Solubility and Net Ionic Equations

Solubility and Net Ionic Equations Courtesy of http://chemistry.bd.psu.edu/jircitano/soluble.html

Net Ionic Equations • What’s the big deal with solubility? • Well, if the ion is soluble, it won’t form a precipitate, and this means it doesn’t react and should be left out of the net ionic equation. • The key is first to write the compound’s chemical formula and then determine if it’s soluble. If it is soluble, then ionize it—if it isn’t, don’t ionize it; leave it as a molecule.

Overview • Molecular equations show species reacting as their molecular formula, with subscripts added to indicate their solid, liquid, gaseous, or aqueous nature. • Ionic equations show species reacting as their ionic components. Subscripts are not needed to describe the state of the matter, because all ions are in aqueous solution. • A net ionic equation is one in which spectator ions are removed. Spectator ions are present in solution but do not participate in the actual precipitation reaction.

Key Terms • salt • An ionic compound that is composed of cations and anions. The constituent ions are held together by ionic bonds, not covalent bonds • electrolyte • a substance that, when dissolved in solution, will enable the solution to conduct electricity • spectator ion • an ion that is present in solution but does not participate in a precipitation reaction

Your new best friend – Memorize it.

Your new best friend – Memorize it. • A Few Provisos, Ah, A Couple of Quid Pro Quo… • The following solubility rules predict the solubility of many ionic compound when applied in order:

Why do some things dissolve? • Intermolecular forces • In process of dissolving, molecules of the solute are inserted into a solvent and surrounded by its molecules. In order for the process to take place, molecular bonds between molecules of solute (ie. sugar) have to be broken and molecular bonds of the solvent also have to be disrupted. Both of these require energy. • Example of sugar • When sugar dissolves in water, new bonds between sugar and water are created. During this process energy is given off. The amount of this energy is sufficient to brake bonds between molecules of sugar and between molecules of water. This example is relevant to any solute and solvent. If the bonds between solvent or solute are to strong and there is not enough energy provided while dissolving to brake them, the solute will not dissolve. • Example of salt (Ionic compounds) • The same energy rule can be applied to salts. They are built of positive and negative ions which are bound together by the force of attraction of their opposite charges. If energy needed to brake their ionic bounds is lower then the energy given off by an interaction of the ions with solvent (ie. water), the salts will dissociate and interact with solvent and thus dissolve.

EXAMPLE 1 • Let's first start with a complete chemical equation and see how the net ionic equation is derived. Take for example the reaction of lead(II) nitrate with sodium chloride to form lead(II) chloride and sodium nitrate, shown below: • Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq)

Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) • This complete equation may be rewritten in ionic form by using the solubility rules. • Rule 2 confirms that lead(II) nitrate is soluble and therefore dissociated. • Rule 1 says the same about NaCl. • As products, sodium nitrate is predicted to be soluble (rules 1 and 2) and will be dissociated. • The lead(II) chloride, however, is insoluble (rule 3).

Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) • This complete equation may be rewritten in ionic form by using the solubility rules. • Rule 2 confirms that lead(II) nitrate is soluble and therefore dissociated. • Rule 1 says the same about NaCl. • As products, sodium nitrate is predicted to be soluble (rules 1 and 2) and will be dissociated. • The lead(II) chloride, however, is insoluble (rule 3). • The above equation written in dissociated form is:

Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) • This complete equation may be rewritten in ionic form by using the solubility rules. • Rule 2 confirms that lead(II) nitrate is soluble and therefore dissociated. • Rule 1 says the same about NaCl. • As products, sodium nitrate is predicted to be soluble (rules 1 and 2) and will be dissociated. • The lead(II) chloride, however, is insoluble (rule 3). • The above equation written in dissociated form is: Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq)  PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq)

Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq)  PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq) • At this point, one may cancel out those ions which have not participated in the reaction. Notice how the nitrate ions and sodium ions remain unchanged on both sides of the reaction.

Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq)  PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq) • At this point, one may cancel out those ions which have not participated in the reaction. Notice how the nitrate ions and sodium ions remain unchanged on both sides of the reaction. • Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq)PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq)

Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq)PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq) • At this point, one may cancel out those ions which have not participated in the reaction. Notice how the nitrate ions and sodium ions remain unchanged on both sides of the reaction. • Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq)  PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq) • What remains is the net ionic equation, showing only those chemical species participating in a chemical process:

Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq)  PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq) • At this point, one may cancel out those ions which have not participated in the reaction. Notice how the nitrate ions and sodium ions remain unchanged on both sides of the reaction. • Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq) PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq) • What remains is the net ionic equation, showing only those chemical species participating in a chemical process: Pb2+(aq) + 2 Cl-(aq) PbCl2(s)

Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq)  PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq) • At this point, one may cancel out those ions which have not participated in the reaction. Notice how the nitrate ions and sodium ions remain unchanged on both sides of the reaction. • Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq) PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq) • What remains is the net ionic equation, showing only those chemical species participating in a chemical process: Pb2+(aq) + 2 Cl-(aq)PbCl2(s) • After a while, you should be able to predict the net ionic equation given only the reactants.

Now it’s your turn! • For instance, suppose you had to determine the net ionic equation resulting from the reaction of solutions of sodium sulfate and barium bromide:

Now it’s your turn! • For instance, suppose you had to determine the net ionic equation resulting from the reaction of solutions of sodium sulfate and barium bromide: BaBr2(aq) + Na2SO4(aq)?

BaBr2(aq) + Na2SO4(aq)? • For instance, suppose you had to determine the net ionic equation resulting from the reaction of solutions of sodium sulfate and barium bromide: BaBr2(aq) + Na2SO4(aq)? • One way to tackle this problem is to examine what ions are found together in solution: Ba2+, Br-, Na+, and SO42-. • We know that barium ions and bromide ions are soluble together (rule 4), but will sodium ions or sulfate ions combine with barium ions to form an insoluble compound?

