The solubility product of Ag 2 CrO 4 is 32 × 1 0 − 12. What is the concentration of CrO 4 − ions in that solution?

Ag 2 CrO 4 ⟶ 2 Ag + + Cr O 4 2 − S K SP = ( 2 s ) 2 s = 4 s 2 ⋅ s = 4 s 3 S = ( 4 K Sp ) 1/3 ⇒ ( 4 32 × 1 0 − 12 ) 1/3 = 2 × 1 0 − 4 M

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minPYQ_2014

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The solubility product of $Ag_{2} CrO_{4}$ is $32 \times 1 0^{-}$ 12. What is the concentration of $CrO_{4}^{-}$ ions in that solution?

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Answer:

Solution:

$Ag_{2} CrO_{4} ⟶ 2 Ag^{+} + Cr O_{4}^{2 -}$ $S$ $K_{SP} = (2 s)^{2} s = 4 s^{2} \cdot s = 4 s^{3}$ $S = (\frac{K _{Sp}}{4})^{1/3} \Rightarrow (\frac{32 \times 1 0 ^{- 12}}{4})^{1/3}$ $= 2 \times 1 0^{- 4} M$

Stream:VITEEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2014

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