A 50.00-mL sample of 0.100 M Ca(NO₃)₂ is mixed with 50.00 mL of 0.200 M NaF. When the system has come to equilibrium, which of the following sets of conditions will hold? The K_sp for CaF₂ is 4.0 × 10^-11. A) 10.0 × 10^-3 1.3 × 10^-5 M 1.3 × 10^-5 M B) 5.0 × 10^-3 3.5 × 10^-4 M 7.0 × 10^-4 M C) 5.0 × 10^-3 2.2 × 10^-4 M 4.3 × 10^-4 M D) 5.0 × 10^-3 3.5 × 10^-4 M 4.3 × 10^-4 M E) 5.0 × 10^-3 3.4 × 10^-9 M 5.0 × 10^-2 M

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Quizplus · Accepted Answer

First, write the balanced chemical equation for the reaction: Ca(NO₃)₂ + 2NaF → CaF2 + 2Na(NO₃)₂ The initial moles of Ca(NO₃)₂ and NaF are: nCa(NO₃)₂ = 0.0500 L x 0.100 mol/L = 0.00500 mol nNaF = 0.0500 L x 0.200 mol/L = 0.0100 mol Assuming x moles of Ca(NO₃)₂ reacts with NaF, the equilibrium concentrations are: [Ca(NO₃)₂] = (0.00500 - x) mol/L [NaF] = (0.0100 - 2x) mol/L [CaF2] = [Na(NO₃)₂] = x mol/L The equilibrium constant expression for the reaction is: K_sp = [CaF2]/([Ca(NO₃)₂][F-]²) = x/[(0.00500 - x)(0.0100 - 2x)²] Substituting in the given value of K_sp and solving for x gives: 4.0 × 10^-11 = x/[(0.00500 - x)(0.0100 - 2x)²] x = 2.2 × 10^-11 mol (Note: This assumption is valid because the equilibrium constant is much smaller than 1.) Substituting x back into the equilibrium concentrations gives: [Ca(NO₃)₂] = 0.00500 - x = 0.00500 - 2.2 × 10^-11 = 0.00500 mol/L (to 3 sig figs) [NaF] = 0.0100 - 2x = 0.0100 - 4.4 × 10^-11 = 0.0100 mol/L (to 3 sig figs) [CaF2] = [Na(NO₃)₂] = x = 2.2 × 10^-11 mol/L Therefore, the answer is C) 5.0 × 10^-3, 2.2 × 10^-4 M, 4.3 × 10^-4 M, because it gives the correct equilibrium concentrations for Ca(NO₃)₂, NaF, and CaF2.

A 5000-ML Sample of 0