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 101) = -(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 [H3O+] of a solution that has a pH = 5.4
[H3O+] = 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:
H2O (l) + H2O(l) = H3O+(aq) + OH-(aq)
According to the Law of Chemical Equilibrium:
Kc = [H3O+] [OH-] / [H2O(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 [H2O]2 that will result in another constant:
Kc [H2O]2 = Kw = [H3O+] [OH-] = 1.0 X 10-14
Kw = 1 X 10-14 = [H3O+] [OH-]
Note: Kw is the ionization equilibrium constant of pure water which always has a value of 1 X 10-14 at 25 degrees Celsius.
Kw = 1 X 10-14 = [H3O+] [ 3.2 X 10-3]
[H3O+] = 1 X 10-14 / 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 H3O+ because for every HCl that breaks apart then one H3O+ is formed and one Cl- is formed. Since strong acids (and bases) ionize 100% then all of the original concentration will be converted to H3O+
Typical strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4
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 + H2O ----> 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
Kw = 1 X 10-14 = [H3O+] [OH-]
[OH-] = 0.04 M (step 2)
1 X 10-14 = [H3O+] [0.04]
[H3O+] = 1 X 10-14/ .04 = 2.5 X 10-13
Determine the pH
pH = -log[H3O+] = -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-] = Kw = 1 X 10-14