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Electrolytes

Electrolytes. Non-electrolytes. Molecular compounds. Do NOT form ions in water. Weak electrolytes. Weak acids. Insoluble salts (insoluble ionic compounds). Form very VERY few ions in water. Electrolytes. Strong electrolytes. Strong acids. Soluble salts (soluble ionic compounds).

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Electrolytes

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  1. Electrolytes Non-electrolytes Molecular compounds Do NOT form ions in water Weak electrolytes Weak acids Insoluble salts (insoluble ionic compounds) Form very VERY few ions in water

  2. Electrolytes Strong electrolytes Strong acids Soluble salts (soluble ionic compounds) Completely break apart into ions

  3. Electrolytes Strong acids? HCl(aq) HNO3(aq) HBr(aq) HClO3(aq) HI(aq) HClO4(aq) H2SO4(aq) ANY other acid is a weak acid (which makes it a weak electrolyte)

  4. Solubility Rules THESE ONLY APPLY TO IONIC COMPOUNDS!!!!! • All ammonium and group 1 metal salts are SOLUBLE (no exceptions) • All nitrate, chlorate, perchlorate, and acetate salts are SOLUBLE (no exceptions) • Most chloride, bromide, and iodide salts are SOLUBLE. EXCEPTIONS! Compound containing silver, mercury (I), and lead (II) • Most fluoride salts are SOLUBLE. EXCEPTIONS! Compounds containing magnesium, calcium, strontium, barium, and lead (II) • Most sulfate salts are SOLUBLE. EXCEPTIONS! Compounds containing strontium, barium, mercury (I), and lead (II) • Most sulfide and hydroxide salts are INSOLUBLE. EXCEPTIONS! Compounds containing calcium, strontium, and barium • Most carbonate, phosphate, oxalate, and chromate salts are INSOLUBLE.

  5. Solubility and Reactions Na2C2O4(aq) + CaCl2(aq) 2 NaCl(aq) + CaC2O4(s) Na2C2O4(aq) + CaCl2(aq) 2 NaCl(aq) + CaC2O4(s) Na2C2O4(aq) + CaCl2(aq) 2 NaCl(aq) + CaC2O4(s) Na2C2O4(aq) + CaCl2(aq) 2 NaCl(aq) + CaC2O4(s) Na2C2O4(aq) + CaCl2(aq) 2 NaCl(aq) + CaC2O4(s) S-E S-E S-E W-E 3 H2SO4(aq) + 2 Fe(OH)3(s) Fe2(SO4)3(aq) + 6 H2O(l) 3 H2SO4(aq) + 2 Fe(OH)3(s) Fe2(SO4)3(aq) + 6 H2O(l) 3 H2SO4(aq) + 2 Fe(OH)3(s) Fe2(SO4)3(aq) + 6 H2O(l) 3 H2SO4(aq) + 2 Fe(OH)3(s) Fe2(SO4)3(aq) + 6 H2O(l) 3 H2SO4(aq) + 2 Fe(OH)3(s) Fe2(SO4)3(aq) + 6 H2O(l) S-E W-E S-E N-E 2 LiCl(aq) + Hg2(C2H3O2)2(aq) 2 LiC2H3O2(aq) + Hg2Cl2(s) 2 LiCl(aq) + Hg2(C2H3O2)2(aq) 2 LiC2H3O2(aq) + Hg2Cl2(s) 2 LiCl(aq) + Hg2(C2H3O2)2(aq) 2 LiC2H3O2(aq) + Hg2Cl2(s) 2 LiCl(aq) + Hg2(C2H3O2)2(aq) 2 LiC2H3O2(aq) + Hg2Cl2(s) 2 LiCl(aq) + Hg2(C2H3O2)2(aq) 2 LiC2H3O2(aq) + Hg2Cl2(s) S-E S-E S-E W-E

  6. Solubility and Reactions 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) W-E S-E W-E N-E K2CO3(aq) + 2 HClO4(aq) 2 KClO4(aq) + H2O(l) + CO2(g) K2CO3(aq) + 2 HClO4(aq) 2 KClO4(aq) + H2O(l) + CO2(g) K2CO3(aq) + 2 HClO4(aq) 2 KClO4(aq) + H2O(l) + CO2(g) K2CO3(aq) + 2 HClO4(aq) 2 KClO4(aq) + H2O(l) + CO2(g) K2CO3(aq) + 2 HClO4(aq) 2 KClO4(aq) + H2O(l) + CO2(g) K2CO3(aq) + 2 HClO4(aq) 2 KClO4(aq) + H2O(l) + CO2(g) S-E S-E S-E N-E N-E 2 NaNO3(aq) + Mg(ClO3)2(aq) 2 NaClO3(aq) + Mg(NO3)2(aq) 2 NaNO3(aq) + Mg(ClO3)2(aq) 2 NaClO3(aq) + Mg(NO3)2(aq) 2 NaNO3(aq) + Mg(ClO3)2(aq) 2 NaClO3(aq) + Mg(NO3)2(aq) 2 NaNO3(aq) + Mg(ClO3)2(aq) 2 NaClO3(aq) + Mg(NO3)2(aq) 2 NaNO3(aq) + Mg(ClO3)2(aq) 2 NaClO3(aq) + Mg(NO3)2(aq) S-E S-E S-E S-E

