Cu+(aq) + 3 X-(aq) ⇌ CuX32-(aq) K = 1.0 x 109
Calculate CuX32- formed from the reactants.
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):
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.
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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.