Solubility Rules & Electrolytes: Chlorides, Bromides, Iodides, Acetates, Nitrates, Hydroxi, Study notes of Chemistry

Download Solubility Rules & Electrolytes: Chlorides, Bromides, Iodides, Acetates, Nitrates, Hydroxi and more Study notes Chemistry in PDF only on Docsity! Section 2.2.2: Solubility Rules for Chlorides, Bromides and Iodides -All chloride, bromide and iodide ionic compounds are soluble in water except those containing silver(I), mercury(I), lead(II) and copper(I). Section 2.2.4: Solubility Rules for Acetate, Chlorate, Perchlorate, Nitrate and Hydroxide Compounds -All acetate, chlorate, perchlorate and nitrate ionic compounds are soluble in water. -All group 1 hydroxide compounds are soluble in water. Magnesium, calcium, strontium and barium hydroxides are only slightly soluble. All other hydroxides are insoluble in water. Section 2.2.6: Solubility Rules for Sulfides -Group 1 sulfide ionic compounds are water soluble while all other sulfide compounds are not. Section 2.2.8: Solubility Rules for Carbonates, Chromates and Sulfates -Group 1 carbonate and chromate ionic compounds are water soluble while all other carbonate and chromate compounds are not. -Most sulfate ionic compounds are soluble except ionic compounds made with Mg, Ca, Sr, Ba, Ra, Hg(I) and Pb(II). Section 2.2.10: Solubility Rules and Dissociation of Ammonium Compounds -All ammonium compounds are water soluble. 2.3.1.b Reaction of CoCl2 (aq) with Na2S (aq) CoCl2(s) is soluble in water. Hence, when mixed with water, this solid breaks down into Co+2(aq) and Cl -(aq) ions according to the ionic equation: CoCl2(s) Ú Co+2(aq) + 2 Cl - (aq) Na2S(s) is soluble in water. Hence, when mixed with water, this solid breaks down into Na +(aq) and S -2(aq) ions according to the ionic equation: Na2S(s) Ú 2 Na +(aq) + S -2(aq) Mixing of these two aqueous solutions then makes it possible for a reaction between these ions to take place. Will a reaction occur and if so, how is it represented? Obviously, Na +(aq) will not react with Cl - (aq) since NaCl is water soluble but Co+2(aq) will form a compound with S -2(aq) since CoS is not water soluble (see solubility rules for sulfides). Mixing the two solutions lead to a dark precipitate of solid cobalt(II) sulfide. Since the sodium and chloride ions remain as free ions in solution and do not form a precipitate, they are spectator ions during the precipitation of CoS. The complete ionic equation between these compounds is written as: 2 Na +(aq) + S -2 (aq) + Co +2(aq) + 2 Cl -(aq) Ú CoS(s) + 2 Na +(aq) + 2 Cl -(aq) Since the Na +(aq) and Cl -(aq) ions are spectator ions (i.e. they do not participate in the reaction), they can be cancelled on both side of the equation. The resulting equation is called the net ionic equation and is written as: Co +2(aq) + S -2 (aq) Ú CoS(s) Section 2.4.1: Electrolytes and Nonelectrolytes -A substance which, when dissolved in water, makes an electrically conducting solution is called an electrolyte. Electrolytes are substances which dissociate into ions when dissolved in water. -A substance which, when dissolved in water, makes an electrically non- conducting solution is called a nonelectrolyte. Nonelectrolytes do not dissociate into ions when dissolved in water. Examples of nonelectrolytes are molecular compounds other than acids and bases. Some electrolytes when dissolved in water dissociate completely into cations and anions. Such compounds are called strong electrolytes. Examples of strong electrolytes are soluble ionic compounds, strong acids and strong bases (see part 2.5). For example, when NaCl(s) is dissolved in water all NaCl dissociates into Na+(aq) and Cl - (aq). NaCl is a strong electrolyte. Similarly, when dissolving HCl gas (hydrogen chloride) in water, we form hydrochloric acid. Hydrochloric acid is written as HCl(aq) and exists as H +(aq) and Cl -(aq) ions. Using the above set-up, a strong electrolyte is characterized by a very brightly lit light bulb. Some electrolytes, when dissolved in water dissociate partially into cations and anions. Partial dissociation means that some of the original compound exists in solution in the undissociated (molecular) form. Substances that dissociate partially in solution are called weak electolytes. Examples of weak electrolytes are weak acids and weak bases (see part 2.5). For instance, acetic acid, HC2H3O2(l), the main component of vinegar dissolves in water and dissociates partially into H +(aq) and C2H3O2 - (aq) ions. However, most of the molecules of HC2H3O2(aq) remain undissociated in solution. Using the above set-up, a weak electrolyte is characterized by a dimly lit light bulb. Section 2.5.2: Arrhenius Acids and Bases . An acid according to Arrhenius is a substance which, when dissolved in water, produces hydrogen ions (protons), H +(aq). A base according to Arrhenius is a substance which, when dissolved in water, produces hydroxide ions, OH -(aq). Rule #3: The oxidation number for oxygen in most oxygen compounds (excluding peroxides and superoxides, a minor fraction of oxygen compounds) is equal to - 2. Rule #4: In all group I compounds, the oxidation number of the metal element is +1. (does not apply to H, since it is not a