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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 CrO4−CrO _{4}^{-}ions in that solution?

A

2×10−4M2 \times 10^{-4} M

B

16×10−4M16 \times 10^{-4} M

C

8×10−4M8 \times 10^{-4} M

D

8×10−8M8 \times 10^{-8} M

Answer

2×10−4M2 \times 10^{-4} M

Explanation

Solution

Ag2CrO4S⟶2Ag+2S+CrO42−S\underset{S}{Ag _{2} CrO _{4}} \longrightarrow \underset{2S}{2 Ag ^{+}}+ \underset{S}{CrO _{4}^{2-}} Ksp=(2s)2s=4s3K_{ sp }=(2 s)^{2} s=4 s^{3} S=(Ksp4)1/3=(32×10−124)1/3S=\left(\frac{K_{ sp }}{4}\right)^{1 / 3}=\left(\frac{32 \times 10^{-12}}{4}\right)^{1 / 3} =2×10−4M=2 \times 10^{-4} M