By Sunil Bhardwaj

5530 Views


In 1908 G.N Lewis introduced the concept of activity and activity coefficient in order to explain the deviation of an electrolyte solution or gas from ideal behaviour. When the solution is highly diluted, it means completely dissociated into ions and all the ions are available for conductance. This solution is said to be ideal solution and for ideal solution effective concentration (observed) is equal to actual concentration (theoretical).

But the electrolyte which is not completely dissociated in that case effective (observed) concentration is always less than actual (theoretical) concentration.

Lewis introduced the term activity coefficient to explain the deviation from ideal behaviour. $$ \boxed { Activity \ coefficient \left( \gamma \right) = \frac { Effective \ Concentration \left( a \right) }{ Actual \ concentration \left( m \right) } } $$ or $$ \boxed { a = m\gamma } \qquad ....(1) $$ Consider an electrolyte $$ { A }_{ x }{ B }_{ y } \rightleftarrows x{ A }^{ + } + y{ B }^{ - } $$ That means one mole of \({ A }_{ x }{ B }_{ y }\) produces x moles of cations \(({ A }^{ + })\) and y moles of anions \(({ B }^{ - })\) in solution.

Therefore activity of above electrolyte is defined as the product of activities of cations and anions, $$ \boxed { a = { \left( { a }_{ + } \right) }^{ x }{ \left( { a }_{ - } \right) }^{ y } } \qquad ....(2) $$ But activity of cation \({ { a }_{ + } } = m_{ + }\gamma _{ + }\) where \(m_{ + }\) is concentration of cation and \(\gamma _{ + }\) is activity coefficient of cation.

As one mole of electrolyte produces x moles of cation, concentration of cation will be x times the concentration of electrolyte (i.e. \(m_{ + } = xm\)). $$ \therefore \boxed { { { a }_{ + } } = xm\gamma _{ + } } \\ similarly \ \boxed { { { a }_{ - } } = ym\gamma _{ - } } \qquad ....(3) $$ Lets put these values on equation (2), we get $$ a = { \left( xm\gamma _{ + } \right) }^{ x }{ \left( ym\gamma _{ - } \right) }^{ y } $$ $$= { \left( { x }^{ x }{ m }^{ x }{ \gamma _{ + } }^{ x } \right) }{ \left( { y }^{ y }{ m }^{ y }{ \gamma _{ - } }^{ y } \right) } $$ $$ = \left( { x }^{ x }{ y }^{ y } \right) { \left( { m }^{ x }{ m }^{ y } \right) }\left( { \gamma _{ + } }^{ x }{ \gamma _{ - } }^{ y } \right) $$ $$ = \left( { x }^{ x }{ y }^{ y } \right) { \left( { m }^{ x+y } \right) }\left( { \gamma _{ \pm } }^{ x+y } \right) $$ where \(\gamma _{ \pm }\) is mean activity coefficient

Thus $$ \boxed { a = { x }^{ x }{ y }^{ y }{ m }^{ x+y }{ \gamma _{ \pm } }^{ x+y } } \qquad ....(4) $$

Case 1: for uni-univalent electrolyte e.g. HCl, NaCl, KCl etc, $$ NaCl \longrightarrow { Na }^{ + } + { Cl }^{ - } $$ Here x = 1 and y = 1, Therefore equation (4) becomes $$ a = { 1 }^{ 1 }{ 1 }^{ 1 }{ m }^{ 1+1 }{ \gamma _{ \pm } }^{ 1+1 } $$ $$ \boxed { a = { m }^{ 2 }{ \gamma _{ \pm } }^{ 2 } } \qquad ....(5) $$

Case 2: for uni-bivalent electrolyte e.g. \({ Na }_{ 2 }{ CO }_{ 3 }, Zn{ Cl }_{ 2 }\), etc. $$ Zn{ Cl }_{ 2 } \longrightarrow { Zn }^{ +2 } + 2{ Cl }^{ - } $$ Here x = 1 y = 2, Therefore equation (4) becomes $$ a = { 1 }^{ 1 }{ 2 }^{ 2 }{ m }^{ 1+2 }{ \gamma _{ \pm } }^{ 1+2 } $$ $$ \boxed { a = 4{ m }^{ 3 }{ \gamma _{ \pm } }^{ 3 } } \qquad ....(6) $$

Case 3: for uni-trivalent electrolyte e.g. \(Al{ Cl }_{ 3 }\) $$ Al{ Cl }_{ 3 } \longrightarrow { Al }^{ +3 } + 3{ Cl }^{ - } $$ Here x = 1 y = 3 Therefore equation (4) becomes $$ a = { 1 }^{ 1 }3^{ 3 }{ m }^{ 1+3 }{ \gamma _{ \pm } }^{ 1+3 } $$ $$ \boxed { a = 27{ m }^{ 4 }{ \gamma _{ \pm } }^{ 4 } } \qquad ....(7)$$

Numerical Problem: Calculate the activity of \(Zn{ Cl }_{ 2 }\) solution which is 0.5 molal. The mean molal ionic activity coefficient \({ \gamma _{ \pm } } = 0.9\)

Ans:

Now \(Zn{ Cl }_{ 2 }\) is uni-bivalent electrolyte therefor x = 1 and y = 2

The equation for activity becomes, $$ a = x^{ x }y^{ y }{ m }^{ x+y }{ \gamma _{ \pm } }^{ x+y } $$ $$ \therefore a = { 1 }^{ 1 }{ 2 }^{ 2 }{ m }^{ 1+2 }{ \gamma _{ \pm } }^{ 1+2 } $$ $$ \boxed { a = 4{ m }^{ 3 }{ \gamma _{ \pm } }^{ 3 } } $$ Lets substitute the values, $$ \therefore a = 4 \times { \left( 0.5 \right) }^{ 3 } \times \left( 0.9 \right) ^{ 3 } $$ $$ = 4 \times { \left( 0.125 \right) } \times \left( 0.729 \right) $$ $$ = 0.3645 $$

Numerical Problem: Calculate the activity of NaCl solution which is 0.05 molal. The mean molal ionic activity coefficient \({ \gamma _{ \pm } } = 0.821\)

Numerical Problem: Calculate the activity of \(Ca{ Cl }_{ 2 }\) solution which is 0.5 molal. The mean molal ionic activity coefficient \({ \gamma _{ \pm } } = 0.510\)

Numerical Problem: Calculate the activity of \(Zn{ SO }_{ 4 }\) solution which is 0.01 molal. The mean molal ionic activity coefficient \({ \gamma _{ \pm } } = 0.387\)

Numerical Problem: Calculate the activity of HCl solution which is 0.001 molal. The mean molal ionic activity coefficient \({ \gamma _{ \pm } } = 0.996\)

Numerical Problem: Calculate the activity of \({ H }_{ 2 }S{ O }_{ 4 }\) solution which is 0.01 molal. The mean molal ionic activity coefficient \({ \gamma _{ \pm } } = 0.543\)