Question

Minimum moles of $NH_{3}$ required to be added to $1L$ solution so as to dissolve $0.1mol$ of $\operatorname{AgCl}\left(\mathrm{K}_{\mathrm{sp}}=1.0 \times 10^{-10}\right)$ 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. $\mathrm{Ag}_{\mathrm{aq}}^{+}+2 \mathrm{NH}_{3 \mathrm{aq}} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2_{\mathrm{aq}}}^{+}\right] \quad ; \quad \mathrm{K}_{\mathrm{f}}$ ____________________________________________ $\mathrm{AgCl}_{(\mathrm{s})}+2 \mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2 \mathrm{aq}}^{+}\right]+\mathrm{Cl}_{\mathrm{aq}}^{-} \quad \mathrm{K}=\mathrm{K}_{\mathrm{sp}} \times K_f$ $(\mathrm{C}-0.2) \mathrm{M} \quad 0.1 \mathrm{M} \quad 0.1 \mathrm{M}$
Equilibrium concentration
$\mathrm{K}=10^{-10} \times 10^{8}=10^{-2}=\frac{\left[\mathrm{Ag}\left[\mathrm{NH}_{3}\right)_{2}^{+}\right]\left[\mathrm{Cl}^{-}\right]}{\left[\mathrm{NH}_{3}\right]^{2}}$
$10^{-2}=\frac{10^{-1} \times 10^{-1}}{(\mathrm{M}-0.2)^{2}}$
$\therefore \mathrm{M}-0.2=1$
$\therefore \mathrm{M}=1.2 \mathrm{M}$