Though the multiple extraction is more beneficial than single step extraction. For analytical chemist it is easy to find out the amount extracted in single extraction, so he follow single extraction rather than multiple extraction because it is tedious. The relation between percentage extraction (E) for single extraction can be derived as follow: We know $$ { W }_{ 1 } = \left[ \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right] \times a \qquad ...(1) $$ where \({ W }_{ 1 }\) is the weight of solute remaining after 1st extraction.
\( { V }_{ W } \)= Volume of Aqueous Phase
\( { V }_{ o } \) = Volume of Organic Phase
\(D\) = Distribution Ratio
\(a\) = Total weight of Solute present initially.
The amount extracted will be a $$ { W }_{ 1 } = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \qquad ...(2) $$ $$ = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) $$ $$ \therefore Percentage \ extraction \ (E) = \left( \frac { a - { W }_{ 1 } }{ a } \right) \times 100 $$ Lets Substitute the values from equation (1) $$ E = \left( \frac { a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) }{ a } \right) \times 100 $$ $$ E = \left( 1 - \left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \right) \times 100 $$ $$ E = \left( \frac { { V }_{ o }D + { V }_{ W } - { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ $$E = \left( \frac { { V }_{ o }D }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ on dividing the numerator and denominator of RHS by \({ V }_{ o }\) we get $$ E = \left( \frac { D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } \right) \times 100 $$ $$ \therefore Percentage extraction (E) = \frac { 100D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } $$ Thus % extraction (E) depends on D as well as \(\frac { { V }_{ W } }{ { V }_{ o } } \).
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