Question
Given Ag(NH3)2+$\rightleftharpoons$ Ag+ + 2NH3, Kc = 7.2 × 10–8 and Ksp of AgCl = 1.8 × 10–10 at 298 K. If ammonia is added to a water solution containing excess of AgCl(s) only, then calculate the concentration of the complex in 1.0 M aqueous ammonia.
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Solution
$AgCl(s) + 2NH_{3}(aq) \rightleftharpoons Ag\left( NH_{3} \right)_{2}^{+}(aq) + Cl^{-}(aq) $$${1.0\text{ M}00 }{\text{Eqn.}(1 - 2x)\text{ M}x\text{ M}x\text{ M}}$$
Now, $K_{eq} = \frac{\left\lbrack Ag\left( NH_{3} \right)_{2}^{+} \right\rbrack\left\lbrack Cl^{-} \right\rbrack}{\left\lbrack NH_{3} \right\rbrack^{2}} \times \frac{\left\lbrack Ag^{+} \right\rbrack}{\left\lbrack Ag^{+} \right\rbrack} = \frac{K_{sp}}{K_{c}}$
$= \frac{1.8 \times 10^{- 10}}{7.2 \times 10^{- 8}} = \frac{1}{400}$
Now, $\frac{x.x}{(1 - 2x)^{2}} = \frac{1}{400} \Rightarrow x = \frac{1}{22} = 0.045\text{ M}$
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