Introduction
The chemical substance which dissolves in water can be classified into two types; non-electrolytes and electrolytes. The non-polar substances which do not produce ions in solution or in the molten state are called non-electrolytes. They do not dissociate into ions and do not conduct electricity. eg. glucose, benzene, urea, alcohol, etc.
The chemical substances which produce ions in solution or in the molten state are called electrolytes. They are good conductors of electricity as they dissociate into ions. eg. HCl, NaCl, H2SO4, CuSO4, etc.
The electrolyte which undergoes complete ionization in an aqueous solution is called the strong electrolyte. eg. HCl, H2SO4, HNO3, NaOH, KOH, NaCl, KCl, CH3COONa, NH4Cl, etc. An electrolyte that is partially ionized in an aqueous solution and partly remains in undissociated molecules is called a weak electrolyte. eg. HCOOH, CH3COOH, HCN, H2CO3, H3PO4, NH4OH, Ca(OH)2, etc.
Arrhenius theory of ionization
To express the conducting part of an electrolyte, Arrhenius proposed a theory in 1887. Followings are the postulates of this theory:
- When an electrolyte is dissolved in water, it splits up into two types of oppositely charged particles called ions and the process is called ionization.
- The positively charged particles are called cations and the negatively charged particles are called anions.
- The sum of positively charged on cation is equal to the sum of negatively charged on anion. So an electrolyte in a whole is electrically neutral.
- The electrical conductivity of an electrolyte in a solution is due to flow of Ions towards oppositely charged electron.
- The electrical conductivity depends upon the nature of ions present in a solution of an electrolyte.
- In a solution of an electrolyte, there exist a dynamic equilibrium between unionised molecules and ions present in the solution.
- The strength of an electrolyte is expressed in terms of degree of ionization (α) which measures the strength of ionization. It is defined as the fraction of total number of moles which undergoes into ionization.
\alpha = \frac{no.\ of\ moles\ split\ into\ ions}{total\ no.\ of\ moles\ dissolved}
- The property of an electrolyte are the properties of ion produced by it in a solution.
Ostwald’s dilution law
According to Arrhenius theory of ionization, there exists a dynamic equilibrium between unionized molecules and ions present in the solution of an electrolyte.
Let us consider an electrolyte AB with concentration c mol/lit. and with a degree of ionization (α) is under dilution, then following equilibrium holds good.
\underset{unionized\ molecule}{AB} \rightleftharpoons A^{+} + B^{-}
The concentration terms at equilibrium is given by:
[AB] = (c-αc)mol/lt = c(1-α)mol/lt
[A+] = αc mol/lt
[B–] = αc mol/lt
Applying the law of mass action:
Rate of ionization = k1[AB] = k1c(1-α)
Rate of combination = k2[A+][B–] = k2αc x αc
where k1 and k2 are the constants for the process of ionization and combination respectively.
At equilibrium,
Rate of ionization = Rate of combination
K_{1}[AB] = K_{2}[A^{+}][B^{-}]\\ \frac{[A^{+}][B^{-}]}{[AB]} = \frac{K_{1}}{K_{2}}=k
where k is another constant called ionization constant.
\frac{\alpha c\ \times\ \alpha c}{c(1-\alpha)}=k\\ \frac{\alpha ^{2}c}{(1-\alpha)}=k\\ For\ a\ weak\ electrolyte, \alpha <<1\\ so, 1-\alpha = 1.\\ then,\\ \frac{\alpha^{2}c}{1}=k\\ \alpha^{2}=k/c\\ \alpha = \sqrt{\frac{k}{c}}\\ \alpha\ \alpha\ \sqrt{\frac{1}{c}}
This is the mathematical expression of Ostwald’s dilution law which shows the concentration dependence of the degree of ionization.
It can be stated as, “The degree of ionization of weak electrolyte is inversely proportional to the square root of molar concentration”.
Limitations
It is only applicable for weak electrolytes. Strong electrolytes ionize completely. For strong electrolytes, 1-α= cannot be equal to 1. So the equation becomes k = α2/0 which is undefined.
Factors affecting degree of ionization
1. Nature of electrolyte: It is the chief factor that influences the degree of ionization. Strong electrolyte undergoes complete ionization. So they have a high degree of ionization. Weak electrolyte undergoes partial ionization. So they have a low degree of ionization.
2. Nature of solution: The degree of ionization of an electrolyte is directly proportional to the dielectric constant of the solvent. For example, the dielectric constant of water is greater than that of dimethyl sulfoxide (DMSO). Therefore the degree of ionization of a given electrolyte is higher in water than in DMSO.
