The number of moles of NH_{3} that must be ad | vedclass

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(C) Initial moles of $Ag^{+} = 0.80 \, M 	imes 2 \, L = 1.6 \, mol$.Let $n$ be the moles of $NH_{3}$ added. The reaction is: $Ag^{+} + 2NH_{3} ight

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$4$

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(C) Initial moles of $Ag^{+} = 0.80 \, M 	imes 2 \, L = 1.6 \, mol$.Let $n$ be the moles of $NH_{3}$ added. The reaction is: $Ag^{+} + 2NH_{3} ightleftharpoons [Ag(NH_{3})_{2}]^{+}$.Since $K_{f}$ is very large $(1.0 	imes 10^{8})$,we assume the reaction goes to completion.$[Ag(NH_{3})_{2}]^{+} \approx 0.80 \, M$ (as all $Ag^{+}$ is converted).Remaining $Ag^{+} = 5.0 	imes 10^{-8} \, M$.$K_{f} = \frac{[[Ag(NH_{3})_{2}]^{+}]}{[Ag^{+}][NH_{3}]^{2}} = 1.0 	imes 10^{8}$.$1.0 	imes 10^{8} = \frac{0.80}{(5.0 	imes 10^{-8})[NH_{3}]^{2}}$.$[NH_{3}]^{2} = \frac{0.80}{5.0 	imes 10^{-8} 	imes 10^{8}} = \frac{0.80}{5.0} = 0.16$.$[NH_{3}] = \sqrt{0.16} = 0.4 \, M$.Total $NH_{3}$ added = $NH_{3}$ consumed + $NH_{3}$ free = $(2 	imes 0.80) + (0.4 	imes 2) = 1.6 + 0.8 = 2.4 \, mol$.Wait,re-evaluating the stoichiometry: $n_{total} = n_{consumed} + n_{free} = (2 	imes 1.6) + (0.4 	imes 2) = 3.2 + 0.8 = 4.0 \, mol$.

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