Net ionic equations and solubility rules PO43– Na+ 2Ca2+ 3Cl– Al3+ Net ionic equations and solubility rules

Review: forming ions Ionic (i.e. salt) refers to +ve ion plus -ve ion Usually this is a metal + non-metal or metal + polyatomic ion (e.g. NaCl, NaClO3, Li2CO3) Polyatomic ions are listed on page 71 (aq) means aqueous (dissolved in water) For salts (aq) means the salt exists as ions NaCl(aq) is the same as: Na+(aq) + Cl–(aq) Acids form ions: HCl(aq) is H+(aq) + Cl–(aq), Bases form ions: NaOH(aq) is Na+ + OH– Q - how is charge determined (+1, -1, +2, etc.)? A - via valences (periodic table or see pg. 71) F, Cl gain one electron, thus forming F–, Cl– Ca loses two electrons, thus forming Ca2+

Background: valences and formulas Charge can also be found via the compound E.g. in NaNO3(aq) if you know Na forms Na+, then NO3 must be NO3– (NaNO3 is neutral) By knowing the valence of one element you can often determine the other valences Q - Write the ions that form from Al2(SO4)3(aq)? Step 1 - look at the formula: Al2(SO4)3(aq) Step 2 - determine valences: Al3 (SO4)2 (Al is 3+ according to the periodic table) Step 3 - write ions: 2Al3+(aq) + 3SO42–(aq) Note that there are 2 aluminums because Al has a subscript of 2 in the original formula

Practice with writing ions Q - Write ions for Na2CO3(aq) A - 2Na+(aq) + CO32–(aq) (from the PT Na is 1+. There are 2, thus we have 2Na+. There is only one CO3. It must have a 2- charge) Notice that when ions form from molecules, charge can be separated, but the total charge (and number of each atom) stays constant. Q - Write ions for Ca3(PO4)2(aq) & Cd(NO3)2(aq) A - 3Ca2+(aq) + 2PO43–(aq) A - Cd2+(aq) + 2NO3–(aq) Q - Write ions for Na2S(aq) and Mg3(BO3)2(aq) A - 2Na+(aq) + S2–(aq), 3Mg2+(aq)+ 2BO33–(aq)

Types of chemical equations Equations can be divided into 3 types (pg. 396) 1) Molecular, 2) Ionic, 3) Net ionic Here is a typical molecular equation: Cd(NO3)2(aq) + Na2S(aq)  CdS(s) + 2NaNO3(aq) We can write this as an ionic equation (all compounds that are (aq) are written as ions): Cd2+(aq) + 2NO3–(aq) + 2Na+(aq) + S2–(aq)  CdS(s) + 2Na+(aq) + 2NO3–(aq) To get the NET ionic equation we cancel out all terms that appear on both sides: Net: Cd2+(aq) + S2–(aq)  CdS(s)

Equations must be balanced There are two conditions for molecular, ionic, and net ionic equations Materials balance Both sides of an equation should have the same number of each type of atom Electrical balance Both sides of a reaction should have the same net charge Q- When NaOH(aq) and MgCl2(aq) are mixed, _______(s) and NaCl(aq) are produced. Write balanced molecular, ionic & net ionic equations Mg(OH)2

First write the skeleton equation NaOH(aq) + MgCl2(aq)  Mg(OH)2(s) + NaCl(aq) Next, balance the equation 2 2 Ionic equation: 2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq)  Mg(OH)2(s) + 2Na+(aq) + 2Cl-(aq) Net ionic equation: 2OH-(aq) + Mg2+(aq)  Mg(OH)2(s) Write balanced ionic and net ionic equations: CuSO4(aq) + BaCl2(aq)  CuCl2(aq) + BaSO4(s) Fe(NO3)3(aq) + LiOH(aq)  ______(aq) + Fe(OH)3(s) Na3PO4(aq) + CaCl2(aq)  _________(s) + NaCl(aq) Na2S(aq) + AgC2H3O2(aq)  ________(aq) + Ag2S(s) LiNO3 Ca3(PO4)2 NaC2H3O2

