Part a: What mass of solid CuCL2 is required to prepare 450.0mL of 0.125M CuCl2 solution??
Part B: What are the concentrations of Cu +2 ions and Cl- ions in the above solutions.
I understand part a, and got the right answer. i need help understanding part b.
For every mole of CuCl2, you have one mole of Cu++ ions, and two moles of Cl- ions.
Thank you i appreciate it.
To calculate the concentrations of Cu+2 ions and Cl- ions in the CuCl2 solution, you need to consider the stoichiometry of the compound.
Part B: Concentration of Cu+2 ions
The stoichiometry of CuCl2 tells us that for every 1 mole of CuCl2, we have 1 mole of Cu+2 ions. Therefore, the concentration of Cu+2 ions will be equal to the concentration of CuCl2.
Given:
Volume of solution = 450.0 mL
Concentration of CuCl2 = 0.125 M (which is the same as the concentration of Cu+2 ions)
So, the concentration of Cu+2 ions in the solution is 0.125 M.
Part B: Concentration of Cl- ions
The CuCl2 compound dissociates in water into Cu+2 ions and 2 Cl- ions. This means that for every 1 mole of CuCl2, we have 2 moles of Cl- ions. Therefore, the concentration of Cl- ions will be twice the concentration of CuCl2.
Given:
Volume of solution = 450.0 mL
Concentration of CuCl2 = 0.125 M
To find the concentration of Cl- ions, we multiply the concentration of CuCl2 by 2:
Concentration of Cl- ions = 0.125 M x 2 = 0.250 M
So, the concentration of Cl- ions in the solution is 0.250 M.
To summarize:
- The concentration of Cu+2 ions in the solution is 0.125 M.
- The concentration of Cl- ions in the solution is 0.250 M.