**Then pH = -log 1 X 10 ^{-1} = -(0 +
-1) = 1 **

**Therefore the **

**One can have a pH that is a negative value
in for example strong acid solutions greater than 1 mole/liter.
**

**For a 10 M H ^{+} **

**pH =-log(1 X 10 ^{1}) = -(0 + 1) = -1
**

**Most acidic and basis substances whose pH
lie between the 0-14 range. Do not confuse the term acid strength
with pH. The ****strength
of an acid**** has to do with the
percentage of the initial protons that are ionized. If a higher
percentage of the original protons are ionized and therefore donated
as hydrated protons (hydronium ions) then the acid will be stonger.
Strong acids are Hydrochloric, Hydrobromic, Nitric, Sulfuric, and
Perchloric acids. In each of these molecular acids the percentage of
ionization is almost 100%. **

**Example: What would be the
[H _{3}O^{+}] of a solution that has a pH =
5.4 **

**[H _{3}O^{+}] =
Antilog (-pH) = Antilog (-5.4) = 3.98 X 10^{-6}
**

**Note: Determining Antilogs is done in the
following manner using a calculator: **

**Enter the number which you wish to take the
Antilog of in this case -5.4 (press the change sign key (+/-) to
change the sign to negative) **

**Depress the second function key sometimes
called the inverse (inv) key **

**Depress the log key **

**Read the display **

**Example: Calculate the
[OH ^{-}] of a solution that has a pOH = 8.2
**

**[OH ^{-}] = Antilog (-pOH) =
Antilog ( - 8.2 ) = 6.31 X 10^{-}9 **

**Example: Calculate the
[H3O ^{+}] when the [OH^{-}] = 3.2 X
10^{-3} **

**Water ionizes only slightly giving the
following equilibrium: **

**H _{2}O (l) + H_{2}O(l) =
H_{3}O^{+}(aq) + OH^{-}(aq) **

**According to the Law of Chemical
Equilibrium: **

**K _{c} =
[H_{3}O^{+}] [OH^{-}] /
[H_{2}O(l)]^{2} **

Note: concentration of pure water is a constant 55.5

**Since the molar concentration of water is
constant we multiply both sides of the above expression by
[H _{2}O]^{2} that will result in another
constant: **

**K _{c}
[H_{2}O]^{2} = K_{w} =
[H_{3}O^{+}] [OH^{-}] = 1.0 X
10^{-14} **

**K _{w} = 1 X 10^{-14} =
[H_{3}O^{+}] [OH^{-}]
**

Note: K_{w}is the ionization equilibrium constant of pure water which always has a value of 1 X 10^{-1}4 at 25 degrees Celsius.

**K _{w} = 1 X 10^{-1}4 =
[H_{3}O^{+}] [ 3.2 X 10^{-3}]
**

**[H _{3}O^{+}] = 1 X
10^{-1}4 / 3.2 X 10^{-3} = 3.125 X 10^{-12}
**

**Yes, if the acid or base is a strong one,
then the dissociation will be 100% **

**For example, HCl is classified as a strong
acid so: **

**If I start with a 0.1 M HCl, then I will
have 0.1 M H _{3}O^{+} because for every HCl that
breaks apart then one H_{3}O^{+} is formed and one
Cl^{-} is formed. Since strong acids (and bases) ionize 100%
then all of the original concentration will be converted to
H_{3}O^{+} **

**Typical strong acids: HCl, HBr, HI,
HNO _{3}, H_{2}SO_{4}, HClO_{4}
**

**Strong Bases: All the Hydroxide compounds of
Group 1 and Group 2 metals **

**LiOH, NaOH, KOH, etc and Be(OH) _{2},
Ca(OH)_{2}, Mg(OH)_{2}, etc **

**What would be the [OH ^{-}]
of a .2 M NaOH solution? **

**.2 M NaOH will produce .2 M OH ^{-}
since the breakdown is 100% **

**What would be the [OH ^{-}]
of a .3 M Ca(OH)_{2} **

**.3 M Ca(OH) _{2} will produce .6M
OH^{-} because for every one Ca(OH)_{2} that breaks
apart TWO OH^{-} ions are produced (twice as much) (look at
the equation and note the coefficients) **

**Now here is an example for you to work out.
**

**Given 0.02M Ba(OH) _{2} solution:
**

**Determine the Hydroxide ion molar
concentration **

**Determine the Hydrogen ion concentration
**

**Determine the pH **

**Determine to pOH **

**Once you have answered the questions in the
above example then you can check the ****correct
answers**** **

**Given 0.02M Ba(OH) _{2} solution:
**

**Determine the Hydroxide ion molar
concentration **

**Determine the Hydrogen ion concentration
**

**Determine the pH **

**Determine to pOH **

**Here is the solution: **

**Write the dissociation Equation **

**Ba(OH) _{2} + H_{2}O ---->
Ba^{+2} + 2OH^{-} **

**Determine the Hydroxide Concentration **

**Since all Hydroxides of Group 2 metals are
strong bases dissociating 100% [Ba(OH) _{2}] = .02 M
**

**According to the balanced equation for every
Ba(OH)2 that dissociates twice as many Hydroxide ions forms
**

**[OH ^{-}] = 2(.02) = .04 M
**

**Determine the Hydrogen ion concentration
**

**K _{w} = 1 X 10^{-14} =
[H_{3}O^{+}] [OH^{-}]
**

**[OH ^{-}] = 0.04 M (step 2)
**

**1 X 10 ^{-14} =
[H_{3}O^{+}] [0.04] **

**[H _{3}O^{+}] = 1 X
10^{-14}/ .04 = 2.5 X 10^{-13} **

**Determine the pH **

**pH =
-log[H _{3}O^{+}] = -log[ 2.5 X
10^{-13}] **

**pH = -[-12.6] = 12.6 **

**Determine pOH **

**pH + pOH = 14 **

**12.6 + pOH = 14 **

**pOH = 1.4 **

**Determine the pH knowing that pH + pOH
= 14 **

**pH = 14 - pOH = 14 - 5 = 9 **

**In a water solvent based solution:
**

**[H ^{+}]
[OH^{-}] = K_{w} = 1 X 10^{-14}
**