What is the concentration of the Cd²⁺(aq)ion in a 0.030 M Cd(NO₃)₂ solution that is also 1.0 M NH₃? At this temperature,K_f for Cd(NH₃)₄²⁺ = 1.0 $\times$ 10⁷. A) M B) M C) M D) M E)none of these

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Quizplus · Accepted Answer

First, write the balanced equation: Cd(NH₃)₂ + S1P0 ⇌ Cd²⁺(aq) Next, write the expression for the equilibrium constant (Kf): Kf = [Cd²⁺(aq)] /([Cd(NH₃)₂] × [S1P0]) Then, substitute the given values: Kf = 1.0 × 10^-10 = [Cd²⁺(aq)] / ((0.030 M) × (1.0 M)) Solve for [Cd²⁺(aq)]: [Cd²⁺(aq)] = 3.0 × 10^-12 M Therefore, the concentration of Cd²⁺(aq) ion in a 0.030 M Cd(NO₃)₂ solution that is also 1.0 M NH2₃ is 3.0 × 10^-12 M, which corresponds to answer choice B.

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