Explain the separation factor?

If a solution contains two or more solutes say A and B, it is observed that when A is extracted, some amount of B is also extracted. The extent of the seperation can be expressed in terms of one factor called Seperation Factor \(\beta\) . This is related to the distribution ratio of A and B. $$ \boxed { \beta = \frac { D_{ A } }{ { D }_{ B } } = \frac { { { C }_{ o(A) } }/{ { C }_{ a(A) } } }{ { { C }_{ o(B) } }/{ { C }_{ a(B) } } } } $$ It is ratio therefore no unit and no dimension.

The larger value is always placed in the numerator. \beta must be made as large as possible by choice of extractant and by adjusting the volume ratio.

When \(D_{ A }\) = 10 and \(D_{ B } \)= 0.1. The \(\beta = \frac { 10 }{ 0.1 } = 100%\)

single extraction in case will remove 91% of A and 9% of B. It can be obtained for A, $$ E = \left[ \frac { 100{ D }_{ A } }{ { D }_{ A } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ for $${ V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 10 }{ 10 + 1 } \right] = \left[ \frac { 1000 }{ 11 } \right] = 90.9% $$ $$ Similarly \ for \ B, \ E = \left[ \frac { 100{ D }_{ B } }{ { D }_{ B } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ $$ for \ { V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 0.1 }{ 0.1 + 1 } \right] = \left[ \frac { 10 }{ 1.1 } \right] = 9.1% $$ The seperation of A is almost complete from B if the seperation factor B is high. It can be only in the case when \({ D }_{ A }\) is large and \({ D }_{ B }\) is small. For a given value of \({ D }_{ A }\) and \({ D }_{ B }\), the seperation effect can be increased by adjusting the volume ratio given by Nush Densen Equation which says. $$ \boxed { \frac { { V }_{ o } }{ { V }_{ W } } = \frac { 1 }{ { \left( { D }_{ A }{ D }_{ B } \right) }^{ { 1 }/{ 2 } } } } $$


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