1. Lesson 2 Ion Concentration

2. 1. What is the concentration of each ion in a 0.300 M AlCl 3 solution? AlCl 3  Al 3+ +3Cl -

3. 1. What is the concentration of each ion in a 0.300 M AlCl 3 solution? AlCl 3  Al 3+ +3Cl - 0.300 M

4. 1. What is the concentration of each ion in a 0.300 M AlCl 3 solution? AlCl 3  Al 3+ +3Cl -0.300 M

5. 1. What is the concentration of each ion in a 0.300 M AlCl 3 solution? AlCl 3  Al 3+ +3Cl - 0.300 M0.300 M0.900 M

6. 2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl 2 in 600.0 mL of water? Molarity=

7. 2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl 2 in 600.0 mL of water? Molarity=80.0 g x 1 mole 111.1 g

8. 2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl 2 in 600.0 mL of water? Molarity=80.0 g x 1 mole 111.1 g 0.600 L

9. 2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl 2 in 600.0 mL of water? Molarity=80.0 g x 1 mole 111.1 g 0.600 L =1.20 M

10. 2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl 2 in 600.0 mL of water? Molarity=80.0 g x 1 mole 111.1 g 0.600 L =1.20 M CaCl 2  Ca 2+ +2Cl -

11. 2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl 2 in 600.0 mL of water? Molarity=80.0 g x 1 mole 111.1 g 0.600 L =1.20 M CaCl 2  Ca 2+ +2Cl - 1.20 M

12. 2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl 2 in 600.0 mL of water? Molarity=80.0 g x 1 mole 111.1 g 0.600 L =1.20 M CaCl 2  Ca 2+ +2Cl - 1.20 M2.40 M

13. 3. If 40.0 mL of 0.400 M Potassium chloride solution is added to 60.0 mL of 0.600 M Calcium nitrate solution, what is the resulting concentration of all ions? KCl  K + +Cl - 40.0 x0.400 M =0.160 M0.160 M 100.0 Dilution Factor Ca(NO 3 ) 2  Ca 2+ +2NO 3 - 60.0 x0.600 M =0.360 M0.720 M 100.0

14. 4. If the [Cl - ] = 0.600 M, calculate the number of grams AlCl 3 that would be dissolved in 3.00 L of water.

15. 4. If the [Cl - ] = 0.600 M, calculate the number of grams AlCl 3 that would be dissolved in 3.00 L of water. AlCl 3  Al 3+ +3Cl - 0.600 M

16. 4. If the [Cl - ] = 0.600 M, calculate the number of grams AlCl 3 that would be dissolved in 3.00 L of water. AlCl 3  Al 3+ +3Cl - 0.200 M0.600 M

17. 4. If the [Cl - ] = 0.600 M, calculate the number of grams AlCl 3 that would be dissolved in 3.00 L of water. AlCl 3  Al 3+ +3Cl - 0.200 M0.200 M0.600 M

18. 4. If the [Cl - ] = 0.600 M, calculate the number of grams AlCl 3 that would be dissolved in 3.00 L of water. AlCl 3  Al 3+ +3Cl - 0.200 M0.200 M0.600 M 3.00 L

19. 4. If the [Cl - ] = 0.600 M, calculate the number of grams AlCl 3 that would be dissolved in 3.00 L of water. AlCl 3  Al 3+ +3Cl - 0.200 M0.200 M0.600 M 3.00 L x 0.200 mole 1 L

20. 4. If the [Cl - ] = 0.600 M, calculate the number of grams AlCl 3 that would be dissolved in 3.00 L of water. AlCl 3  Al 3+ +3Cl - 0.200 M0.200 M0.600 M 3.00 L x 0.200 mole x 133.5 g = 1 L 1 mole

21. 4. If the [Cl - ] = 0.600 M, calculate the number of grams AlCl 3 that would be dissolved in 3.00 L of water. AlCl 3  Al 3+ +3Cl - 0.200 M0.200 M0.600 M 3.00 L x 0.200 mole x 133.5 g = 80.1 g 1 L 1 mole

22. 5. If the [SO 4 2- ] = 0.900 M in 20.0 mL of Ga 2 (SO 4 ) 3, determine the [Ga 3+ ] and the molarity of the solution.

23. Ga 2 (SO 4 ) 3  2Ga 3+ +3SO 4 2-

24. 5. If the [SO 4 2- ] = 0.900 M in 20.0 mL of Ga 2 (SO 4 ) 3, determine the [Ga 3+ ] and the molarity of the solution. Ga 2 (SO 4 ) 3  2Ga 3+ +3SO 4 2- 0.900 M

25. 5. If the [SO 4 2- ] = 0.900 M in 20.0 mL of Ga 2 (SO 4 ) 3, determine the [Ga 3+ ] and the molarity of the solution. Ga 2 (SO 4 ) 3  2Ga 3+ +3SO 4 2- 0.600 M0.900 M

26. 5. If the [SO 4 2- ] = 0.900 M in 20.0 mL of Ga 2 (SO 4 ) 3, determine the [Ga 3+ ] and the molarity of the solution. Ga 2 (SO 4 ) 3  2Ga 3+ +3SO 4 2- 0.300 M0.600 M0.900 M

27. 5. If the [SO 4 2- ] = 0.900 M in 20.0 mL of Ga 2 (SO 4 ) 3, determine the [Ga 3+ ] and the molarity of the solution. Ga 2 (SO 4 ) 3  2Ga 3+ +3SO 4 2- 0.300 M0.600 M0.900 M Do not use the 20.0 mL!

28. 6. Write the formula, complete, and net ionic equation for the reaction: Ca (s) +H 2 SO 4(aq) →

29. 6. Write the formula, complete, and net ionic equation for the reaction: Ca (s) +H 2 SO 4(aq) → H 2(g) +CaSO 4(s) low solubility

30. 6. Write the formula, complete, and net ionic equation for the reaction: Ca (s) +H 2 SO 4(aq) → H 2(g) +CaSO 4(s) low solubility Ca (s) +2H + + SO 4 2- → H 2(g) +CaSO 4(s) only break up aqueous!

31. 6. Write the formula, complete, and net ionic equation for the reaction: Ca (s) +H 2 SO 4(aq) → H 2(g) +CaSO 4(s) low solubility Ca (s) +2H + + SO 4 2- → H 2(g) +CaSO 4(s) only break up aqueous! Ca (s) +2H + + SO 4 2- → H 2(g) +CaSO 4(s) nothing cancels!