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6-2.Equilibrium-II (Ionic Equilibrium)
hard
Calculate the molar solubility of $AgCl$ at $25\,^oC$ in $3.0\ M\ NH_3$ $(K_{sp}$ of $AgCl = 2.0 \times 10^{-10} , K_f$ of $[Ag(NH_3)_2]^+ = 1.25 \times 10^7)$
A
$0.15\ M$
B
$0.05\ M$
C
$0.136\ M$
D
$0.10\ M$
Solution
$AgCl + \mathop {2N{H_3}}\limits_{(3 – 2x)} \to \mathop {{{[Ag{{(N{H_3})}_2}]}^ + }}\limits_x + \mathop {C{l^ – }}\limits_x $
$\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{sp}} \times \mathrm{K}_{\mathrm{f}}$
$\mathrm{K}_{\mathrm{c}}=2 \times 10^{-10} \times 1.25 \times 10^{7}=2.5 \times 10^{-3}$
$\mathrm{K}_{\mathrm{e}}=\frac{\mathrm{x}^{2}}{(3-2 \mathrm{x})^{2}}=2.5 \times 10^{-3}$
$\frac{x}{3-2 x}=\sqrt{2.5 \times 10^{-3}}=0.05$
$x=\frac{0.15}{1.1}=0.136\, \mathrm{M}$
$(c)$ is correct answer.
Standard 11
Chemistry
