Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

The solubility product of Ag2CrO4Ag _{2} CrO _{4} is 32×10−1232 \times 10^{-12}. What is the concentration of CrO42−CrO _{4}^{2-} ions in that solution (in gg ions L−1)\left.L^{-1}\right) ?

A

2×10−42\times 10^{-4}

B

8×10−48\times 10^{-4}

C

8×10−88\times 10^{-8}

D

16×10−416\times 10^{-4}

Answer

2×10−42\times 10^{-4}

Explanation

Solution

First find relationship between solubility product and solubility after writing dissociation reaction of Ag2CrO4Ag_{2}CrO_{4} . Then substitute the given values to find the answer. Let the solubility of Ag2CrO4=xg/LAg _{2} CrO _{4}=x g / L Ag2CrO4→2Ag++Cr2O42−A g_{2} CrO _{4} \rightarrow 2 Ag ^{+}+ Cr _{2} O _{4}^{2-} Conc. x 2x x Ksp=[Ag2+][CrO42−]x\, 2 x\, x\, K _{s p}=\left[ Ag ^{2+}\right]\left[ CrO _{4}{ }^{2-}\right] =(2x)2(x)=(2 x)^{2}(x) or Ksp=4x3K_{s p}=4 x^{3} ∴x=Ksp43\therefore x=\sqrt[3]{\frac{K_{s p}}{4}} Given, Ksp=32×10−12K_{s p}=32 \times 10^{-12} ∴x=32×10−1243\therefore x=\sqrt[3]{\frac{32 \times 10^{-12}}{4}}