Pure water being a weak electrolyte under goes self ionization to a small extent as follows:
The equilibrium constant for this reaction is:
The concentration of unionized water is taken as constant because the degree on ionization of water is very small. So we can write this equation as:
where Kw is a constant and is known as the ionic product of water whose value is 1.008 x 10-14 mol2 L-2at 298 K. In pure water the concentration of H3O+ and OH- are equal and so we can write,
[H3O+] = [OH-]
If, Kw = [H3O+] [OH-] = 1.008 x 10-14 mol2 L-2 then,
[H3O+] [OH-] = 1.008 x 10-14
[H3O+]2 = 1.008 x 10-14
Thus in pure water [H3O+] = [OH-] = 1.0 x 10-7 mol L-1 at 298 K
Effect of temperature on K
The value of Kw varies with the change in temperature. The values of [H3O+] and [OH-] are always equal to each other at all temperatures but the values of Kw are different at different temperatures. The value of Kw increases with the rise in temperature. This is because increase in temperature will shift the equilibrium in the forward direction producing large concentrations of [H3O+] and [OH-] ions (Le Chatelier's principle).
Hence, Kw increases with rise in temperature.
In acidic solution
When an acidic solution of HCl is added to a pure neutral solution of water, the concentration of [H3O+] becomes larger than 1.0 x 10-7 mol L-1. The dissociation equilibrium of water shifts in the reverse direction (Le Chatelier's principle). The excess [OH-] ions combine with hydronium ions to form undissociated water molecules so that the value of Kw remains constant in the solution. The concentration of [OH-] ions will then be equal to
The concentration of [H3O+] is more than the concentration of the [OH-] ions in acidic solution.When a few drops of a base like NaOH is added to pure water, the concentration of [OH-] increases and that of hydrogen ions decreases. The concentration of [H3O+] can be calculated as:
Thus in basic solution the concentration of [OH-] will be greater than of [H3O+].
It can be concluded that the hydronium and hydroxyl ions are always present in solution whether they are acidic or basic. However their concentrations differ.
6. Calculate the hydronium and hydroxyl ion concentrations in (i) 0.01 M HCl (ii) 0.001 M NaOH solution at 298 K.
(i) HCl completely ionizes as:
The concentration of hydronium ions is equal to that of hydrochloric acid,[H3O+] = [HCl]
[HCl] = 0.01M = 1 x 10-2 mol L-1The ionic product of water is Kw = [H3O+] [OH-]
Thus [H3O+] = 1 x 10-2 mol L-1 and [OH-] = 1 x 10-12 mol L-1
(ii) NaOH ionizes completely as:NaOH(aq) Na+(aq) + OH-(aq)
The concentration of [OH-] is equal to that of NaOH i.e.,[NaOH] = [OH-]
[NaOH] = 0.001 M = 1 x 10-3 mol L-1[OH-] = 1 x 10-3 mol L-1
Kw = [H3O+] [OH-]
Thus, [OH-] = 1 x 10-3 mol L-1and [H3O+] = 1 x 10-11 mol L-1.
Ph and ph scale
In 1909, Sorensen introduced a term for expressing the concentration of hydrogen ions, which give an idea about the acidic and basic characters of the aqueous solution. This term was called 'pH' which means the 'power of hydrogen ions'. The pH is defined as "the negative logarithm of the H3O+ ion concentration in moles per litre".
For neutral solution at 298 K,
[H3O+] = [OH-] = 1.0 x 10-7 mol L-1so that, pH = -log [H3O+] = -log (1.0 x 10-7) = 7
Substituting different values for [H3O+] in the above relation we have,For acidic solution pH <>
For basic solution pH > 7For neutral solution pH = 7
A scale called as the pH scale is devised to express the acidic and basic properties of solution in terms of the pH value.
Fig: 8.1 - The pH scaleFrom the scale it is clear that for solutions with
pH between 0 to 2 strongly acidicpH between 2 to 4 moderately acidic
pH between 4 to 7 weakly acidicpH between 7 to 10 weakly basic
pH between 10 to 12 moderately basicpH between 12 to 14 strongly basic.
6. Calculate the pH value of (i) 0.001 M HCl and (ii) 0.01 M NaOH
(i) Since HCl is a strong acid, it completely ionizes and therefore, H3O+ ions concentration is equal to that of the acid itself i.e.,[H3O+] = [HCl] = 0.001 M = 1 x 10-3 M
now, pH = -log [H3O+]pH = -log [1 x 10-3]
= -(-3) log 10 = 3 (log 10 =1)(ii) Since NaOH is a strong base, it completely ionizes and therefore, OH- ions concentration is equal to that of the base itself i.e.,
[OH-] = [NaOH] = 0.01 M = 1 x 10-2 MKw = [H3O+] [OH-]
pH = -log [H3O+]
pH = -log [1 x 10-12]= -(-12) log 10 = 12
7. Calculate the pH of a solution whose hydronium ion concentration is 6.2 x 10-9 mol L-1.
[H3O+] = 6.2 x 10-9 MpH = -log [H3O+]
pH = -log [6.2 x 10-9]= -(log 6.2 - 9 log 10)
= -log 6.2 + 9 x 1 (log 6.2 = 0.79)= 9 - 0.79 = 8.21
8. Acid A, B, C and D have the following pKa values: A = 1.5, B = 3.5, C = 2.0, D = 5.0. Arrange these acids in the increasing order of acid strength.
We know that,pKa = -log Ka or Ka = 10-pKa
Therefore, for the given acids,Ka (A) = 10-1.5 Ka (B) = 10-3.5
Ka (C) = 10-2.0 Ka (D) = 10-5.0Since, 10-5.0 <>-3.5 <>-2.0 <>-1.5
Hence, the strength of acids follows the order, D <>9. The value of Kw is 9.55 x 10-14 at a certain temperature. Calculate the pH of water at this temperature.
Kw = 9.55 x 10-14For water [H3O+] = [OH-]
If, Kw = [H3O+] [OH-] = 9.55 x 10-14 then,[H3O+] [H3O+] = 9.55 x 10-14
[H3O+]2 = 9.55 x 10-14
pH = -log [H3O+]pH = -log [3.09 x 10-7]
= -(log 3.09 + log 10-7)= -(0.49 - 7) = 6.51
10. What is the pH of a solution whose hydrogen ion concentration is 0.005 x 10-3 kg dm-3?
In the solution [H+] = 0.005 x 10-3 kg dm-3 = 0.005 x 10-3x 103Lm-3= 0.005 g dm-3 = 0.005 mol dm-3
11. The pH of blood is maintained at 7.4 due to the presence of HCO-3 and H2CO3. If Ka of H2CO3 in blood is 8 x 10-7 calculate the ratio [HCO-3]:[H2CO3] in blood.
pH = 7.4= -log[H+]log[H+]= -7.4 = 8.6 [H+]= 3.98 x 10-8 4 x 10-8 M