Question

Calculate the solubility of silver chloride in a solution that is 0.110 M in NH3. Express your answer using two significant figures.

Answer 1

AgCl_(s) <======> Ag⁺_(aq) + Cl^-_{(aq)    $K_{sp} = \frac{[Ag^{+}][Cl^{-}]}{[AgCl]}=1.77 \times 10^{-10}$}

Ag⁺_(aq) + 2NH_3(aq) <=====> Ag(NH₃)₂⁺_(aq)   $K_{f} = \frac{[Ag(NH_{3})_{2}^{+}]}{[Ag^{+}][NH_{3}]^{2}}=1.7 \times 10^{7}$

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AgCl(s) + 2NH_3(aq) <======> [Ag(NH₃)₂]⁺_(aq) + Cl^-

I ---- 0.110 M 0 0 Using ICE table

C ----- -2x +x +x let us suppose 'x' moles [Ag(NH₃)₂]⁺ formed

E ---- 0.110 - 2x +x +x from x moles AgCl. Since AgCl is a solid we neglect moles of AgCl.

$K_{eq} = \frac{[Ag(NH_{3})_{2}^{+}][Cl^{-}]}{[AgCl][NH_{3}]^{2}}$

$K_{eq} = K_{f} \times K_{sp}= \frac{[Ag(NH_{3})_{2}^{+}][Cl^{-}]}{[AgCl][NH_{3}]^{2}}$

$K_{eq} = (1.77 \times10^{-10})\times(1.7\times 10^{7}) = \frac{[Ag(NH_{3})_{2}^{+}][Cl^{-}]}{[AgCl][NH_{3}]^{2}}$

$3.01 \times10^{-3} = \frac{[Ag(NH_{3})_{2}^{+}][Cl^{-}]}{[AgCl][NH_{3}]^{2}}$

$3.01 \times10^{-3} = \frac{[x][x]}{[0.110-2x]^{2}}$

$3.01 \times10^{-3} = \frac{[x]^{2}}{[0.110-2x]^{2}}$

$\sqrt{3.01 \times10^{-3}} = \frac{[x]}{[0.110-2x]}$

$0.055 = \frac{[x]}{[0.110-2x]}$

Concentration of Ag(NH₃)₂⁺ formed =x= 0.0054 M = Concentration of AgCl soluble = x = 5.4x10^-3 M