5401 Views
9386 Views
5232 Views
4129 Views
12404 Views
5116 Views
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 \({ K }_{ sp }\).
4467 Views
5279 Views
4477 Views
5001 Views
3804 Views
Apply to join the passionate instructors who share their expertise and knowledge with the world. You'll collaborate with some of the industry's best producers, directors, and editors so that your content is presented in the best possible light..