Question

A solution is 0.10 M in Ag+ and 0.10 M in Au 3+, which one will precipitate first as sodium chloride is added
Ksp(AgCl)= 1.8X10^-10
Ksp(AuCl3)= 3.2X10^-25

Answer 1

Answer on Question #74883 – Chemistry – Organic Chemistry

Task:

A solution is 0.10 M in $\mathrm{Ag^{+}}$ and 0.10 M in $\mathrm{Au}^{3+}$, which one will precipitate first as sodium chloride is added.

$$\mathrm{K_{sp}(AgCl)} = 1.8 \times 10^{-10};$$ $$\mathrm{K_{sp}(AuCl_3)} = 3.2 \times 10^{-25}.$$

Solution:

1) $AgCl(s) = Ag^{+} + Cl^{-}$;

$$K_{sp}(AgCl) = [Ag^{+}] \cdot [Cl^{-}];$$ $$[Cl^{-}]_1 = \frac{K_{sp}(AgCl)}{[Ag^{+}]} = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \, M.$$

2) $AuCl_3(s) = Au^{3+} + 3Cl^{-}$;

$$K_{sp}(AuCl_3) = [Au^{3+}] \cdot [Cl^{-}]^3;$$ $$[Cl^{-}]^3 = \frac{K_{sp}(AuCl_3)}{[Au^{3+}]} = \frac{3.2 \times 10^{-25}}{0.10} = 3.2 \times 10^{-24};$$ $$[Cl^{-}]_2 = \sqrt[3]{3.2 \times 10^{-24}} = 1.47 \times 10^{-8} \, M.$$ $$[Cl^{-}]_1 < [Cl^{-}]_2;$$ $$1.8 \times 10^{-9} \, M < 1.47 \times 10^{-8} \, M.$$

The lower the concentration of chloride ions, the faster the precipitate precipitates.

Therefore,

AgCl will precipitate first as sodium chloride is added.

Answer: AgCl will precipitate first as sodium chloride is added.

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