# Problem: A solution is formed by mixing 50.0 mL of 10.0 M NaX with 50.0 mL of 2.0 X 10  -3 M CuNO3. Assume that Cu+ form complex ions with X - as follow:Cu+(aq) + X -(aq) ⇌ CuX(aq)             K  1 = 1.0 X 10 2CuX(aq) + X -(aq) ⇌ CuX2-(aq)          K 2 = 1.0 X 104CuX2-(aq) + X -(aq) ⇌ CuX32-(aq)     K 3 = 1.0 X 103with an overall reactionCu+(aq) + 3X-(aq) ⇌ CuX32-(aq)       K = 1.0 X 10 9Calculate the following concentration at equilibrium.c. Cu+

###### FREE Expert Solution

Cu+(aq) + 3 X-(aq) ⇌ CuX32-(aq)       K = 1.0 x 109

Calculate CuX32- formed from the reactants.

$\overline{)\mathbf{M}\mathbf{=}\frac{\mathbf{mol}}{\mathbf{L}}}$

50.0 mL of 10.0 M NaX

NaX(aq) → Na+(aq) + X-(aq)

moles CuX32- = 0.1667 mol

50.0 mL of 2.0 X 10-3 M CuNO3

CuNO3(aq) → Cu+(aq) + NO3-(aq)

moles CuX32- = 1.0x10-4 mol

CuNO3 → produces a smaller amount of the product → limiting reactant

Calculate X-(aq) left in the reaction:

moles X- initial = 0.500 mol

moles X- reacted = 3.0x10-4 mol

moles X- left = 0.500 mol - 3.0x10-4 mol

moles X- left = 4.997 mol

Calculate the concentrations of CuX32-(aq) and X-(aq):

final volume = 0.100 L

[CuX32-] = 1.00x10-3 M

[X-] = 4.997 M

Calculate the concentrations of Cu+(aq):

95% (454 ratings)
###### Problem Details

A solution is formed by mixing 50.0 mL of 10.0 M NaX with 50.0 mL of 2.0 X 10  -3 M CuNO3. Assume that Cu+ form complex ions with X - as follow:

Cu+(aq) + X -(aq) ⇌ CuX(aq)             K  1 = 1.0 X 10 2

CuX(aq) + X -(aq) ⇌ CuX2-(aq)          K 2 = 1.0 X 104

CuX2-(aq) + X -(aq) ⇌ CuX32-(aq)     K 3 = 1.0 X 103

with an overall reaction

Cu+(aq) + 3X-(aq) ⇌ CuX32-(aq)       K = 1.0 X 10 9

Calculate the following concentration at equilibrium.

c. Cu+