Silver chloride, $AgCl (K_{s p} = 1.8 \times 10^{- 10})$ can be dissolved in solutions containing ammonia due to the formation of the soluble complex ion $Ag {({NH}_{3})}_{2}^{+}$ $(K_{f} = 1.0 \times 10^{8})$
What is the minimum amount of ${NH}_{3}$ that would need to be added to dissolve $0.010 mol AgCl$ in 1.00 L of solution?

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0.020 mol

0.0945 mol

0.13 mol

0.010 mol

answer is C.

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Detailed Solution

$\underline{\underset{\underset{}{0.01}}{AgCl}} + \underset{a - 0.02}{\underset{a}{2 {NH}_{3}}} ⇌ \underset{0.01}{\underline{Ag {({NH}_{3})}_{2}^{+}}} + {\underset{0.01}{\underline{Cl}}}^{-} K_{e q} = K_{s p} \times K_{f}$

$\Rightarrow \frac{(0.01)^{2}}{(a - 0.02)^{2}} = 1.8 \times 10^{- 10} \times 10^{8}$

$\Rightarrow a = 0.0945 mole$

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