  7. A solution of chromium (III) sulfate is added to a solution of cesium phosphate. Give the net ionic equation for any reaction that happens. Step 1: Write a balanced chemical equation (molecular equation) Cr2(SO4)3 + Cs3PO4  CrPO4 + Cs2SO4 Cr2(SO4)3(aq) + 2 Cs3PO4(aq)2 CrPO4(s)+ 3 Cs2SO4(aq) Cr2(SO4)3(aq) + 2 Cs3PO4(aq)2 CrPO4(s)+ 3 Cs2SO4(aq) Cr2(SO4)3(aq) + 2 Cs3PO4(aq)2 CrPO4(s)+ 3 Cs2SO4(aq) Cr2(SO4)3(aq) + 2 Cs3PO4(aq) 2 CrPO4(s)+ 3 Cs2SO4(aq) Cr2(SO4)3(aq) + 2 Cs3PO4(aq)2 CrPO4(s)+ 3 Cs2SO4(aq) Cr2(SO4)3(aq) + 2 Cs3PO4(aq)2 CrPO4(s)+ 3 Cs2SO4(aq) Cr2(SO4)3(aq) + 2 Cs3PO4(aq)2 CrPO4(s)+ 3 Cs2SO4(aq)

  8. A solution of chromium (III) sulfate is added to a solution of cesium phosphate. Give the net ionic equation for any reaction that happens. Step 1: Write a balanced chemical equation (molecular equation) Cr2(SO4)3(aq) + 2 Cs3PO4(aq) 2 CrPO4(s)+ 3 Cs2SO4(aq) SE SE WE SE Step 2: Break STRONG ELECTROLYTES ONLY into ions 2 Cr+3(aq) + 3 SO42-(aq) + 6 Cs+(aq) + 2 PO4-3(aq) 2 CrPO4(s) + 6 Cs+(aq) + 3 SO42-(aq) 2 Cr+3(aq) + 3 SO42-(aq) 2 Cr+3(aq) + 3 SO42-(aq) + 6 Cs+(aq) + 2 PO4-3(aq) 2 Cr+3(aq) + 3 SO42-(aq) + 6 Cs+(aq) + 2 PO4-3(aq) 2 CrPO4(s)

  9. A solution of chromium (III) sulfate is added to a solution of cesium phosphate. Give the net ionic equation for any reaction that happens. Step 1: Write a balanced chemical equation (molecular equation) Cr2(SO4)3(aq) + 2 Cs3PO4(aq) 2 CrPO4(s)+ 3 Cs2SO4(aq) SE SE WE SE Step 2: Break STRONG ELECTROLYTES ONLY into ions 2 Cr+3(aq) + 3 SO42-(aq) + 6 Cs+(aq) + 2 PO4-3(aq) 2 CrPO4(s) + 6 Cs+(aq) + 3 SO42-(aq) Step 3: Cross out all spectator ions N.I.E 2 Cr+3(aq) + 2 PO4-3(aq) 2 CrPO4(s) 2 Cr+3(aq) + 2 PO4-3(aq) 2 CrPO4(s)

  10. A zinc acetate solution is combined with 3.50 M perchloric acid. Give the net ionic equation. Step 1: Write a balanced chemical equation (molecular equation) HClO4 + Zn(C2H3O2)2  HC2H3O2 2 HClO4(aq) + Zn(C2H3O2)2(aq)2 HC2H3O2(aq)+ Zn(ClO4)2(aq) 2 HClO4(aq) + Zn(C2H3O2)2(aq)2 HC2H3O2(aq)+ Zn(ClO4)2(aq) 2 HClO4(aq) + Zn(C2H3O2)2(aq)2 HC2H3O2(aq)+ Zn(ClO4)2(aq) 2 HClO4(aq) + Zn(C2H3O2)2(aq)2 HC2H3O2(aq)+ Zn(ClO4)2(aq) 2 HClO4(aq) + Zn(C2H3O2)2(aq)2 HC2H3O2(aq)+ Zn(ClO4)2(aq) 2 HClO4(aq) + Zn(C2H3O2)2(aq)2 HC2H3O2(aq)+ Zn(ClO4)2(aq) 2 HClO4(aq) + Zn(C2H3O2)2(aq) 2 HC2H3O2(aq)+ Zn(ClO4)2(aq)

