A 50.0-mL sample of 2.0 *10^-⁴ M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN)₃²^-: The concentration of Cu⁺ at equilibrium is

Question

Quizplus · Accepted Answer

The initial moles of Cu⁺ can be calculated as: moles = concentration x volume = (2.0 F1F1F1S1F1F1F10 10^-⁴ M) x (50.0 mL / 1000 mL/1 L) = 0.100 mol The initial moles of NaCN can also be calculated: moles of NaCN = concentration x volume = (4.0 M) x (50.0 mL / 1000 mL/1 L) = 0.200 mol Assuming all the Cu⁺ reacts with NaCN, the final concentration of Cu⁺ can be calculated as the moles of Cu⁺ remaining after the reaction divided by the total volume of the solution: moles of Cu⁺ remaining = initial moles - moles of Cu⁺ reacted with NaCN moles of Cu⁺ reacted with NaCN = 0.200 mol (since Cu⁺ and NaCN are in a 1:1 molar ratio) moles of Cu⁺ remaining = 0.100 mol - 0.200 mol = -0.100 mol Since it is not possible for the moles of Cu⁺ remaining to be negative, we know that all of the Cu⁺ has reacted with NaCN. Therefore, the final concentration of Cu⁺ is 0 M. However, Cu(CN)₃²^- has been formed. Since the molar ratio between Cu(CN)₃²^- and Cu⁺ is also 1:1, the final concentration of Cu(CN)₃²^- is also 0.100 mol / 0.100 L = 1.0 F1F1F1S1F1F1F10 10^-¹⁴ M. Therefore, the answer is (B) 1.2 F1F1F1S1F1F1F10 10^-¹⁴ M.

A 500-ML Sample of 2