Question

What mass of AgI will dissolve in 0.1 L of 1.0 M NH₃? Neglect change in conc. of NH₃ .
(Atomtextc weight of Ag & I are 108 amu and 127 amu respectively)
$\left[{\text{Given : K}}_{\text{sp}}\left(\text{AgI}\right)=\text{1.5}\times 10^{-16}{\left(\text{m}\text{o}\text{l}/\text{L}\right)}^2{;\; K}_{\text{f}}\left[\text{Ag}{\left({\text{NH}}_3\right)}_2^{+}\right]=\text{1.6}\times 10^7\right]$.

• A. $4.9\times 10^{-5}g$
• B. $\mathrm{0.0056\; g}$
• C. $\mathrm{0.035\; g}$
• D. $\mathrm{0.00115\; g}$

Answer 1

Answer: (D)

Due to higher value of K_f , mostly Ag⁺ converted into complex.
$\text{AgI}\left(\text{s}\right)\rightleftharpoons {Ag}^{+}\left(\text{aq.}\right)+{\text{I}}^{-}\left(\text{aq.}\right){;\; K}_{\text{sp}}$
${\text{Ag}}^{+}\left(\text{aq.}\right)+2{\text{NH}}_3\left(\text{aq.}\right)\rightleftharpoons \text{Ag}{\left({\text{NH}}_3\right)}_2^{+}\left(\text{aq.}\right);K_{\text{f}}$
______________________________________
overal reaction is
$\text{AgI}\left(\text{s}\right)+2{\text{NH}}_3\left(\text{aq.}\right)\rightleftharpoons Ag{\left({\text{NH}}_3\right)}_2^{+}\left(\text{aq.}\right)+{\text{I}}^{-}\left(\text{aq.}\right)$
${\text{K}}_{\text{eq}}={\text{K}}_{\text{sp}}·{\text{K}}_{\text{f}}$
${\text{K}}_{\text{sp}}·{\text{K}}_{\text{f}}=\frac{{\text{x}}^2}{1^2}$
$\therefore x=\text{4.9}\times 10^{-5}mol/litre$
So, mass of AgI required is
$=\text{4.9}\times 10^{-5}\times 235=\text{0.0115 g/L}$
$\mathrm{I}\mathrm{n}\mathrm{}0.1\mathrm{L},\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{A}\mathrm{g}\mathrm{I}\mathrm{}\mathrm{}\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{w}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{}\mathrm{b}\mathrm{e}=\frac{0.0115}{10}\mathrm{}\mathrm{g}=1.15\times 10{-3}^{}\mathrm{g}\mathrm{}\mathrm{}$