Qualitative Analysis- Solubility and Precipitation A.S. 91162 (2.2) Year 12 Chemistry.

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Presentation transcript:

Qualitative Analysis- Solubility and Precipitation A.S (2.2) Year 12 Chemistry

Ions in Solution An Ion is an atom which has gained or lost electrons Cations –Have a positive charge Anions –Have a negative charge

SodiumNa + Lead (II)Pb 2+ PotassiumK+K+ ZincZn 2+ SilverAg + Iron (II)Fe 2+ ammoniumNH 4 + Manganese (II)Mn 2+ MagnesiumMg 2+ aluminiumAl 3+ CalciumCa 2+ chromiumCr 3+ BariumBa 2+ Iron (III)Fe 3+ Copper (II)Cu 2+ Cations

ChlorideCl - chromateCrO 4 2- BromideBr - sulfateSO 4 2- IodideI-I- OxideO 2- hydroxideOH - SulfideS 2- nitrateNO 3 - CarbonateCO 3 2- permanganateMnO 4 - dichromateCr 2 O 7 2- Anions

Solubility of Ionic Compounds Some ionic compounds are soluble others insoluble depending if they dissolve in water or not. How do we know which are which?

Predicting precipitation reactions using solubility rules The solubility rules are used to determine whether precipitation reactions occur or not. A precipitate is an insoluble compound formed during a chemical reaction in solution.

The solubility rules Nitrates NO 3 - Chlorides Cl - Sulfates SO 4 2- Carbonates CO 3 2- Hydroxides OH - All soluble All soluble except All insoluble except AgClPbSO 4 Na 2 CO 3 NaOH PbCl 2 CaSO 4 K 2 CO 3 KOH BaSO 4 (NH 4 ) 2 CO 3

You do not have to memorize these rules, as you will be provided with a flow chart. You must know how to use the flowchart in order to identify unknown ions in solution.

Molecular, ionic and net ionic equations Molecular Equations-complete formulas are written for all the reactants and products, no ions are written. Ionic equations-only involve those chemical species which are involved in a chemical reaction. All spectator ions are eliminated. Spectator ions- do not participate in the chemical reaction but are present in the reaction mixture.

1)molecular equation: CaCl 2(aq) + Na 2 CO 3(aq)  CaCO 3(s) +2NaCl (aq) ionic equation: Ca 2+ (aq) + 2 Cl - (aq) + 2 Na + (aq) + CO 3 2- (aq)  CaCO 3(s) + 2 Na + (aq) + 2 Cl - (aq) ionic equation: Ca 2+ (aq) + CO 3 2- (aq)  CaCO 3(s) Write the ionic equation for the reaction of an aqueous solution of CaCl 2 and an aqueous solution of Na 2 CO 3. Molecular, ionic and net ionic equations

KCl (aq) + AgNO 3(aq)  KNO 3 + AgCl Look at the solubility rules to see if any of the products are insoluble in water. AgCl is insoluble - all chlorides are soluble except for Ag + and Pb 2+ Balanced molecular equation: KCl (aq) + AgNO 3(aq)  KNO 3(aq) + AgCl (s ) The net ionic equation is: Cl - (aq) + Ag + (aq)  AgCl (s) KCl (aq) + AgNO 3(aq)  KNO 3(aq) + AgCl (s) If we add a solution of KCl to a AgNO 3 solution will a precipitate form? First write an equation.

If we add a solution of NaNO 3 to an NH 4 Cl solution will a precipitate form? NaNO 3(aq) + NH 4 Cl (aq)  NaCl + NH 4 NO 3 Solubility rules: Nitrates and most chlorides are soluble so mixing these two solutions gives no precipitates. What will result if we add a solution of Pb(NO 3 ) 2 to a solution of KCl? Molecular, ionic and net ionic equations

Complex Ions Formed if a precipitate disappears when excess reagent is added. Metal cations with several ligands attached. Ligands have a pair of non-bonding electrons e.g. H 2 0, NH 3, OH -, SCN - Usually* the number of ligands is twice the charge on the cation. E.g. Cu 2+ forms [Cu(NH 3 ) 4 ] 2+ * Except Al 3+

Things to know about complexes!! The charge of the complex ion is the sum of the charges in the reactants Spectator ions are omitted The quantity of the reactants can be worked out by memorizing the final complex ion (You don’t need to know the chemistry behind this as it is Y13 material, just memorize the complex ions)

Iron Thiocyanate [FeSCN] 2+ Silver diamine[Ag(NH 3 ) 2 ] + Aluminium tetrahydroxide [Al(OH) 4 ] - Lead tetrahydroxide [Pb(OH) 4 ] 2- zinc tetrahydroxide [Zn(OH) 4 ] 2- Zinc tetraamine[Zn(NH 3 ) 4 ] 2+ Copper tetraamine [Cu(NH 3 ) 4 ] 2+ Complex ions

Ammonia in solution You will have noticed that some precipitates form when ammonia solution is added to another solution. It isn’t the ammonia creating the precipitate, it is the hydroxide. How? When ammonia is put into solution it act as a base and water as an acid (hydrolysis). This causes this reaction. NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq)

So when a few drops of ammonia is added to a copper solution, the ionic equation should read: Cu 2+ (aq) + 2OH - (aq)  Cu(OH) 2(s) When excess ammonia is added then the complex ion forms: Cu 2+ (aq) + 4NH 3(aq)  [Cu(NH 3 ) 4 ] 2+ (aq)

The problem with silver When silver reacts with hydroxide, the reaction doesn’t stop when the silver hydroxide is formed as silver hydroxide is unstable. The reaction continues on until the brown silver oxide precipitate is formed. *you will need to remember this fact So the ionic equation is: 2Ag + (aq) + 2OH - (aq)  Ag 2 O (s) + H 2 O (l)