What is the molarity of each ion present if 18.0 g of Al2(SO4)3 is present in 201 mL of solution?
_____ M Al3+
_____M SO42-
See Anonymous on Friday, October 28, 2016 at 4:38pm.
To find the molarity of each ion, we need to first determine the number of moles of Al3+ and SO42- present in the given compound, Al2(SO4)3.
Step 1: Calculate the number of moles of Al2(SO4)3
The molar mass of Al2(SO4)3 can be calculated by adding the molar masses of individual elements in the compound.
Molar mass of Al2(SO4)3 = (2 * atomic mass of Al) + (3 * atomic mass of S) + (12 * atomic mass of O)
= (2 * 26.98 g/mol) + (3 * 32.06 g/mol) + (12 * 16.00 g/mol)
= 342.15 g/mol
To convert the given mass of Al2(SO4)3 into moles, divide the mass by its molar mass:
Number of moles of Al2(SO4)3 = 18.0 g / 342.15 g/mol
≈ 0.0527 mol
Step 2: Determine the molarity of each ion
Since Al2(SO4)3 dissociates into 2 Al3+ ions and 3 SO42- ions, we need to multiply the number of moles of Al2(SO4)3 by the respective stoichiometric coefficients to find the moles of individual ions.
Moles of Al3+ = 0.0527 mol * 2 (stoichiometric coefficient)
= 0.1054 mol
Moles of SO42- = 0.0527 mol * 3 (stoichiometric coefficient)
= 0.1581 mol
Step 3: Calculate the molarity of each ion
Molarity is defined as moles of solute divided by the volume of the solution in liters.
To find the molarity of each ion, we need to convert the given volume of solution (201 mL) into liters by dividing by 1000:
Volume of solution = 201 mL / 1000
= 0.201 L
Molarity of Al3+ = Moles of Al3+ / Volume of solution
= 0.1054 mol / 0.201 L
≈ 0.523 M
Molarity of SO42- = Moles of SO42- / Volume of solution
= 0.1581 mol / 0.201 L
≈ 0.786 M
Therefore, the molarity of Al3+ ion is approximately 0.523 M, and the molarity of SO42- ion is approximately 0.786 M.