You have a 0.850 M solution of Na₂CrO₄ at a given temperature.At what concentration will silver ions need to be added in order for a participate to form? The K_sp for Ag₂CrO₄ at this temperature is 2.00 × 10^-12. A) 2.35 × 10^-12 M B) 1.37 × 10^-5 M C) 5.02 × 10^-10 M D) 1.53 × 10^-6 M

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The expression for the solubility product (Ksp) for Ag2CrO4 is:Ksp = [Ag+]^2[CrO42-]We can use this expression to calculate the concentration of silver ions that need to be added in order for a precipitate to form.Ksp = 2.00 × 10^-12[CrO42-] = 0.850 M[Ag+] = ?Substituting these values into the Ksp expression, we get:2.00 × 10^-12 = [Ag+]^2(0.850)[Ag+]^2 = 2.35 × 10^-12[Ag+] = 1.53 × 10^-6 MTherefore, the concentration of silver ions that need to be added in order for a precipitate to form is 1.53 × 10^-6 M.

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