  11. A zinc acetate solution is combined with 3.50 M perchloric acid. Give the net ionic equation. Step 1: Write a balanced chemical equation (molecular equation) 2 HClO4(aq) + Zn(C2H3O2)2(aq) 2 HC2H3O2(aq)+ Zn(ClO4)2(aq) SE SE WE SE Step 2: Break STRONG ELECTROLYTES ONLY into ions 2 H+(aq) + 2 ClO4-(aq) + Zn2+(aq) + 2 C2H3O2-(aq) 2 HC2H3O2(aq) + Zn2+(aq) + 2 ClO4-(aq) 2 H+(aq) + 2 ClO4-(aq) + Zn2+(aq) + 2 C2H3O2-(aq) 2 HC2H3O2(aq) 2 H+(aq) + 2 ClO4-(aq) + Zn2+(aq) + 2 C2H3O2-(aq) 2 H+(aq) + 2 ClO4-(aq)

  12. A zinc acetate solution is combined with 3.50 M perchloric acid. Give the net ionic equation. Step 1: Write a balanced chemical equation (molecular equation) 2 HClO4(aq) + Zn(C2H3O2)2(aq) 2 HC2H3O2(aq)+ Zn(ClO4)2(aq) SE SE WE SE Step 2: Break STRONG ELECTROLYTES ONLY into ions 2 H+(aq) + 2 ClO4-(aq) + Zn2+(aq) + 2 C2H3O2-(aq) 2 HC2H3O2(aq) + Zn2+(aq) + 2 ClO4-(aq) Step 3: Cross out all spectator ions 2 H+(aq) + 2 C2H3O2-(aq) 2 HC2H3O2(aq) 2 H+(aq) + 2 C2H3O2-(aq) 2 HC2H3O2(aq) N.I.E

  13. Net Ionic Equations Start with a balanced chemical equation: Na2C2O4(aq) + CaCl2(aq) 2 NaCl(aq) + CaC2O4(s) Na2C2O4(aq) + CaCl2(aq) 2 NaCl(aq) + CaC2O4(s) M.E. S.E. S.E. S.E. W.E. Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq) 2Na+(aq) + 2Cl-(aq) + CaC2O4(s) 2Na+(aq) + C2O4-2(aq) 2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq) 2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq) 2Na+(aq) + 2Cl-(aq)

  14. Net Ionic Equations Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq) 2Na+(aq) + 2Cl-(aq) + CaC2O4(s) Cross out every thing that is EXACTLY the same on both sides: 2Na+(aq) + C2O4-2(aq) + Ca2+(aq) + 2Cl-(aq) 2Na+(aq) + 2Cl-(aq) + CaC2O4(s) C2O4-2(aq) + Ca2+(aq) CaC2O4(s) N.I.E.

  15. Net Ionic Equations Start with a balanced chemical equation (including subscripts): M.E. 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) 2 HF(aq) + Ba(OH)2(aq) BaF2(s) + 2 H2O(l) W.E. S.E. W.E. N.E. Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2 HF(aq) + Ba2+(aq) + 2OH-(aq) BaF2(s) 2 HF(aq) 2 HF(aq) + Ba2+(aq) + 2OH-(aq) 2 HF(aq) + Ba2+(aq) + 2OH-(aq) BaF2(s) + 2 H2O(l)

  16. Net Ionic Equations Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2 HF(aq) + Ba2+(aq) + 2OH-(aq) BaF2(s) + 2 H2O(l) Cross out every thing that is EXACTLY the same on both sides: 2 HF(aq) + Ba2+(aq) + 2OH-(aq) BaF2(s) + 2 H2O(l) N.I.E. 2 HF(aq) + Ba2+(aq) + 2OH-(aq) BaF2(s) + 2 H2O(l)

  17. Net Ionic Equations Start with a balanced chemical equation (including subscripts): M.E. Fe2(SO4)3(aq) + 3 PbCl2(s) 2 FeCl3(aq) + 3 PbSO4(s) Fe2(SO4)3(aq) + 3 PbCl2(s) 2 FeCl3(aq) + 3 PbSO4(s) Fe2(SO4)3(aq) + 3 PbCl2(s) 2 FeCl3(aq) + 3 PbSO4(s) Fe2(SO4)3(aq) + 3 PbCl2(s) 2 FeCl3(aq) + 3 PbSO4(s) Fe2(SO4)3(aq) + 3 PbCl2(s) 2 FeCl3(aq) + 3 PbSO4(s) S.E. W.E. S.E. W.E. Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s) 2Fe+3(aq) + 6Cl-(aq) 2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s) 2Fe+3(aq) + 3SO4-2(aq) 2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s) 2Fe+3(aq) + 6Cl-(aq) + 3PbSO4(s)

