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Writing Net Ionic Equations

Writing Net Ionic Equations. Writing ionic equations. A soluble compound will dissociate into ions. An insoluble compound does not dissociate . Precipitate- insoluble solid formed in a reaction. It looks cloudy. NO 3 1–. Pb 2+. Pb 2+. NO 3 1–. NO 3 1–. NO 3 1–.

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Writing Net Ionic Equations

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  1. Writing Net Ionic Equations

  2. Writing ionic equations • A soluble compound will dissociate into ions. • An insoluble compound does not dissociate . • Precipitate- insoluble solid formed in a reaction. It looks cloudy.

  3. NO31– Pb2+ Pb2+ NO31– NO31– NO31– • Pb(NO3)2(s) Pb(NO3)2(aq) • Pb2+(aq) + 2 NO31–(aq) • add water • dissociation: “splitting into ions”

  4. Double Replacement Reaction AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) Ag+ Na+ NO3– Cl–  Ag+ NO3– Na+ Cl–

  5. Pb(NO3)2(aq) + 2NaI(aq)g PbI2i+ 2NaNO3(aq) Molecular Equation Ionic Equation: Pb2+ + 2NO3- + 2Na+ + 2I-g PbI2i + 2Na+ + 2NO3- Spectator ions appear on both sides of the equation without changing. Net ionic equations ignore spectator ions. Net ionic: Pb2+(aq)+ 2I-(aq) g PbI2(s)

  6. Try this • _FeCl3(aq) + _Cu(NO3)2(aq)  ? • If each product is soluble, no reaction occurs (NR)

  7. For double-replacement reactions, reaction will occur if any product is: water, a gas, a precipitatedriving forces _Pb(NO3)2(aq) + _KI(aq)  _PbI2(s) + _KNO3(aq) _KOH(aq) + _H2SO4(aq)  _K2SO4(aq) + _H2O(l)

  8. Summary • Check for solubility of each product. • Write any insoluble products as the product in the net ionic equation. • The reactants will be the ions from which the precipitate formed.

  9. Molarity moles of solute L of solution

  10. One liter of 5.0M HCl contains how many moles of HCl? 1 L x 5.0 mole= 5.0 mole L 55ml of 2.4M glucose contains how many moles of glucose? .055L x 2.4 mole = 0.132 mole L V∙M = mole

  11. How could we make a .100M solution of CuSO4∙5H2O? We need .100 mole of the hydrate in 1.00L of … ? solution (not solvent) How do we measure .100 mole? Molar mass of CuSO4∙5H2O 1 Cu 63.5 1 S 32.1 9 O 144.0 10 H 10.0 249.6

  12. .100 mole x 249.6 g= 24.96 g 1.00 mole in enough water to make 1.00 L

  13. What is the molarity of 250ml solution containing 9.46g CsBr? ggmole 9.46g x 1 mole CsBr= 0.0444 mole 213 g CsBr 250 ml = .250 L M = 0.0444 mole= 0.178 M CsBr .250L

  14. Dilutions How can we make 500 ml of 1.00 M HCl from 6.00 M HCl? Do the number of moles of solute change? No! Since V∙M = mole, then V1M1 = V2M2 V1= ? M1= 6.00 M V2= .500L M2=1.00M V1(6.00M) = (.500L)(1.00M) V1 = .500L∙M= .0833L or 83.3ml 6.00M

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