— Physical Chemistry Chemistry Question
Calculation of concentration
What is the concentration of Cu2+ (in mol dm-3) in a solution prepared by mixing of 1.345 g CuCl2 with 50.00 cm3 of CuSO4 solution with a mass concentration 31.9 g dm-3 and adjusting the volume to 500 cm3 by a HCl solution (c = 0.01 mol dm-3)?
Model Answer
M(CuCl2) = 134.45 g mol-1
n(Cu2+) in 1.345 g of CuCl2 = 1.345 g / 134.45 g mol-1 = 0.0100 mol
M(CuSO4) = 159.62 g mol-1
n(Cu2+) in 50.00 cm3 of CuSO4 solution:
n(Cu2+) = n(CuSO4) = m(CuSO4) / M(CuSO4) = / M(CuSO4) = (31.9 g dm-3 × 0.050 dm3) / 159.62 g mol-1 = 0.0100 mol
Therefore, the total amount of substance of Cu2+ is 0.0100 + 0.0100 = 0.0200 mol
Since 500 cm3 of the resulting solution contains 0.0200 mol Cu2+, the concentration of Cu2+ in the solution is 0.0400 mol dm-3.
Calculate whether a precipitate will be formed if 25.00 cm3 aliquot of the resulting solution in 16.1 is adjusted to pH 8.0 with NaOH and the final volume is 100.0 cm3. Note: Ksp(Cu(OH)2) = 4.8×10-20
Model Answer
The concentration of Cu2+ in the resulting solution is: 0.04 mol dm-3 / 4 = 0.01 mol dm-3
pH = 8.0, implying that [OH-] = 1×10-6
[Cu2+][OH-]2 = (1×10-2) × (1×10-6)2 = 1×10-14 is greater than Ksp(Cu(OH)2), i.e. 4.8×10-20, and, therefore, the precipitate of Cu(OH)2 is formed.