  18. Net Ionic Equations Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s) 2Fe+3(aq) + 6Cl-(aq) + 3PbSO4(s) Cross out every thing that is EXACTLY the same on both sides: 2Fe+3(aq) + 3SO4-2(aq) + 3PbCl2(s) 2Fe+3(aq) + 6Cl-(aq) + 3PbSO4(s) N.I.E. 3SO4-2(aq) + 3PbCl2(s) 2Cl-(aq) + 3PbSO4(s) 3SO4-2(aq) + 3PbCl2(s) 6Cl-(aq) + 3PbSO4(s)

  19. Net Ionic Equations Start with a balanced chemical equation (including subscripts): M.E. NaNO3(aq) + LiCl(aq) NaCl(aq) + LiNO3(aq) NaNO3(aq) + LiCl(aq) NaCl(aq) + LiNO3(aq) NaNO3(aq) + LiCl(aq) NaCl(aq) + LiNO3(aq) NaNO3(aq) + LiCl(aq) NaCl(aq) + LiNO3(aq) NaNO3(aq) + LiCl(aq) NaCl(aq) + LiNO3(aq) S.E. S.E. S.E. S.E. Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. Na+(aq) + NO3-(aq) + Li+(aq) + Cl-(aq) Na+(aq) + NO3-(aq) Na+(aq) + NO3-(aq) + Li+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) Na+(aq) + NO3-(aq) + Li+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + Li+(aq) + NO3-(aq)

  20. Net Ionic Equations Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. Na+(aq) + NO3-(aq) + Li+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + Li+(aq) + NO3-(aq) Cross out every thing that is EXACTLY the same on both sides: Na+(aq) + NO3- (aq) + Li+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + Li+(aq) + NO3-(aq) No reaction

  21. Net Ionic Equations Start with a balanced chemical equation (including subscripts): M.E. 2 HC2H3O2(aq) + Ba(OH)2(aq) Ba(C2H3O2)2(aq) + 2 H2O(l) 2 HC2H3O2(aq) + Ba(OH)2(aq) Ba(C2H3O2)2(aq) + 2 H2O(l) 2 HC2H3O2(aq) + Ba(OH)2(aq) Ba(C2H3O2)2(aq) + 2 H2O(l) 2 HC2H3O2(aq) + Ba(OH)2(aq) Ba(C2H3O2)2(aq) + 2 H2O(l) 2 HC2H3O2(aq) + Ba(OH)2(aq) Ba(C2H3O2)2(aq) + 2 H2O(l) W.E. S.E. S.E. N.E. Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2 HC2H3O2(aq) + Ba2+(aq) + 2 OH-(aq) 2 HC2H3O2(aq) 2 HC2H3O2(aq) + Ba2+(aq) + 2 OH-(aq) Ba2+(aq) + 2 C2H3O2-(aq) 2 HC2H3O2(aq) + Ba2+(aq) + 2 OH-(aq) Ba2+(aq) + 2 C2H3O2-(aq) + 2 H2O(l)

  22. Net Ionic Equations Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2 HC2H3O2(aq) + Ba2+(aq) + 2 OH-(aq) Ba2+(aq) + 2 C2H3O2-(aq) + 2 H2O(l) Cross out every thing that is EXACTLY the same on both sides: 2 HC2H3O2(aq) + Ba2+(aq) + 2OH-(aq) Ba2+(aq) + 2C2H3O2-(aq) + 2 H2O(l) 2 HC2H3O2(aq) + 2OH-(aq) 2C2H3O2-(aq) + 2 H2O(l) HC2H3O2(aq) + OH-(aq) C2H3O2-(aq) + H2O(l) N.I.E.

  23. Net Ionic Equations Start with a balanced chemical equation (including subscripts): M.E. H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l) H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l) H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l) H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l) H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l) S.E. S.E. S.E. N.E. Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq) 2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq) 2Na+(aq) + SO4-2(aq) 2H+(aq) + SO4-2(aq) 2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq) 2Na+(aq) + SO4-2(aq) + 2H2O(l)

  24. Net Ionic Equations Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone: I.E. 2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq) 2Na+(aq) + SO4-2(aq) + 2H2O(l) Cross out every thing that is EXACTLY the same on both sides: 2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq) 2Na+(aq) + SO4-2(aq) + 2H2O(l) H+(aq) + OH-(aq) H2O(l) 2 H+(aq) + 2 OH-(aq) 2 H2O(l) N.I.E.

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