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.

M=molL


50.0 mL of 10.0 M NaX

NaX(aq) → Na+(aq) + X-(aq)

moles CuX32-=50.00 mL×10-3 L1 mL×10.0 mol X-L×1 mol CuX32-3 mol X-

moles CuX32- = 0.1667 mol


50.0 mL of 2.0 X 10-3 M CuNO3

CuNO3(aq) → Cu+(aq) + NO3-(aq)

moles CuX32-=50.00 mL×10-3 L1 mL×2.0×10-3 mol Cu+L×1 mol CuX32-1 mol Cu+

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=50.00 mL×10-3 L1 mL×10.0 mol X-L

moles X- initial = 0.500 mol

moles X- reacted=1.0×10-4 mol CuX32-×3 mol X-1 mol CuX32-

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=50.0 mL+50.0 mLfinal volume=100.0 mL×10-3 L1 mL

final volume = 0.100 L


[CuX32-]=1.0×10-4 mol0.100 L

[CuX32-] = 1.00x10-3 M


[X-]=0.4497 mol0.100 L

[X-] = 4.997 M


Calculate the concentrations of Cu+(aq):

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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+

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.