Silver ions are added to the solution with:
$[B r^{-}]$ = $[C l^{-}]$ = $[C O_{3}^{2 -}]$ = $[A s O_{4}^{3 -}]$ = $0.1 M$

Which compound will precipitate at the lowest $[A g^{+}]$ ?

Question

Silver ions are added to the solution with: [ B r − ] = [ C l − ] = [ C O 3 2 − ​ ] = [ A s O 4 3 − ​ ] = 0.1 M Which compound will precipitate at the lowest [ A g + ] ?

TestHub · Accepted Answer

To determine which compound precipitates first, we calculate the required $[A g^{+}]$ for each to reach saturation. The compound requiring the lowest $[A g^{+}]$ will precipitate first.

For AgBr: $K_{s p} = [A g^{+}] [B r^{-}] \Rightarrow [A g^{+}] = \frac{5 \times 1 0 ^{- 13}}{0.1} = 5 \times 1 0^{- 12} M$

For AgCl: $K_{s p} = [A g^{+}] [C l^{-}] \Rightarrow [A g^{+}] = \frac{1.8 \times 1 0 ^{- 10}}{0.1} = 1.8 \times 1 0^{- 9} M$

For Ag₂CO₃: $K_{s p} = [A g^{+}]^{2} [C O_{3}^{2 -}] \Rightarrow [A g^{+}] = \frac{8.1 \times 1 0 ^{- 12}}{0.1} = 8.1 \times 1 0^{- 11} \approx 9 \times 1 0^{- 6} M$

For Ag₃AsO₄: $K_{s p} = [A g^{+}]^{3} [A s O_{4}^{3 -}] \Rightarrow [A g^{+}] = 3 \frac{1 0 ^{- 22}}{0.1} = 3 1 0^{- 21} = 1 0^{- 7} M$

Comparing the $[A g^{+}]$ values, $5 \times 1 0^{- 12} M$ (for AgBr) is the lowest. Therefore, AgBr will precipitate at the lowest $[A g^{+}]$ .

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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