Question

What is the \[{K_{sp}} \] expression for lead (II) chloride?

Answer 1

Hint: -Solubility product constant (\[{K_{sp}} \])
The \[{K_{sp}} \] expression for a salt is the product of the ion concentrations, each concentration raised to the power of the ion's coefficient in the balanced equation for solubility equilibrium.

Complete answer:
The solubility product constant, \[{K_{sp}} \], denotes the equilibrium constant for the dissolution of a solid substance in an aqueous solution. It is the concentration of a solute in solution at which it dissolves. A substance's \[{K_{sp}} \] value increases in proportion to its solubility.
Let’s write down the equation for lead (II) chloride.
That is \[PbC{l_2}\underset {} \leftrightarrows P{b^{2 + }}(aq) + \,2C{l^ - }(aq)\]
Thus, Between the solid lead (II) chloride and the dissolved ions, an equilibrium governed by the solubility, product constant, \[{K_{sp}} \], will be established. At 298 K, the solubility product constant, \[{K_{sp}} \], of lead (II) chloride was determined to be \ [1.59 \times {10^{ - 5}} \pm 6.0 \times {10^{ - 7}}\].
Hence \[{K_{sp}} \] will be written as
\[{K_{sp}} \] =\ [\left\lfloor {P{b^{2 + }}} \right\rfloor + {\left[ {C{l^ - }} \right]^2}\]
Therefore, this is the final answer \[{K_{sp}} \] =\[1.59 \times {10^{ - 5}} \pm 6.0 \times {10^{ - 7}}\].
Also, to find the\[{K_{sp}} \], multiply the molarities or concentrations of the products. If any of the products have coefficients in front of them, the product must be raised to the coefficient power (and also multiply the concentration by that coefficient).
Solids are omitted from equilibrium constant expressions because their concentrations have no effect on the expression; any change in their concentrations is therefore insignificant. As a result, \[{K_{sp}} \] denotes the maximum extent to which a solid can dissolve in solution. Instead of concentrations found in slightly soluble solutions, ionic activities must be found for highly soluble ionic compounds.

Note:
Common Ion Effect: The solubility of the reaction is reduced by the common ion. For a given equilibrium, a reaction with a common ion present has a lower \[{K_{sp}} \] ​, and the reaction without the ion has a greater \[{K_{sp}} \].
Salt Effect (diverse ion effect): Having an opposing effect on the \[{K_{sp}} \] value compared to the common ion effect, uncommon ions increase the \[{K_{sp}} \] value. Uncommon ions are ions other than those involved in equilibrium.
Ion Pairs: With an ionic pair (a cation and an anion), the \[{K_{sp}} \] value calculated is less than the experimental value due to ions involved in pairing. To reach the calculated \[{K_{sp}} \] value, more solute must be added.