what is the total concentration of all ions in a solution formed by dissolving 1.50 g Al2(SO4)3 in 100.0 ml of solution?

Question

what is the total concentration of all ions in a solution formed by dissolving 1.50 g Al2(SO4)3 in 100.0 ml of solution?

Answer 1

To determine the total concentration of all ions in a solution formed by dissolving 1.50 g of Al2(SO4)3 in 100.0 ml of solution, we need to calculate the moles of Al2(SO4)3 and then determine the concentration of each ion.

1. Calculate the moles of Al2(SO4)3:

The molar mass of Al2(SO4)3 is:

2(Al) + 3(S) + 12(O) = 2(26.98 g/mol) + 3(32.06 g/mol) + 12(16.00 g/mol) = 342.15 g/mol

Moles of Al2(SO4)3 = Mass / Molar mass = 1.50 g / 342.15 g/mol

2. Calculate the concentration of each ion:

Al2(SO4)3 dissociates into 2 Al3+ ions and 3 SO4^2- ions.

Since 1 mole of Al2(SO4)3 gives 2 moles of Al3+ ions and 3 moles of SO4^2- ions, the moles of each ion are:

Moles of Al3+ ions = 2 * (moles of Al2(SO4)3)

Moles of SO4^2- ions = 3 * (moles of Al2(SO4)3)

3. Calculate the concentration (molarity) of each ion:

The concentration of each ion is determined by dividing the moles of that ion by the volume of the solution in liters.

Volume of solution = 100.0 ml = 100.0 ml * (1 L / 1000 ml) = 0.1000 L

Concentration of Al3+ ions = Moles of Al3+ ions / Volume of solution

Concentration of SO4^2- ions = Moles of SO4^2- ions / Volume of solution

Now you can substitute the values you calculated in steps 1 and 3 to find the concentrations of the Al3+ and SO4^2- ions in the solution.