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 \({ K }_{ sp }\).