Asked by Jane
How many moles of NH3 must be added to 1.00 L of solution to dissolve 0.979 g of CuCO3? For CuCO3, Ksp = 2.3 × 10-10.
Answers
Answered by
DrBob222
You didn't give Kf for the copper ammine complex. Be sure to look it up in your text/notes.
mols CuCO3 = approx 0.979/123 = about 0.008 but you need to redo this, as well as all of the other calculations. And be sure to find the Kf for Cu complex in your text; texts often have different Ka, Kb, Kf values.
CuCO3 ==> Cu^2+ + CO3^2- Ksp = 2.3E-10
Cu^2+ + 4NH3 ==> [Cu(NH3)4]^2+ Kf = 5E13
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Add equations and K is Ksp*Kf = 1.2E4; again that's an estimate.
CuCO3(s)+4NH3 =>[Cu(NH3)4]^2+ + CO3^2-
E.........4x......0.008.........0.008
Then substitute the equilibrium line into Keq expression and solve for x = M NH3. Convert that to mols NH3 in 1 L solution.
mols CuCO3 = approx 0.979/123 = about 0.008 but you need to redo this, as well as all of the other calculations. And be sure to find the Kf for Cu complex in your text; texts often have different Ka, Kb, Kf values.
CuCO3 ==> Cu^2+ + CO3^2- Ksp = 2.3E-10
Cu^2+ + 4NH3 ==> [Cu(NH3)4]^2+ Kf = 5E13
--------------------------------------
Add equations and K is Ksp*Kf = 1.2E4; again that's an estimate.
CuCO3(s)+4NH3 =>[Cu(NH3)4]^2+ + CO3^2-
E.........4x......0.008.........0.008
Then substitute the equilibrium line into Keq expression and solve for x = M NH3. Convert that to mols NH3 in 1 L solution.
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