Asked by joe
If 2.50 g of CuSO4 are dissolved in 8.9 *102 mL of 0.21 M NH3, what are the concentrations of Cu(NH3)4^2+, NH3 and Cu^2+ at equilibrium?
Answers
Answered by
DrBob222
mols CuSO4 initially = g/molar mass = 2.50/159.5 = 0.0157. millimols = 15.7 .millimols NH3 = mL x M = 890 x 0.21 = about 187
.............CuSO4 + 4NH3 ==> [Cu(NH3)4]^2+ + [SO4]^2-
I................15.7......187................0.........................0
C.............-15.7.....-62.8.........+15.7.....................15.7
E.................0..........124.2........15.7.....................15.7
So your solution now is 0 mmols CuSO4, 124.2 mmols NH3 and 15.7 mmols Cu complex. You want concentrations which is M = mmols/mL = ?
You do the rest.
NOTE: I rounded here and there and estimated most of the molar masses so you should check everything and round to your satisfaction. Post your work if you get stuck.
.............CuSO4 + 4NH3 ==> [Cu(NH3)4]^2+ + [SO4]^2-
I................15.7......187................0.........................0
C.............-15.7.....-62.8.........+15.7.....................15.7
E.................0..........124.2........15.7.....................15.7
So your solution now is 0 mmols CuSO4, 124.2 mmols NH3 and 15.7 mmols Cu complex. You want concentrations which is M = mmols/mL = ?
You do the rest.
NOTE: I rounded here and there and estimated most of the molar masses so you should check everything and round to your satisfaction. Post your work if you get stuck.
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