This topic teaches students basic principles required to solve problems involving PH calculation of strong acid and strong alkali.
It also cosiders why solution of some salts may have PH value > 7 or < 7, as well as making students to know why water and ammonia are bases.
Please find solutions to the problems as an attachment.
- PH is the negative logarithm to the base 10 of hydrogen ion or hydrozonium ion concentration.
- PH is a measure of the strength of an acid or an alkali.
Consider the reversible ionization or dissociation of water into hydrogen and hydroxyl ions:
H2O H+ + OH-
Kw is the dissociation constant of water. By the law of mass action it is defined as the product of the concentrations of hydrogen and hydroxyl ions divided by the concentration of water.
Kw = [H+][OH-]
[H2O] is very negligible, therefore:
Kw = [H+] [OH-] --------- (i)
Take the negative logarithm of the expression in equation (i)
-log10Kw = -log10([H+] [OH-])
-log10Kw = -log10([H+] + (-log10[OH-] ---------- (ii)
-log10Kw = PKw ------------ (iii)
-log10([H+] = PH ----------- (iv)
-log10[OH-] = POH ------- (v)
Substituting equations (i), (ii) and (iii) in equation (ii), we have
PKw = PH + POH
Recall that, Kw = [H+] [OH-] ------ (i)
When water dissociates, equal concentration of hydrogen ion and hydroxyl ion are produced, hence,
[H+] = [OH-] ------ (ii)
Substituting [H+] for [OH-] in equation (i), it becomes
Kw = [H+] [H+]
Kw = [H+]2
[H+] = (Kw)1/2
[H+] = (10-14 mol2/dm6)1/2
= 10-14x1/2 mol2x1/2/dm6x1/2
PH = -log10([H+]
= -7(-log1010), but log1010
= -7 x -1
Formula of tetraoxosulphate (VI) acid is H2SO4(aq)
H2SO4(aq) 2H+(aq) + SO42-(aq)
At t=0 0.01M 0 0
At t=t 0 2 x 0.01M 0.01M
Stoichiometrically, 2 moles of hydrogen ions are produced from one mole of the acid
t = 0 refers to the beginning of the reaction, before dissociation commenced while t = t refers to the complete dissociation of the acid.
The equation shows that the acid dissociated completely to produce twice the number of mole of the acid started with.
From the equation, [H+] = 2 x 0.01 mol/dm3 = 0.02 mol/dm3
PH = ...
The expert examines PH calculations for basic principles. Ammonia and water pH values are determined.