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Solubility rules: Answers to lab

Solubility rules: Answers to lab. Answers 1 - 8. Alkali metals (IA) 2. NH 4 + NO 3 – 4. Br – , Cl – , SO 4 2– NO 3 – 6. Na + or K + or NH 4 + Ag 3 PO 4 AlPO 4 Ba 3 (PO 4 ) 2 Ca 3 (PO 4 ) 2 Co 3 (PO 4 ) 2 Cu 3 (PO 4 ) 2 FePO 4 Ni 3 (PO 4 ) 2

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Solubility rules: Answers to lab

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  1. Solubility rules: Answers to lab

  2. Answers 1 - 8 • Alkali metals (IA) 2. NH4+ • NO3– 4. Br –, Cl –, SO42– • NO3– 6. Na+ or K+ or NH4+ • Ag3PO4 AlPO4 Ba3(PO4)2 Ca3(PO4)2 Co3(PO4)2 Cu3(PO4)2 FePO4 Ni3(PO4)2 • Ca(NO3)2 Soluble rule 2 FeCl3 Soluble rule 3 Ni(OH)2 Insoluble rule 5 AgNO3 Soluble rule 2 BaSO4 Insoluble rule 4 CuCO3 Insoluble rule 6

  3. Answer 9 You have a mixture of Ba2+, Pb2+, Cu2+, Na+ Which can be precipitated without the others? Use Cl –, Br –, or I – to precipitate Pb2+ (rule 3) Filtering leaves Ba2+, Cu2+, Na+ Use SO42– to precipitate Ba2+ (rule 4) Filtering leaves Cu2+, Na+ Use OH – (rule 5), or any of PO43–, CO32–, SO32–, S2– (rule 6) to precipitate Cu2+ Since all Na+ compounds are soluble, this will stay in solution

  4. Determining net ionic reactions Step 1: Write formula for reactants and products using valences. Products are determined by switching +ve and –ve. Step 2: Determine if any products are insoluble (use solubility rules). Note: all reactants must be soluble (i.e. aq) in order to mix. If all products are aqueous: “no reaction” Step 3: Balance the equation Step 4: Write the ionic equation Step 5: Write the net ionic equation E.g. sodium phosphate + calcium chloride

  5. Answers: 10 i) and ii) i) (NH4)2S(aq) + Pb(NO3)2(aq)  PbS(s) + 2NH4NO3(aq) 2NH4+(aq) + S2–(aq) + Pb2+(aq) + 2NO3–(aq)  PbS(s) + 2NH4+(aq) + 2NO3–(aq) Net: S2–(aq) + Pb2+(aq)  PbS(s) ii) Co(NO3)2(aq) + Na2SO4(aq)  No reaction Step 1: Write correct formulas Step 2: Determine if any products are insoluble Step 3: Balance the equation Step 4: Write the ionic equation Step 5: Write the net ionic equation

  6. Answers: 10 iii) - vi) iii) AgNO3(aq) + NaI(aq)  AgI(s) + NaNO3(aq) Ag+(aq) + NO3–(aq) + Na+(aq) + I–(aq)  AgI(s) + Na+(aq) + NO3–(aq) Net: Ag+(aq) + I–(aq)  AgI(s) iv)Al(NO3)3(aq) + 3KOH(aq)  Al(OH)3(s) + 3KNO3(aq) Al3+(aq) + 3NO3–(aq) + 3K+(aq) + 3OH–(aq)  Al(OH)3(s) + 3K+(aq) + 3NO3–(aq) Net: Al3+(aq) + 3OH–(aq)  Al(OH)3(s) v)Zn(NO3)2(aq) + K2CO3(aq)  ZnCO3(s) + 2KNO3(aq) Zn2+(aq) + 2NO3–(aq) + 2K+(aq) + CO32–(aq)  ZnCO3(s) + 2K+(aq) + 2NO3–(aq) Net: Zn2+(aq) + CO32–(aq)  ZnCO3(s) vi) KCl(aq) + NaNO3(aq)  No reaction For more lessons, visit www.chalkbored.com

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