Net: SO42–(aq) + Ba2+(aq)  BaSO4(s) Cu2+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq)  Cu2+(aq) + 2Cl–(aq) + BaSO4(s) Net: SO42–(aq) + Ba2+(aq)  BaSO4(s) Fe3+(aq) + 3NO3–(aq) + 3Li+(aq) + 3OH–(aq)  3Li+(aq) + 3NO3–(aq) + Fe(OH)3(s) Net: Fe3+(aq) + 3OH–(aq)  Fe(OH)3(s) 2Na3PO4(aq) + 3CaCl2(aq) Ca3(PO4)2(s)+ 6NaCl(aq) 6Na+(aq) + 2PO43–(aq) + 3Ca2+(aq) + 6Cl–(aq)  Ca3(PO4)2(s)+ 6Na+(aq) + 6Cl–(aq) Net: 2PO43–(aq) + 3Ca2+(aq)  Ca3(PO4)2(s) 2Na+(aq) + S2–(aq) + 2Ag+(aq) + 2C2H3O2–(aq)  2Na+(aq) + 2C2H3O2–(aq) + Ag2S(s) Net: S2–(aq) + 2Ag+(aq)  Ag2S(s)

Solubility Precipitation refers to the formation of a solid from ions. A precipitate is “insoluble” Soluble and insoluble are general terms to describe how much of a solid dissolves. Solubility can be predicted from rules (pg.399) These are general rules, based on observation To determine solubility, follow the rules in order Note: in rule 4 that sulfate = SO42- You will not have to memorize these rules, you will have to use the rules to predict solubility Read over example 11.2 (pg. 400) Do 11.26 (435) (list the relevant rule for each) Do PE 5 (pg. 400) and 11.28 (pg. 435)

AB(aq) + CD(aq)  AD(s) + CB(aq) Precipitation Precipitation refers to the formation of a solid from ions Metathesis refers to double displacement: AB + CD  AD + CB A metathesis involving ions going to one or more solids, is called precipitation: AB(aq) + CD(aq)  AD(s) + CB(aq) Recall: aq indicates the compound is aqueous (as ions)

Solubility - 11.26 a) Ca(NO3)2 - Soluble rule 2 (salts containing NO3- are soluble) b) FeCl2 - Soluble rule 3 (all chlorides are soluble) c) Ni(OH)2 - Insoluble rule 5 (all hydroxides are insoluble) d) AgNO3 - Soluble e) BaSO4 - Insoluble rule 4 (Sulfates are soluble, except … Ba2+) f) CuCO3 - Insoluble rule 6 (containing CO32- are insoluble)

Solubility PE 5 (a) - pg. 400 Ionic: Ag+(aq) + NO3-(aq) + NH4+(aq) + Cl-(aq) Note: combine, in your head, the positive and negative ions. If together a pair is insoluble, they will form a precipitate (s). In this case AgCl is insoluble (rule 3)  AgCl(s) + NO3-(aq) + NH4+(aq) Net ionic: Ag+(aq) + Cl-(aq)  AgCl(s) If no solid is formed then write N.R.

Solubility PE 5 (b), (c) - pg. 400 Ionic: 2Na+(aq) + S2– (aq) + Pb2+(aq) + 2C2H3O2–(aq) In this case PbS is insoluble (rule 6) 2Na+(aq) + S2–(aq) + Pb2+(aq) + 2C2H3O2–(aq)  PbS(s) + 2Na+(aq) + 2C2H3O2–(aq) Net ionic: Pb2+(aq) + S2–(aq)  PbS(s) Ba2+(aq) + 2Cl–(aq) + NH4+(aq) + NO3–(aq) In this case all combinations are soluble  Ba2+(aq) + 2Cl–(aq) + NH4+(aq) + NO3–(aq) Net ionic: N.R. (all ions cancel) For more lessons, visit www.chalkbored.com