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Ionic Equations

Ionic Equations. The reaction of lead nitrate and potassium iodide to form bright yellow lead (II) iodide was dramatic. Let’s look at another example of this type of reaction in more detail. Ni +2 SO 4 -2. Na +1 CO 3 -2. The Reactants. sodium carbonate.

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Ionic Equations

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  1. Ionic Equations The reaction of lead nitrate and potassium iodide to form bright yellow lead (II) iodide was dramatic. Let’s look at another example of this type of reaction in more detail.

  2. Ni +2 SO4-2 Na+1 CO3-2 The Reactants sodium carbonate nickel (II) sulfate = NiSO4 = Na2CO3

  3. Visualize the Reactants nickel (II) sulfate sodium carbonate Na2CO3 (aq) There are ions everywhere! NiSO4 (aq) SO4-2 Na+1 CO3-2 Ni +2 Na+1

  4. Are they really Ionic? NiSO4 (aq) water Na2CO3 (aq)

  5. Ni +2 and CO3-2 = NiCO3 (s) This is insoluble by rule #5. Na+1and SO4-2 = Na2SO4 (aq) This is soluble by rules #1 & 4. Predict the Products NiSO4 (aq) + Na2CO3 (aq) -->

  6. Balance the Equation NiSO4 (aq) + Na2CO3 (aq) -->NiCO3 (s) + Na2SO4 (aq) Its balanced as written!

  7. Ionic & Net Equations NiSO4 (aq) + Na2CO3 (aq) -->NiCO3 (s) + Na2SO4 (aq) Write the ionic equation by showing all soluble compounds as separated ions. Ni+2 + SO4-2 + 2Na+1 + CO3-2 --> NiCO3 (s) + 2Na+1+ SO4-2 Identify the spectator ions. Remove the spectator ions. What’s left is the net ionic equation. Ni+2 + CO3-2 --> NiCO3 (s)

  8. Watch the Reaction No, Beaker, not that reaction!

  9. Watch the Reaction

  10. Nature of the Products The material at the bottom can be filtered out: it’s a solid. The liquid above the solid conducts a current and must contain ions in solution.

  11. Visualize the Products THAT’S what is in the Beaker ?! For the ions calculate molarity. SO4-2 Na+1 Na+1 NiCO3 (s) For the solid calculate mass.

  12. 0.100 L x 0.60 mol. Na2CO3 x 2 Na+1 = 0.12 moles Na+1 1 liter solution 1 Na2CO3 0.300 L x 0.20 mol. NiSO4 x 1 SO4-2 = 0.060 moles SO4-2 1 liter solution 1 NiSO4 Calculate Ion Concentrations The Na+1 ions came from the Na2CO3 (aq). There was 100. mL of this 0.60M solution. The SO4-2 ions came from the NiSO4 (aq). There was 300. mL of this 0.20M solution.

  13. 0.60 mol. Na2CO3 = X moles X = .060 moles Na2CO3 0.10 liter soln. 0.060 mol Na2CO3 x 2Na+1 = 0.12 moles Na+1 0.20 mol. NiSO4 = X moles X = 0.060 moles NiSO4 0.300 liter soln 0.060 mol. NiSO4 x 1 SO4-2 = 0.060 moles SO4-2 Alternate for Calculating Ion Concentrations The Na+1 ions came from the Na2CO3 (aq). There was 100. mL of this 0.60M solution. The SO4-2 ions came from the NiSO4 (aq). There was 300. mL of this 0.20M solution.

  14. For Na+1: M = 0.12 moles = 0.30M 0.400 L For SO4-2: M = 0.060 moles = 0.15M 0.400 L Combining 100 mL + 300 mL creates a solution of 400 mL. Super Beaker Solves the problem!

  15. 0.100 L x 0.60 mol. Na2CO3 x 1 CO3 = 0.060 moles CO3 1 liter solution 1 Na2CO3 0.300 L x 0.20 mol. NiSO4 x 1 Ni = 0.060 moles Ni 1 liter solution 1 NiSO4 CalculatePpt. Mass The CO3 in NiCO3 came from the Na2CO3 (aq). There was 100. mL of this 0.60M solution. 0.060 moles CO3 x 60.0 grams = 3.60 grams CO3 1 mole The Ni in NiCO3 came from the NiSO4 (aq). There was 300. mL of this 0.20M solution. 0.060 moles Ni x 58.7 grams = 3.5 grams Ni 1 mole 3.60 g + 3.5 g = 7.1 grams NiCO3

  16. 0.100 L x 0.60 mol. Na2CO3 x 1 NiCO3 x 118.7 g NiCO3 = 1 liter solution 1 Na2CO3 1 mole NiCO3 7.1 grams NiCO3 Ppt. Mass: alternate You can also solve this as a stoichiometry problem from the balanced equation: NiSO4 (aq) + Na2CO3 (aq) -->NiCO3 (s) + Na2SO4 (aq)

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