What are the respective concentrations (M) of Cu2+ and Cl- afforded by dissolving 0.733mol
CuCl2 in water and diluting to 306mL?
To determine the concentrations of Cu2+ and Cl- in the solution, we first need to calculate the number of moles of CuCl2 dissolved in 306mL of water.
Given:
- Moles of CuCl2 = 0.733 mol
- Volume of solution = 306 mL = 0.306 L
1. Calculate the molarity of CuCl2:
Molarity (M) = moles of solute / volume of solution (L)
Molarity of CuCl2 = 0.733 mol / 0.306 L = 2.40 M
2. Since CuCl2 dissociates into Cu2+ and 2 Cl-, the concentration of Cu2+ will be equal to the molarity of CuCl2 (2.40 M), and the concentration of Cl- will be twice the molarity of CuCl2 (2.40 M * 2 = 4.80 M)
Therefore, the concentrations of Cu2+ and Cl- in the solution are:
- [Cu2+] = 2.40 M
- [Cl-] = 4.80 M