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Acid Base Equilibrium

Certain substances such as sodium chloride, potassium hydroxide conduct electricity in their aqueous solutions while some substances like aqueous solution urea, sugar cannot conduct electricity.

The substances which conduct electricity in their aqueous solution are termed as electrolytes while those which do not conduct electricity are termed as non-electrolyte. The dissolution of electrolyte in water split it into charged particles called as ions and the process is known as ionization or dissociation. Some electrolytes ionized completely in solution and known as strong electrolytes. For example, sodium chloride, hydrogen chloride etc are examples of strong electrolytes.

On the other hand, weak electrolytes are weakly ionized in their solutions like ammonium hydroxide, acetic acid. In solutions of weak electrolytes, the ions produced by dissociation of electrolytes are in equilibrium with undissociated molecules of electrolyte. For example, the dissociation of acetic acid in acetate ion and hydronium ion is written as follow.

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Electrolytes may be acids, bases or salts. In these ionic solids; ions are held together by strong electrostatic force of attraction. When such solids dissolved in solvent, the electrostatic interactions between ions reduce which facilitates the free movement of ions in the solution and get stabilized by their solvation with solvent molecules.

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Degree of Dissociation

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Acids like perchloric acid (HClO4), hydrochloric acid (HCl), nitric acid (HNO3) and sulfuric acid (H2SO4) are strong electrolytes as they are almost completely dissociates into their constituents ions in their aqueous solution and thereby acting as proton donors(H+). Similarly strong bases like potassium hydroxide (KOH), sodium hydroxide (NaOH) almost completely ionized in an aqueous medium giving hydroxyl ions.

Weak electrolytes like acetic acid (CH3COOH), ammonium hydroxide (NH4OH) are partially ionized. Therefore the ions produced as a result of dissociation of weak electrolytes are present in dynamic equilibrium with the undissociated molecules.

The fraction of total number of total number of molecules of electrolytes dissolved, that ionizes at equilibrium is known as degree of dissociation or degree of ionization.
For example, the ionization of some weak electrolytes AB in water can be written as given below.

AB(aq) A+(aq) + B-(aq)

Let ‘C’ be the concentration of electrolyte in the solution and α be the degree of
dissociation. The initial and equilibrium concentration can be shown as follow;

AB(aq) ⇔ A+(aq) + B-(aq)

Initial concentration C - -
Equilibrium concentration C-Cα Cα Cα

Hence, the equilibrium constant

K = $\frac{[A^+][B^-]}{[AB]}$ = $\frac{[Cα][Cα]}{C(1-α)}$


K = $\frac{C^2\ α^2}{C(1-α)}$ = $\frac{C\ α^2}{[1-α]}$

For weak electrolytes, under normal concentration, the degree of dissociation, α is very small as compared to unity and hence (1-α) can be taken as 1 in the denominator of equation (1).

So the equilibrium constant K = Cα2

or α2 = K/C

or α =

Hence, degree of dissociation is inversely proportional to concentration of given electrolyte, thus as the concentration decreases, the degree of dissociation increases. As the concentration C approaches to zero or dilution approaches to infinity, the degree of dissociation reaches to unity that is maximum value of that. This generalization is called as Ostwald dilution law. The concentration of hydronium ions can be calculated by using the relation between concentration and degree of dissociation.

[H3O+] = Cα

Ionic Product of Water

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Water is a weak electrolyte and weakly ionized to give H+(aq) and OH-(aq) ions.

Η2Ο(aq) Η+(aq) + ΟΗ-(aq)

Hence, the ionization constant of water can be written as

Kα(H2O) =$\frac{[H^+][OH^-]}{[H_2O}$

The concentration of water is very high and a small fraction of it undergoes ionization, therefore the concentration of water [H2O] may be taken as constant and combine with Kα(H2O). The sum of Kα(H2O) and [H2O] is called as ionic product of water , represented as Kw.

Kα(H2O) [H2O] = [H+] [OH-]
Kw = [H+] [OH-]

Experimentally at 298 K temperature, the concentration of H+ ion in pure water is found to be 1.0 x 10-7 mol/L. Since in pure water, the concentration of H+ ion is equal to the concentration of OH- ions as the dissociation of water produces equal number of H+ ions, therefore

[H+] = [OH-] = 1.0 x 10-7 mol/L
Kw = [H+] [OH-]
or Kw = [H3O+] [OH-]

Kw = [1.0 x 10-7 mol/L] [1.0 x 10-7 mol/L]
=1.0 x 10-14 mol2/L2

Since, the concentration of H+ and OH- ions in pure water is 10-7 at 298 K temperature and [H2O] is 55.55 M, therefore from equation (2)

Kα(H2O) = $\frac{[10^{-7}] [10^{-7}]}{55.55}$

= 1.8 x 10-16

Hence, the ionization constant of water is differ from ionic product of water and can be written as

Kα(H2O) [H2O] = Kw

The degree of dissociation of water, α= [H+] / C

α = 10-7 mol/L / 55.55 mol/L
= 1.8 x 10-9

The ionic product of water increases with temperature. However at all temperature the concentration of H+ ions remains equal to the concentration of OH- ions in pure water. The following table listed the value of ionic product of water at different temperatures.

Temperature (K)
Ionic product of water (Kw)
373 (56) x 10-14
5.474 x 10-14
2.917 x 10-14
1.008 x 10-14
0.687 x 10-14
0.292 x 10-14
0.113 x 10-14

In pure water the concentration of H+ and OH- ions is same but on the addition of some acid or base to water these concentrations no longer remain equal, but the value of ionic product of water remains same at a particular temperature.

