A solution of copper(II) sulfate that contains 15.00% CuSO4 by mass has a density of 1.169 g/mL. A 25.0 mL portion of this solution was reacted with excess concentrated ammonia to form a dark blue solution. When cooled, filtered and dried, 6.127 g of a dark blue solid were obtained. A 0.195 g sample of the solid was analyzed for ammonia by titrating with 0.1036 M hydrochloric acid solution, requiring 30.63 mL to reach the equivalence point. A 0.150 g sample was analyzed for copper (II) by titrating with 0.0250 M EDTA, (which reacts with Cu2+ in a 1:1 ratio). The endpoint was reached after 24.43 mL of the EDTA were added. A 0.200 g sample was heated at 110 ˚C to drive off water, producing 0.185 g of the anhydrous material.

The part a of the question is at

http://www.enotes.com/homework-help/solution-copper-ii-sulfate-that-contains-15-00-442608

b. Find the number of moles of Cu2+ in the 25.0 mL portion

### 1 Answer | Add Yours

Number of moles of Cu2+ ions in 25 mL portion of the solution can be obtained in two ways.

An approximate estimation from mass-density considerations: Mass of 25.0 mL solution =V*d=25*1.169=29.225 g

Amount of CuSO4 in this solution = 15*29.225/100=4.38375 g.

Number of moles of CuSO4 ions in 25 mL portion of the solution =4.38375/159.5=0.0275

Since one mole of complex contains 1 mole of Cu2+, number of moles of Cu2+ ions in 25 mL portion of the solution will also be 0.0275.

A more accurate method involves analysis based on titration results. On reaction with aq. NH3, CuSO4 forms a blue tetra ammonium copper complex of varying hydrate composition. Let the molar mass of the complex be M.

During analysis of NH3, moles of the complex taken, by weighing= 0.195/M

Moles of NH3 obtained by titration with HCl = 30.63*0.1036*10^-3

Since one mole of complex contains 4 moles of NH3, moles of the complex obtained by titration = (30.63*0.1036*10^-3)/4

Therefore, 0.195/M=(30.63*0.1036*10^-3)/4

`rArr` M=245.8 g/Mole

Similarly, during analysis of Cu2+, moles of the complex taken, by weighing= 0.150/M

Moles of NH3 obtained by titration with EDTA = 24.43*0.025*10^-3

Since one mole of complex contains 1 mole of Cu2+, moles of the complex obtained by titration = (24.43*0.025*10^-3)

Therefore, 0.150/M=(24.43*0.025*10^-3)

`rArr` M=245.6 g/Mole

and again, during analysis of H2O, moles of the complex taken, by weighing= 0.200/M

Moles of H2O evaporated from it = (0.200-0.185)/18

Therefore, 0.200/M=0.015/18

`rArr` M=240 g/Mole

Therefore, correct molar mass of the complex is 245.5, which corresponds to the formula

[Cu(NH3)4(H2O)]SO4, i.e. tetraammonium Copper (II) complex, monohydrate.

25 mL portion of the CuSO4 solution produced 6.125 g of the complex.

Number of moles of the complex = 6.125/245.5=0.025 moles.

Since one mole complex contains one mole of Cu2+ ions, number of moles of Cu2+ ions in 25 mL portion of the solution was thus, 0.025.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes