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Ionic Equilibrium — Chemistry STD 12 — Question

Tamilnadu BoardEnglish MediumSTD 12ChemistryIonic Equilibrium3 Marks
Question
  1. 1. Point out the differences between ionic product and solubility product.
    2. The solubllity of AgCl in water at 298 K is $1.06 \times 10^{-5}$ mole per litre. Calculate is solubility product at this temperature.
✓

Answer

1. Ionic product
1. It is applicable to all types of solutions.
2. Its value changes with the change in con centration of the ions.
Solubility product
1. It is applicable to the saturated solutions.
2. It has a definite value for an electrolyte at a constant temperature.
2. The solubility equilibrium in the saturated solution is $AgCl ( s ) \rightleftharpoons Ag ^{+}( aq )+ Cl ^{-}( aq )$
The solubility of $AgCl$ is $1.06 \times 10^{-5}$ mole per litre.
${\left[ Ag ^{+}( aq )\right]=1.06 \times 10^{-5} mol L ^{-1}}$
${\left[ Cl ^{-}( aq )\right]=1.06 \times 10^{-5} mol L ^{-1}}$
$K _{ sp }=\left[ Ag ^{+}( aq )\right]\left[ Cl ^{-}( aq )\right]$
$=\left(1.06 \times 10^{-5} mol L ^{-1}\right) \times\left(1.06 \times 10^{-5} mol L ^{-1}\right)$
$=1.12 \times 10^{-2} mol ^2 L ^{-2}$

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