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Lesson 2 Ion Concentration

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

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Lesson 2 Ion Concentration

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  1. Lesson 2 Ion Concentration

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

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

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

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

  6. 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? Molarity =

  7. 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? 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 CaCl2 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 CaCl2 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 CaCl2 in 600.0 mL of water? Molarity = 80.0 g x 1 mole 111.1 g 0.600 L = 1.20 M CaCl2 Ca2+ + 2Cl-

  11. 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? Molarity = 80.0 g x 1 mole 111.1 g 0.600 L = 1.20 M CaCl2 Ca2+ + 2Cl- 1.20 M

  12. 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? Molarity = 80.0 g x 1 mole 111.1 g 0.600 L = 1.20 M CaCl2 Ca2+ + 2Cl- 1.20 M 2.40 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 x 0.400 M = 0.160 M 0.160 M 100.0 Dilution Factor Ca(NO3)2 Ca2+ + 2NO3- 60.0 x 0.600 M = 0.360 M 0.720 M 100.0

  14. 4. If the [Cl-] = 0.600 M, calculate the number of grams AlCl3 that would be dissolved in 3.00 L of water.

  15. 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- 0.600 M

  16. 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 0.600 M

  17. 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

  18. 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

  19. 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 1 L

  20. 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 = 1 L 1 mole

  21. 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

  22. 5. If the [SO42-] = 0.900 M in 20.0 mL of Ga2(SO4)3, determine the [Ga3+] and the molarity of the solution.

  23. 5. If the [SO42-] = 0.900 M in 20.0 mL of Ga2(SO4)3, determine the [Ga3+] and the molarity of the solution. Ga2(SO4)3 2Ga3+ + 3SO42-

  24. 5. If the [SO42-] = 0.900 M in 20.0 mL of Ga2(SO4)3, determine the [Ga3+] and the molarity of the solution. Ga2(SO4)3 2Ga3+ + 3SO42- 0.900 M

  25. 5. If the [SO42-] = 0.900 M in 20.0 mL of Ga2(SO4)3, determine the [Ga3+] and the molarity of the solution. Ga2(SO4)3 2Ga3+ + 3SO42- 0.600 M 0.900 M

  26. 5. If the [SO42-] = 0.900 M in 20.0 mL of Ga2(SO4)3, determine the [Ga3+] and the molarity of the solution. Ga2(SO4)3 2Ga3+ + 3SO42- 0.300 M 0.600 M 0.900 M

  27. 5. If the [SO42-] = 0.900 M in 20.0 mL of Ga2(SO4)3, determine the [Ga3+] and the molarity of the solution. Ga2(SO4)3 2Ga3+ + 3SO42- 0.300 M 0.600 M 0.900 M Do not use the 20.0 mL!

  28. 6. Write the formula, complete, and net ionic equation for the reaction: Ca(s) + H2SO4(aq)→

  29. 6. Write the formula, complete, and net ionic equation for the reaction: Ca(s) + H2SO4(aq) → H2(g) + CaSO4(s) low solubility

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

  31. 6. Write the formula, complete, and net ionic equation for the reaction: Ca(s) + H2SO4(aq) → H2(g) + CaSO4(s) low solubility Ca(s) + 2H+ + SO42-→ H2(g) + CaSO4(s) only break up aqueous! Ca(s) + 2H+ + SO42-→ H2(g) + CaSO4(s) nothing cancels!

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