Calculate the concentrations of all ions present in the following solutions of strong electrolytes :
2.0 times 10-3 M AL2(SO4)3
There are 2 mols Al^3+ in 1 mol Al2(SO4)3 so (Al^3+) is twice that of the salt.
There are 3 mols SO4^2- in 1 mol Al2(SO4)3 so (SO4)^2- is three times that of the salt.
To calculate the concentrations of all ions present in a solution of a strong electrolyte like AL2(SO4)3, you need to consider the dissociation of the compound into its respective ions.
The formula for aluminum sulfate (AL2(SO4)3) indicates that it dissociates into 2 aluminum ions (Al3+) and 3 sulfate ions (SO4^2-). Since AL2(SO4)3 is a strong electrolyte, it fully dissociates in water.
So, based on this, the concentration of aluminum ions (Al3+) in the solution will be twice the concentration of aluminum sulfate, which is 2.0 × 10^(-3) M × 2 = 4.0 × 10^(-3) M.
Similarly, the concentration of sulfate ions (SO4^2-) will be three times the concentration of aluminum sulfate, which is 2.0 × 10^(-3) M × 3 = 6.0 × 10^(-3) M.
Therefore, the concentrations of all ions present in the AL2(SO4)3 solution are as follows:
Aluminum ions (Al3+): 4.0 × 10^(-3) M
Sulfate ions (SO4^2-): 6.0 × 10^(-3) M