Question

Minimum moles of $N{H}_3$ required to be added to $1L$ solution so as to dissolve $0.1mol$ of $\mathrm{AgCl}\left({\mathrm{K}}_{\mathrm{sp}}=1.0\times 10^{-10}\right)$ by the reaction is:

$\text{AgCl(s)}+2{\left(\text{NH}\right)}_3\rightleftharpoons {\left(\left[\text{Ag}{\left({\left(\text{NH}\right)}_3\right)}_2\right]\right)}^{+}+{\left(\text{Cl}\right)}^{-}$

Given ${\mathrm{K}}_{\mathrm{f}}$ of ${\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}=10^8$

• A. $0.5mol$
• B. $1.0mol$
• C. $1.1mol$
• D. $1.2mol$

Answer 1

Answer: (D)

${\mathrm{AgCl}}_{(\mathrm{s})}+\mathrm{aq}\rightleftharpoons {\mathrm{Ag}}_{(\mathrm{aq})}^{+}+{\mathrm{Cl}}_{\mathrm{aq}}^{-};{\mathrm{K}}_{\mathrm{sp}}$ Sol. ${\mathrm{Ag}}_{\mathrm{aq}}^{+}+2{\mathrm{NH}}_{3\mathrm{aq}}\rightleftharpoons \left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_{2_{\mathrm{aq}}}^{+}\right];{\mathrm{K}}_{\mathrm{f}}$ ____________________________________________ ${\mathrm{AgCl}}_{(\mathrm{s})}+2{\mathrm{NH}}_3\rightleftharpoons \left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_{2\mathrm{aq}}^{+}\right]+{\mathrm{Cl}}_{\mathrm{aq}}^{-}\mathrm{K}={\mathrm{K}}_{\mathrm{sp}}\times K_f$ $(\mathrm{C}-0.2)\mathrm{M}0.1\mathrm{M}0.1\mathrm{M}$
Equilibrium concentration
$\mathrm{K}=10^{-10}\times 10^8=10^{-2}=\frac{\left[\mathrm{Ag}{\left[{\mathrm{NH}}_3\right)}_2^{+}\right]\left[{\mathrm{Cl}}^{-}\right]}{{\left[{\mathrm{NH}}_3\right]}^2}$
$10^{-2}=\frac{10^{-1}\times 10^{-1}}{(\mathrm{M}-0.2{)}^2}$
$\therefore \mathrm{M}-0.2=1$
$\therefore \mathrm{M}=1.2\mathrm{M}$