BaBr2(aq) + Na2SO4(aq)? • Barium ions and sodium ions, both being positive in charge will repel each other, so no compound is expected to form between them. • On the other hand, sulfate ions and barium ions would form barium sulfate (insoluble; rule 6). • The sodium ions must therefore combine with bromide ions to form sodium bromide. • According to the solubility rules, sodium bromide should be soluble (rule 1).

BaBr2(aq) + Na2SO4(aq)? • Barium ions and sodium ions, both being positive in charge will repel each other, so no compound is expected to form between them. • On the other hand, sulfate ions and barium ions would form barium sulfate (insoluble; rule 6). • The sodium ions must therefore combine with bromide ions to form sodium bromide. • According to the solubility rules, sodium bromide should be soluble (rule 1). • Now we can write a complete balanced equation:

BaBr2(aq) + Na2SO4(aq)? • Barium ions and sodium ions, both being positive in charge will repel each other, so no compound is expected to form between them. • On the other hand, sulfate ions and barium ions would form barium sulfate (insoluble; rule 6). • The sodium ions must therefore combine with bromide ions to form sodium bromide. • According to the solubility rules, sodium bromide should be soluble (rule 1). • Now we can write a complete balanced equation: • BaBr2(aq) + Na2SO4(aq)BaSO4(s) + 2 NaBr(aq)

NOW… Let’s write that ionic equation! • As before, the above equation can be rewritten showing the soluble species as ions in solution: Ba2+(aq) + 2 Br-(aq) + 2 Na+(aq) + SO42-(aq) BaSO4(s) + 2 Na+(aq) + 2 Br-(aq)

NOW… Let’s write that ionic equation! • As before, the above equation can be rewritten showing the soluble species as ions in solution: Ba2+(aq) + 2 Br-(aq) + 2 Na+(aq) + SO42-(aq) BaSO4(s) + 2 Na+(aq) + 2 Br-(aq) • Next, cross out any species which have not changed on both sides of the reactions; these are sometimes called spectator ions:

NOW… Let’s write that ionic equation! • As before, the above equation can be rewritten showing the soluble species as ions in solution: Ba2+(aq) + 2 Br-(aq) + 2 Na+(aq) + SO42-(aq) BaSO4(s) + 2 Na+(aq) + 2 Br-(aq) • Next, cross out any species which have not changed on both sides of the reactions; these are sometimes called spectator ions: Ba2+(aq) + 2 Br-(aq) + 2 Na+(aq) + SO42-(aq) BaSO4(s) + 2 Na+(aq) + 2 Br-(aq)

NOW… Let’s write that ionic equation! • As before, the above equation can be rewritten showing the soluble species as ions in solution: Ba2+(aq) + 2 Br-(aq) + 2 Na+(aq) + SO42-(aq) BaSO4(s) + 2 Na+(aq) + 2 Br-(aq) • Next, cross out any species which have not changed on both sides of the reactions; these are sometimes called spectator ions: Ba2+(aq) + 2 Br-(aq) + 2 Na+(aq) + SO42-(aq) BaSO4(s) + 2 Na+(aq) + 2 Br-(aq) • What remains is the balanced, net ionic equation:

NOW… Let’s write that ionic equation! • As before, the above equation can be rewritten showing the soluble species as ions in solution: Ba2+(aq) + 2 Br-(aq) + 2 Na+(aq) + SO42-(aq) BaSO4(s) + 2 Na+(aq) + 2 Br-(aq) • Next, cross out any species which have not changed on both sides of the reactions; these are sometimes called spectator ions: Ba2+(aq) + 2 Br-(aq) + 2 Na+(aq) + SO42-(aq) BaSO4(s) + 2 Na+(aq) + 2 Br-(aq) • What remains is the balanced, net ionic equation: Ba2+(aq) + SO42-(aq)BaSO4(s)

Ready for Homework? • Or more practice?

Hydrochloric acid and sodium hydroxide are mixed. Write the net ionic equation.

Hydrochloric acid and sodium hydroxide are mixed. Write the net ionic equation. • This is a mixture of a strong acid and a strong base, so each ionizes completely. H+ + Cl- + Na+ + OH- 

Hydrochloric acid and sodium hydroxide are mixed. Write the net ionic equation. • This is a mixture of a strong acid and a strong base, so each ionizes completely. H+ + Cl- + Na+ + OH-  • The two possible compounds formed are sodium chloride, which is soluble, and water, which is molecular; thus water is the only product in our net ionic equation. H+(aq)+ OH-(aq) H2O(l)

Chlorine gas is bubbled into a solution of potassium iodide; write the net ionic equation. • This one is a single replacement, so you need to consider the activity series. Since halogens are involved, you can determine their activity by using the periodic table: Cl is more active than I.

Chlorine gas is bubbled into a solution of potassium iodide; write the net ionic equation. • This one is a single replacement, so you need to consider the activity series. Since halogens are involved, you can determine their activity by using the periodic table: Cl is more active than I. Cl2 + K+ + I- 

Chlorine gas is bubbled into a solution of potassium iodide; write the net ionic equation. • This one is a single replacement, so you need to consider the activity series. Since halogens are involved, you can determine their activity by using the periodic table: Cl is more active than I. Cl2 + K+ + I-  • Remember that halogens are diatomic and that all potassium compounds are soluble. The resulting compound is also soluble, so K+ is a spectator and is left out of the final equation. Cl2(g)+ I-(aq) I2(s)+ Cl-(aq)

Solubility and Net Ionic Equations