Ion Concentration

1. What is the concentration of each ion in a 0.300 M AlCl3 solution? 0.3 mole AlCl3 Al3+ + 3Cl- Al3+ Cl- 0.3 mole 0.3 mole 0.300 M 0.300 M 0.900 M Cl- 0.3 mole

2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl2 in 600.0 mL of water? 1 mol 80.0 g x 111.1 g [ CaCl2 ] = = 1.20 M 0.6000 L CaCl2 Ca2+ + 2Cl- 1.20 M 1.20 M 2.40 M

The molarity decreases when water is added by a factor of: V1/V2 3. If 1.25 L of 0.560 M Aluminum sulphate solution is added to 4.75 L of water, what is the resulting concentration of each ion? M1V1 = M2V2 (0.560 M)(1.25 L) = M2(6.00 L) M2 = 0.11666 M Al2(SO4)3 2Al3+ + 3SO42- 0.11666 M 0.233 M 0.350 M x 3/2 Shortcut The molarity decreases when water is added by a factor of: V1/V2 Al2(SO4)3 → 2Al3+ + 3SO42- 1.25 6.00 0.560 M x 2 0.233 M x 3/2 0.350 M

4. If the [SO42-] = 0.100 M in 20.0 mL of Ga2(SO4)3, determine the [Ga3+] and the molarity of the solution. Ga2(SO4)3 2 Ga3+ + 3SO42- 0.0333 M 0.0667 M 0.100 M

5. If the [Cl-] = 0.400 M, calculate the number of grams of AlCl3 that would be dissolved in 3.00 L of water. AlCl3 Al3+ + 3Cl- 0.13333 M 0.400 M 3.00 L 0.13333 mol 133.5 g x x = 53.4 g 1 L 1 mol

6. If 40. 0 mL of 0. 400 M potassium chloride solution is added to 60 6. If 40.0 mL of 0.400 M potassium chloride solution is added to 60.0 mL of 0.600 M calcium chloride, what is the resulting concentration of each ion. v1 KCl K+ + Cl_ 40.0 100.0 0.400 M 0.160 M 0.160 M v2 v1 CaCl2 Ca2+ + 2Cl_ 60.0 100.0 0.600 M 0.360 M 0.720 M v2 [Cl_] = 0.720 M + 0.160 M = 0.880 M