metal) Na in NaCl(s) or in Na2SO4(s) has an oxidation number of +1 Rule #5: In all group II compounds, the oxidation number of the metal is +2. Ca in CaCO3 and in Ca(NO3)2 has an oxidation number of +2 Rule #6: In all fluorine compounds, the oxidation number of fluorine is -1. The oxidation number of F in NaF, CaF2 and AlF3 is always -1. Rule #7: In all Cl, Br, I compounds, the oxidation number of the halogen is -1, unless the halogen is combined with oxygen or with another halogen. In NaCl, Cl has an oxidation number of -1. In BrCl, Br has an oxidation number of +1 and Cl has an oxidation number of -1. Note the second halogen in the formula (Cl) plays the role of a nonmetal (oxidation number is -1 in this case, as for F another halogen), while the first halogen in the formula (Br) plays the role of a metal cation (positive oxidation number). Rule #8: In a compound, the oxidation number of hydrogen is +1 if H is bonded to a nonmetal. H in NH3, in CH4, in H2O and in HCN has the same oxidation number of +1. Rule #9: In a compound, the oxidation number of hydrogen is -1 if H is bonded to a metal. (Note: in this case, H behaves like an anion and is called hydride) H in NaH (sodium hydride), in CaH2 (calcium hydride) has the oxidation number of -1. Rule #10: The sum of the oxidation numbers of all elements in a compound is equal to 0 (the charge of the compound). For example in NO2, oxidation #(nitrogen) + 2 x oxidation #(oxygen) = 0. We will write this symbolically as: “N” + 2 “O” = 0 Rule #11: The sum of the oxidation numbers of all elements in a polyatomic ion is equal to the charge of the polyatomic ion. For example in CO3-2, oxidation #(carbon) + 3 x oxidation #(oxygen) = -2. We will write this symbolically as: “C” + 3 “O” = -2 STEP 1: Balance the element being oxidized or reduced. Obviously, you need to be sure which element is being oxidized or reduced. In the above half-reaction, Mn is being reduced, since its oxidation number decreases from the value of +7 in MnO4-(aq) to the value of +2 in Mn+2(aq). The half-reaction: MnO4-(aq) Ú Mn+2(aq) is already balanced for Mn. STEP 2: Balance the oxidation number by adding the appropriate number of electrons on the appropriate side of the half-reaction. Manganese has an oxidation number of +7 on the left-hand side and an oxidation number of +2 on the right hand side. Hence, this reaction is a reduction half-reaction and 5 electrons ((+7) – (+2) = 5) must be added on left-hand side of the half-reaction. MnO4-(aq) + 5 e - Ú Mn+2(aq) + 7 - 5 = + 2 STEP 3: Balance the charges by adding the appropriate number of H+ on the appropriate side of the reaction when the reaction is run in acidic conditions, or, add the appropriate number of OH – on the appropriate side of the reaction when the reaction is run in basic conditions. Whether a reaction is run under acidic or basic conditions is always given in the problem of interest. The reaction balanced up to step 2 is such that there are 6 negative charges on the LHS of the equation and 2 positive charges on the RHS of the equation. Acidic Conditions: We add H + to balance the charges. Since there is a differential of 8 charges between the LHS and the RHS, we must add 8 H + on the left- hand side of the equation to balance the charges. The oxidation half-reaction becomes: MnO4-(aq) + 8 H +(aq) + 5 e - Ú Mn+2(aq) Basic Conditions: We add OH - to balance the charges. Since there is a differential of 8 charges between the LHS and the RHS, we must add 8 OH - on the right- hand side of the equation to balance the charges. The oxidation half- reaction becomes: MnO4-(aq) + 5 e - Ú Mn+2(aq) + 8 OH -(aq) STEP 4: Balance the elements hydrogen and oxygen by adding the appropriate number of H2O molecules on the appropriate side of the half- reaction. If adding the appropriate number of H2O molecules does not lead to a fully balanced reaction, then, you must have made a mistake along the way. Acidic Conditions: You need to add 4 H2O molecules on the RHS of the equation to get the fully balanced half-reaction: MnO4-(aq) + 8 H +(aq) + 5 e - Ú Mn+2(aq) + 4 H2O(l) Basic Conditions: You need to add 4 H2O molecules on the LHS of the equation to get the fully balanced half-reaction: MnO4-(aq) + 4 H2O(l) + 5 e - Ú Mn+2(aq) + 8 OH -(aq) PART IV: Combine the OXIDATION and the REDUCTION HALF-REACTIONS. Here, we combine the oxidation and the reduction half-reactions to obtain the equation for the full redox reaction. To do so, we need to ensure that the same number of electrons is produced by the oxidation half-reaction and used up in the reduction half-reaction. Since 5 electrons are used up by the reduction half-reaction, the oxidation half-reaction must be multiplied by 5 so that it produces 5 electrons. 5 Fe+2(aq) Ú 5 Fe +3 (aq) + 5 e – 5 e- released by oxidation MnO4 - (aq) + 8 H + (aq) + 5 e - Ú Mn+2(aq) + 4 H2O(l) 5 e- used up in reduction We can now add the two half-reactions and obtain: MnO4-(aq) + 8 H +(aq) + 5 e - + 5 Fe+2(aq) Ú 5 Fe+3(aq) + 5 e - + Mn+2(aq) + 4 H2O(l) The redox equation is then simplified to yield the fully balanced equation (for each element and for the charges): MnO4-(aq) + 8 H +(aq)+ 5 Fe+2(aq) Ú 5 Fe+3(aq) + Mn+2(aq) + 4 H2O(l)