3. Dilution: The degree of ionization increases with an increase in dilution or decrease in concentration.
4. Temperature: On increasing in temperature, the degree of ionization increases.
5. Common Ion effect: The presence of a strong electrolyte in the solution of a weak electrolyte having a common ion suppresses the ionization of the weak electrolyte. For example, ionization of weak acid (CH3COOH) is decreased in presence of the strong electrolyte (CH3COONNa). This effect is simply called the common Ion effect.
Strength of acid and base in terms of Ka and Kb
The strength of acid and base is defined as the capacity to give H+ and OH- ions in an aqueous solution respectively.
Consider an acid HA in an aqueous solution,
HA + H_{2}O\ \rightleftharpoons\ H_{3}O^{+} + A^{-}
Applying the law of mass action, the equilibrium constant is given by:
k = \frac{[H_{3}O^{+}][A^{-}]}{[HA][H_{2}O]}
\frac{[H_{3}O^{+}][A^{-}]}{[HA]}=k[H_{2}O]=k_{a}
Here, Ka is called the dissociation constant for acid.
For example, Ka at 25°C for CH3COOH is 1.8 x 10-5 and HCN is 6.2 x 10-10. The higher the value of Ka, the stronger is the acid. So CH3COOH is stronger than HCN.
Consider an acid B in an aqueous solution,
B + H_{2}O\ \rightleftharpoons\ BH^{+} + OH^{-}
Applying the law of mass action, the equilibrium constant is given by:
k=\frac{[BH^{+}][OH^{-}]}{[B][H_{2}O]}\\ \frac{[BH^{+}][OH^{-}]}{[B]}=k[H_{2}O]=k_{b}
Here, Kb is the dissociation constant for the base. For example, Kb for NH4OH is 1.81 x 10-5 and for C6H5NH2 is 4.7 x 10-10. Higher the value of Kb, the stronger is the base. So, C6H5NH2 is a weaker base than NH4OH.
pKa and pKb values
pKa and pKb can be calculated by the formula like pH value. Thus,
pKa = -log[Ka] and pKb = -log[Kb]
The pH of a solution depends upon pK values of acid and base.
i. When pKa = pKb, then pH = 7, the solution is neutral.
ii. When pKa > pKb, then pH > 7, the solution is basic.
iii. When pKa < pKb, then pH < 7, the solution is acidic.
The acid with a lower value of pKa will be stronger acid and the base with a lower value of pKb will be a stronger base.
Ionic product of water (Ionization constant of water)
Water is a weak electrolyte. In water, the following equilibrium holds good:
H2O ⇌ H+ + OH–
Applying law of mass action,
k=\frac{[H^{+}][OH^{-}]}{[H_{2}O]}
Being an electrolyte, only a small extent of water undergoes ionization. So the concentration of the unionized part of the water is taken as constant.
\begin{align*} [H_{2}O]&= constant\\ K &=\frac{[H^{+}]][OH^{-}]}{constant}\\ [H^{+}]][OH^{-}]&=constant \times K\\ [H^{+}]][OH^{-}]&=K_{w} \end{align*}
Kw is another constant called the ionic product of water.
pKa and pKb values
pKa and pKb can be calculated by the formula like pH value. Thus,
pKa = -log[Ka] and pKb = -log[Kb]
The pH of a solution depends upon pK values of acid and base.
i. When pKa = pKb, then pH = 7, the solution is neutral.
ii. When pKa > pKb, then pH > 7, the solution is basic.
iii. When pKa < pKb, then pH < 7, the solution is acidic.
Solved Numerical Examples
1. What will be the H+ ion concentration of a solution having pH 5.5?
Solution:
pH = -log[H+]
[H+] = antilog[-pH] = antilog[-5.5] = 3.16 x 10-6 M
2. 10-2 mole of KOH is dissolved in 10 litres of water. What will be the pH of solution?
solution:
\begin{align*} Molarity\ of\ KOH &=\frac{no.\ of\ moles}{Vol.\ in\ litre}\\ &= \frac{10^{-2}}{10}=10^{-3}M \end{align*}
KOH ionizes as:
\underset{10^{-3}M}{KOH}\ \rightleftharpoons K^{+}+ \underset{10^{-3}M}{OH^{-}}
pOH = -log[OH–] = -log[10-3] = 3
pH = 14 – pOH = 14 – 3 = 11
3. Calculate the pH of aqueous solution of 10-7 moles of NaOH ?
solution:
\underset{10^{-7}M}{NaOH}\ \rightleftharpoons Na^{+}+ \underset{10^{-7}M}{OH^{-}}
Here the ionization of water should not be neglected. so,
\underset{10^{-7}M}{H_{2}O}\ \rightleftharpoons H^{+}+ \underset{10^{-7}M}{OH^{-}}
Total concentration of OH– = 10-7M + 10-7M = 2 x 10-7M
pOH = -log[OH–] = -log[2 x 10-7] = 6.698
pH = 14 – pOH = 14 – 6.698 = 7.302