For example, the addition of acid to water increases the concentration of hydronium ions, therefore the equilibrium of dissociation of water gets shifted in the reverse direction and hydronium ions H3O+ would combine with hydroxyl ion OH- to form undissociated water molecules and the value of Kw in solution may remain the same as that in pure water.

Hence, the addition of acid to pure water decreases the hydroxyl ions concentration to the relation.

[OH-] = Kw/[H3O+]

Similarly, the addition of base increases the concentration of hydroxyl ions and decreases the hydronium ions in solution to the relation.
[H3O+] = Kw /[OH-]

Acid Base Equilibrium Problems

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Below you could see problems

Solved Examples

Question 1: Calculate the degree of dissociation and concentration of hydronium ions of 0.01 M solution of CH3COOH,
K of CH3COOH is 4.8 x 10-10.
The ionization of acetic acid can be represented as

CH3COOH(aq) + H2O(l) ⇔ CH3COO- (aq) + H3O+(aq)

The degree of dissociation of acetic acid is α and concentration is ‘C’
then the equilibrium concentration of all species would be

[CH3COOH] =C(1-α)
[CH3COO-]= Cα
[H3O+]= Cα

The dissociation constant of acid,
K = $\frac{[CH3COO^-] [H_3O^+]}{[CH_3COOH]}$
= Cα2
= (1-α)
Since α is very small compare to unity, therefore 1-α  may be taken as 1 and  K = Cα2.

α =√ K/C   =√4.8 x 10-10/0.01    = 2.2 x 10-4
3Ο+] = Cα = 0.01 x 2.2 x 10-4 = 2.2 x 10-6 mol/L


Question 2: What would be the degree of ionization of 0.02 M acetic acid (Ka = 1.8 x 10-5M).
Calculate the degree of dissociation of same acid, if the solution also contains 0.01 M CH3COONa (sodium acetate).
Degree of dissociation
α =  √ Ka/C =  √ 1.8 x 10-10/0.02 = 0.03
Sodium acetate is a strong electrolyte and would be completely ionized in solution.
Assume  ‘x’ mol/L of acetic acid be ionized.

CH3COOH(aq) CH3COO-(aq) + H+(aq)
(0.02 –x)M                   x M                   x M
CH3COONa(aq) → CH3COO-(aq) + Na+(aq)
0.01 M            0.01 M
[H+]    = x mol/L,
[CH3COO-]   = (x+ 0.01)mol/L = 0.01 mol/L
[CH3COOH] = (0.02 –x) mol/L=0.02 mol/L
Ka = $\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$
1.8 x 10-5 = X x 0.01/ 0.02 
X = [H+] = 3.6 x 10-5 M

Degree of dissociation, α = x / 0.02 = 3.6 x 10-5/0.02 = 1.8 x 10-4


Question 3: Calculate the concentration of hydronium ions and hydroxyl ions in

a) 0.01 M solution of HCl.
b) 0.01 M solution of NaOH at 298 K temperature.

Assuming that HCl and NaOH are completely ionized under given conditions.
a) ΗCl(aq) + H2O(l)  ⇔  H3O+(aq) + Cl-(aq)

Since HCl is a strong electrolyte, therefore it will be completely ionized and the concentration of acid and hydronium ions would be same.

[HCl] = [ H3O+] = 0.01M or 1 x 10-2 mol/L
[OH-] = [Kw]/[H3O+]
= 1 x 10-14/1 x 10-2 = 1 x 10-12 mol/L

b) NaOH (aq) → Na+ (aq)  + OH- (aq)

Since NaOH is a strong electrolyte, therefore

[OH-] = [NaOH] = 0.01M or 1 x 10-2 mol/L
[H3O+]  = [Kw]/ [OH-]
= 1 x 10-14/1 x 10-2 = 1 x 10-12 mol/L


Question 4: Calculate the concentration of hydronium ions in 0.001 M solution of barium hydroxide (CsOH) at 298 K temperature,
assuming that CsOH is completely ionized under the given conditions.
CsOH(aq) → Cs+(aq)   + OH-(aq)
Since cesium hydroxide is completely ionized, so one molecule of base will give one hydroxyl ion.

Hence [OH-] = [CsOH] = 0.001 = 1 x  10-3 mol/L
[H3O+] = Kw/[OH-] = 1 x 10-14/1x 10-3 = 1 x 10-11 mol/L


Question 5: The degree of dissociation of a 0.1 M acetic acid solution is 0.132.
What would be the concentration of H+ ion and ionization constant K of acetic acid.
Acetic acid is a weak electrolyte and dissociate in acetate ion and hydrogen ion.

CH3COOH ⇔  CH3COO-   + H+

If the initial concentration of acetic acid is taken as, C then initial and equilibrium concentration of all species would be;
                                    CH3COOH  ⇔  CH3COO+ H+

                                   Initial concentration               C               0               0                               
Equilibrium concentration       C-Cα        Cα          Cα
 Where α is degree of  dissociation given as 0.132.
Dissociation constant K = Cα2/(1-α)  or Cα2= 0.1 x (0.132)2 = 1.32 x 10-2 M
Concentration of H+; [H+] = Cα = 0.1 x 0.132 = 1.32 x 10-2 M


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