Auto dissociation of Water. Water molecules allow protons to be transferred between molecules & an equilibrium is established. HOH (l) + HOH (l) H 3 O + ( aq ) + OH - ( aq ) Overall: 2H 2 O (l) H 3 O + ( aq ) + OH - ( aq )
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Water molecules allow protons to be transferred between molecules & an equilibrium is established.
HOH (l) + HOH(l) H3O+ (aq) + OH-(aq)
Overall: 2H2O (l) H3O+ (aq) + OH-(aq)
Keq = Kw = [H3O+] [OH-] = 1.0 x 10-14 at 250C
Anytime H3O+ions or OH- ions are present, this equilibrium is automatically established & this regulates the allowable concentrations of H3O+ & OH-.
Thus, Kw = [H3O+] [OH-] can be used to calculate [H3O+] & [OH-] in acidic & basic solutions. In a neutral solution [H3O+] [OH-] = 1 x 10-7 M (i.e. pH=7)
1) In a certain container of H2O it was found that the [OH-] was 1.0 x 10-4. What is [H3O+]? What is the pH?
2) A 0.0001 M HCl (a strong acid) solution will have a pH of what? What would be [OH-] be?
3) Ca(OH)2 is a strong base. If a 0.000 05M solution dissociates in water, what will the pH be?
[H3O+] = 10-pH M
Ex) solution where [H3O+] = 0.001 M (10-3M). Find the pH.
[H3O+] = 0.001 M or 1 x 10-3 M
so pH = 3
Ex) Find the pH of a 0.0001 M solution of HCl.
We need to find the [H3O+].
Since HCl is a strong acid, it completely dissociates.
HCl + H2O H3O+ + Cl-
Initial: 0001M 0 M so [H3O+] = 0.0001 M = 1 x 10-4
0 M .0001 M pH = 4
pH – the pH of a solution is defined as the negative of the log of the hydronium ion concentration
Ex) [H3O+] = 3.25 x 10-3 M, find the pH.
Answer: pH = -log 3.25 x 10-3=
Ex) [H3O+] = 1.44 x 10-10 M, find the pH.
Answer: pH = -log 1.44 x 10-10 =
Ex) pH = 2.80, find [H3O+].
We must work in reverse (take the “antilog” of the negative pH).
pH = 2.80 [H3O+] = antilog -2.8 =
Ex) pH = 11.4 [H3O+] = antilog -11.4 =
The pOH is the pH value of a solution using the concentration of the OH- ions.
pH + pOH = 14
Ex) What is the pH of a 3.41 x 10-4 M NaOH solution?