What is the final concentration of aluminum cation?You have to be able to determine the number of moles of Al+3 that each solution contributes to the final solution: 1. aluminum chloride: AlCl3 Multiply the volume (in L) by the molarity to determine the number of moles of each compound dissolved in solution. 0.0431 L x 0.279 M = 0.0120 mol AlCl3 in solution When the AlCl3 in placed in the water, it dissociates to form the following ions: AlCl3 -- Al+3 + 3Cl- So for every one mole of AlCl3 placed in the water, one mole of aluminum ions will dissociate. Therefore, since we have 0.0120 mole of AlCl3 in the solution, that means that the aluminum chloride will contribute 0.0120 Al+3 ions to the final solution. 2. Aluminum sulfate: Al2(SO4)3 Multiply the volume (in L) by the molarity to determine the number of moles of each compound dissolved in solution. 0.0146 L x 0.464 M = 0.00677 mol Al2(SO4)3 in solution When the Al2(SO4)3 in placed in the water, it dissociates to form the following ions: Al2(SO4)3 -- 2Al+3 + 3(SO4)-2 So for every one mole of Al2(SO4)3 placed in the water, two mole of aluminum ions will dissociate. Use the mole ratios of the dissociation reaction to determine the number of moles of Al+3 ions that the aluminum sulfate contributes to the solution. 0.00677 mol Al2(SO4)3 x (2 mol Al+3 ions / 1 mol Al2(SO4)3) = 0.01354 mol Al+3 ions Therefore, the total number of Al+3 ions in solution is the sum: 0.01354 + 0.0120 = 0.02554 mol Al+3 ions The last piece of information needed to determine the concentration of the final solution is the volume of the final solution. Since the two volumes were mixed, the volume of the final solution will be the sum of the two solutions. 0.0431 L + 0.0146 L = 0.0577 L Therefore, to calculate molarity: Molarity = moles of solute / liters of solution Molarity = 0.02554 mol / 0.0577 L = 0.443 M Al+3Other related question