Lesson 2 Ion Concentration.

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Presentation transcript:

Lesson 2 Ion Concentration

1. What is the concentration of each ion in 2. 0 L of a 0. 300 M AlCl3 1. What is the concentration of each ion in 2.0 L of a 0.300 M AlCl3 solution? AlCl3  Al3+ + 3Cl- 0.300 M 0.300 M 0.900 M

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 mole 80.0 g x 111.1 g Molarity = = 1.20 M 0.600 L CaCl2  Ca2+ + 2Cl- 1.20 M 1.20 M 2.40 M

3. If 40. 0 mL of 0. 400 M AlCl3 solution is added to 60. 0 mL of 0 3. If 40.0 mL of 0.400 M AlCl3 solution is added to 60.0 mL of 0.600 M CaCl2 solution, what is the resulting concentration of all ions? AlCl3  Al3+ + 3Cl- 40.0 x 0.400 M 0.160 M 0.480 M 100.0 CaCl2  Ca2+ + 2Cl- 60.0 x 0.600 M 0.360 M 0.720 M 100.0 [Cl-] = 0.480 + 0.720 M = 1.200 M

4. If the [Cl-] = 0.600 M, calculate the number of grams AlCl3 that would be dissolved in 3.00 L of water. AlCl3  Al3+ + 3Cl- 0.200 M 0.200 M 0.600 M 3.00 L x 0.200 mole x 133.5 g = 80.1 g 1 L 1 mole

5. Write the formula, complete, and net ionic equation for the 5. Write the formula, complete, and net ionic equation for the reaction: Ca(s) + H2SO4(aq) → H2(g) + CaSO4(s) only break up aqueous! low solubility Ca(s) + 2H+ + SO42- → H2(g) + CaSO4(s) Ca(s) + 2H+ + SO42- → H2(g) + CaSO4(s)