By Sunil Bhardwaj

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Definition: The products of activity of ions of sparingly soluble salt at a given temp and in a saturated solution is constant which is known as solubility product of sparingly soluble salt.

e.g. AgCl; $${ K }_{ sp } = { a }_{ { Ag }^{ + } } \times { a }_{ { Cl }^{ - } }$$and for dilute solution $${ K }_{ sp } = C_{ { Ag }^{ + } } \times { C }_{ { Cl }^{ - } }$$ and $${ K }_{ sp } = { S }^{ 2 },$$ where S is solubility. $$\therefore S = \sqrt { { K }_{ sp } }$$ One of the important application of emf is that with the help of it we can find out the solubility product and solubility of sparingly soluble salt.

Procedure: Take two beakers of 50ml capacity or 100ml as available. Take 50ml of 0.1N $$AgN{ O }_{ 3 }$$, in one beaker and 50ml of 0.1M KCl in the other beaker. That beaker which contains KCl in the same you add few drops of$$AgN{ O }_{ 3 }$$, then $$KCl + AgN{ O }_{ 3 } \longrightarrow AgCl\downarrow + KN{ O }_{ 3 }$$ We get turbidity of AgCl. Now this AgCl will dissociate sparingly into its ions. AgCl $$\longrightarrow { Ag }^{ + } + { Cl }^{ - }$$ If we insert the Ag plate (wire) in the two beakers and we connect the two beakers by $$KN{ O }_{ 3 }$$, salt bridge, we get cell as follows- $$\circleddash Ag | \underset { { a }_{ 1 } }{ AgCl (s) } | \underset { 0.1M }{ KCl } || \underset { { a }_{ 2 }, 0.1M }{ AgN{ O }_{ 3 } } | Ag\oplus$$

Working: at L.H.E. $$Ag \longrightarrow { Ag }_{ { a }_{ 1 } }^{ + } + { e }^{ - } (Ox)$$ at R.H.E. $${ Ag }_{ { a }_{ 2 } }^{ + } + { e }^{ - } \longrightarrow Ag (Red)$$ Net cell reaction. $${ Ag }_{ { a }_{ 2 } }^{ + } \longrightarrow { Ag }_{ { a }_{ 1 } }^{ + }$$ $$\therefore K = \frac { { a }_{ { Ag }^{ + } }(1) }{ { a }_{ { Ag }^{ + } }(2) }$$ On applying the Nernst equation we get, $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { RT }{ F } \ln { \left( \frac { { a }_{ { Ag }^{ + } }(1) }{ { a }_{ { Ag }^{ + } }(2) } \right) }$$ $${ E }_{ cell } = 0 - \frac { RT }{ F } \ln { \left( \frac { { a }_{ { Ag }^{ + } }(1) }{ { a }_{ { Ag }^{ + } }(2) } \right) }$$ As both are same electrode $${ E }_{ cell } = \frac { RT }{ F } \ln { \left( \frac { { a }_{ { Ag }^{ + } }(2) }{ { a }_{ { Ag }^{ + } }(1) } \right) }$$ $${ E }_{ cell } = 0.059\log { \left( \frac { { a }_{ { Ag }^{ + } }(2) }{ { a }_{ { Ag }^{ + } }(1) } \right) }$$ $${ E }_{ cell } = 0.059\left[ \log { { a }_{ { Ag }^{ + } }(2) } - \log { { a }_{ { Ag }^{ + } }(1) } \right]$$ $$\frac { { E }_{ cell } }{ 0.059 } = \log { { a }_{ { Ag }^{ + } }(2) } - \log { { a }_{ { Ag }^{ + } }(1) }$$ $$\therefore \log { { a }_{ { Ag }^{ + } }(1) } = \log { { a }_{ { Ag }^{ + } }(2) } - \frac { { E }_{ cell } }{ 0.059 }$$ Or $$\log { { a }_{ { Ag }^{ + } }(1) } = \log { { \left( { m }_{ 2 }{ \gamma }_{ 2 } \right) } } - \frac { { E }_{ cell } }{ 0.059 }$$ If we know molality m and $$\gamma$$ of Ag on RHS and $${ E }_{ cell }$$ then by using above equation we can find out $${ a }_{ { Ag }^{ + } }$$(1) which are involved in AgCl.

Then, $${ K }_{ sp } = { a }_{ { Ag }^{ + } } \times { a }_{ { Cl }^{ - } } or { K }_{ sp }$$ $$= { a }_{ { Ag }^{ + } } \times { m }_{ { Cl }^{ - } } \times { \gamma }_{ { Cl }^{ - } }$$ As all the terms are known to us, then by putting the values we can find out $${ K }_{ sp }$$.