Question

What mass of $AgI$ will dissolve in $1.0L$ of $1.0M$ $N{H_3}$ ? Neglect change in conc. of $NH_3^ - $ [ Given: ${K_{sp}}(AgI)$$ = 1.5 \times {10^{ - 16}}$ ; ${K_f}[Ag(N{H_3})_2^ + ] = 1.6 \times {10^7}]$
(At Wt $ = 108;I = 127$ )

Answer 1

Hint: The equilibrium constant for the dissolution of a solid substance into an aqueous solution is the solubility product constant. The symbol ${K_{sp}}$ is used to represent it. When 1.00 mole of a nonvolatile nonionizing solute dissolves in one kilogram of solvent is called ${K_f}$(molal freezing point depression constant). It represents how many degrees the freezing point of the solvent changes.
Formula used:
${K_{eq}} = {K_{sp}} \times {K_f}$
${K_{eq}}$= Equilibrium constant
${K_{sp}}$ = Solubility product
${K_f}$ = Molal freezing point depression constant

Complete answer:
Given: ${K_{sp}}(AgI)$$ = 1.5 \times {10^{ - 16}}$
${K_f}[Ag(N{H_3})_2^ + ] = 1.6 \times {10^7}$
To find: Mass of $AgI$ that will dissociate
Now,
$AgI(s) \rightleftharpoons A{g^ + }(aq) + {I^ - }(aq)$
$A{g^ + }(aq) + 2N{H_3}(aq) \rightleftharpoons Ag(N{H_3})_2^ + (aq)$
The overall reaction is as follows:
$AgI + 2N{H_3} \rightleftharpoons Ag(N{H_3})_2^ + + {I^ - }$
The substituted values of equilibrium are as follows:
$AgI(s) + 2N{H_3}(aq)$$ = 1 - x$
$Ag{(N{H_3})_2}$$ = x$
${I^ - }(aq)$$ = x$
Therefore,
${K_{eq}} = {K_{sp}} \times {K_f}$
${K_{sp}} \times {K_f} = \dfrac{{{x^2}}}{{{1^2}}}$
$1.5 \times {10^{ - 16}} \times 1.6 \times {10^7} = {x^2}$
$x = 4.9 \times {10^{ - 5}}mol{L^{ - 1}}$
Mass of $AgI$ required $ = 4.9 \times {10^{ - 5}} \times $ Molecular mass of $AgI$
Mass of $AgI$ required $ = 4.9 \times {10^{ - 5}} \times 235$
Mass of $AgI$ required $ = 0.011g$
And hence the required mass of AgI will be 0.11g.

Additional information: The equilibrium constant (${K_{eq}}$) for a solid material dissolving in an aqueous solution is the solubility product constant, ${K_{sp}}$ . It denotes the concentration at which a substance dissolves in water. The greater the ${K_{sp}}$ value of a material, the more soluble it is. The following are some major elements that influence the solubility product constant: The action of the common ion (the presence of a common ion lowers the value of ${K_{sp}}$). The ion-diversity effect (if the ions of the solutes are uncommon, the value of ${K_{sp}}$ will be high). There are ion-pairs present.

Note:
The solubility product is a type of equilibrium constant whose value is temperature dependent. Due to increasing solubility, ${K_{sp}}$ normally rises as the temperature rises. ${K_{sp}}$ is used to describe solutes that are just slightly soluble in solution and do not entirely dissolve. ${K_{sp}}$ is the amount of solute that will dissolve in solution.