Question

Find the solubility product of a saturated solution of ${\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$ in water at 298 K if the emf of the cell ${\text{Ag}}\left| {{\text{A}}{{\text{g}}^{\text{ + }}}} \right.$ (satd. ${\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$ soln.)$\left\| {{\text{A}}{{\text{g}}^{\text{ + }}}} \right.\left( {0.1{\text{M}}} \right)\left| {{\text{Ag}}} \right.$ is $0.164{\text{V}}$ at 298K.

Answer 1

Hint: The solubility product of a salt at a given temperature is calculated from the product of molar concentrations of the ions (formed in the saturated solution at a given temperature) raised to the power equal to the number of times each ion occurs in the equation for solubility equilibrium.
The concentrations of the ions, in turn, are calculated from the Nernst equation for equilibrium.

Complete step by step answer:
The cell reaction which corresponds to the given cell ${\text{Ag}}\left| {{\text{A}}{{\text{g}}^{\text{ + }}}} \right.$ (satd. ${\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$ soln.)$\left\| {{\text{A}}{{\text{g}}^{\text{ + }}}} \right.\left( {0.1{\text{M}}} \right)\left| {{\text{Ag}}} \right.$ is as shown below.
\[{\text{Ag}}\left( {\text{s}} \right) + {\text{A}}{{\text{g}}^ + }\left( {{\text{aq}}} \right) \rightleftharpoons {\text{A}}{{\text{g}}^ + }\left( {{\text{aq}}} \right) + {\text{Ag}}\left( {\text{s}} \right)\]
The Nernst equation for this cell reaction at equilibrium is:
${{\text{E}}^{\text{0}}}_{{\text{cell}}} = \dfrac{{0.0591}}{{\text{n}}}\log {{\text{K}}_{\text{c}}}$
Here, ‘n’ is the number of electrons involved in the cell reaction and ${{\text{K}}_{\text{c}}}$ is the equilibrium constant for the reaction.
In this reaction, the number of electrons involved is 1 and so n is equal to 1. Hence, the Nernst equation for this cell reaction is:
${{\text{E}}^{\text{0}}}_{{\text{cell}}} = \dfrac{{0.0591}}{1}\log \dfrac{{{{\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]}_{{\text{RHS}}}}}}{{{{\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]}_{{\text{LHS}}}}}}$
According to the given question, the emf of the cell is equal to $0.164{\text{V}}$ . Hence, the equation becomes
${\text{0}}{\text{.164}} = \dfrac{{0.0591}}{1}\log \dfrac{{0.1}}{{{{\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]}_{{\text{LHS}}}}}}$
The concentration of silver ions in the left hand side gives the concentration of the silver ions of saturated solution of ${\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$ .
Hence, we have:
$
  \dfrac{{{\text{0}}{\text{.164}}}}{{0.0591}} = \log \dfrac{{0.1}}{{{{\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]}_{{\text{LHS}}}}}} \\
   \Rightarrow 2.77 = - 1 - \log {\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]_{{\text{LHS}}}} \\
   \Rightarrow \log {\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]_{{\text{LHS}}}} = - 3.77 \\
   \Rightarrow {\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]_{{\text{LHS}}}} = 1.6 \times {10^{ - 4}}{\text{M}} \\
 $
Now, the saturated solution of ${\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$ ionizes as follows.
${\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}} \rightleftharpoons 2{\text{A}}{{\text{g}}^ + } + {\text{Cr}}{{\text{O}}_{\text{4}}}^{2 - }$
Thus, the concentration of ${\text{Cr}}{{\text{O}}_{\text{4}}}^{2 - }$ ion is half of the concentration of the silver ions.
Thus, the concentration of ${\text{Cr}}{{\text{O}}_{\text{4}}}^{2 - }$ is $ = \dfrac{{1.66 \times {{10}^{ - 4}}}}{2} = 8.3 \times {10^{ - 5}}{\text{M}}$ .
Now, the solubility product of ${\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$ is
$
  {\text{ = }}{\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]^{\text{2}}}\left[ {{\text{Cr}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right] \\
   = {\left( {1.66 \times {{10}^{ - 4}}} \right)^2}\left( {8.3 \times {{10}^{ - 5}}} \right) \\
   = 2.287 \times {10^{ - 12}}{\text{mo}}{{\text{l}}^{\text{3}}}{{\text{L}}^{{\text{ - 3}}}} \\
 $

Note:
Solubility and solubility products are two different terms. The term solubility is applicable for all types of solutes but the solubility product is only applicable for electrolytes. The solubility product has a constant value for an electrolyte at a given temperature while solubility of an electrolyte can be reduced by the addition of common ions, i.e., an ion already present in the solution due to dissociation of the electrolyte. The essential condition for the precipitation of an electrolyte is that the ionic product should be greater than the solubility product of the substance.