You try: 2) HC 7H 6O 2 3) N 2O 5. 5) HClO 4. 7) Rb 2C 2O 4 8) H 3PO 4 9) AgI 10) Sr(OH) 2. What kind of compound is it? NON ELECTROLYTE (NE)

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1 Solubility: Solubility is the measure of how much of a solute will dissolve in a solvent. In general chemistry, we usually talk about water as the solvent, so we are talking about what compounds will dissolve in water. We also talk a lot about ionic compounds, so the goal is to determine which ionic compounds are soluble in water and which aren t. There are many factors to consider in the solubility of ionic compounds, none of which you need to worry about. Instead, you will be given solubility rules which you must apply. If it turns out that something is soluble (it does dissolve in water), then you need to put a subscript (aq) next to it (i.e. NaCl (aq)). The (aq) stands for aqueous and implies that you have a solution, a mixture of some compound in water. If the compound is insoluble (does not dissolve in water), then you need to put a subscript (s) next to it. You will be given a list of the rules for the exam and the final, but I recommend memorizing the first rule, because it has no exceptions and it takes precedence over all of the other rules. If rule #1 does not apply, look up the anion in the remaining rules. Solubility rules apply ONLY to ionic compounds (and only in water), NOT to acids or molecular compounds!!!!!!! Example: calcium chromate Calcium is not mentioned in the first rule, so you need to find chromate in the remaining rules. Rule #7 says that chromates are insoluble so it is CaCrO 4(s) Example: rubidium oxide rubidium is a group 1 metal, so you need look no further; all group 1 metal containing compounds are soluble, so it is Rb 2O (aq) Example: lead (II) bromide according to rule #3, most bromide salts are soluble, but there are exceptions (silver, mercury (I) and lead (II). This means that lead (II) bromide is going to be insoluble: PbBr 2(s) State the solubility of each of the following compounds in water: 1) (NH 4) 2S 2) BaO 3) Na 2CO 3 4) Hg 2SO 4 5) Pb(ClO 4) 2 6) SrF 2 7) MgC 2O 4 8) Li 3PO 4 9) AgI 10) Mn(OH) 4 Electrolytes: Electrolytes are broken down into 3 different types; non-electrolytes, weak electrolytes, and strong electrolytes. You will have to be able to differentiate between them. It is NOT that hard, it just takes a second to memorize. Seriously, there are no tricks here, simple memorization and you have it. Non-electrolytes: Any molecular compound (compounds without metals). Weak electrolytes: Any weak acid/bases and all insoluble salts. Strong electrolytes: Any strong acid or any soluble salt That s it. You can already tell between insoluble salts and soluble salts (solubility rules). You learned was a molecular compound was in the second week of class and since I am going to tell you what the strong acids are, you can tell the difference between strong and weak acids (the 7 strong acids are HCl (aq), HBr (aq), HI (aq), HNO 3(aq), HClO 3(aq), HClO 4(aq), and H 2SO 4(aq)). You try: 1) (NH 4) 2S 2) HC 7H 6O 2 3) N 2O 5 4) Hg 2SO 4 5) HClO 4 6) SrF 2 7) Rb 2C 2O 4 8) H 3PO 4 9) AgI 10) Sr(OH) 2 Ionic compound Soluble in water (solubility rules) Insoluble in water (solubility rules) What kind of compound is it? Molecular compound NON ELECTROLYTE (NE) Strong Acid (one of the 7) Acid Weak Acid (not one of the 7) STRONG ELECTROLYTE (SE) WEAK ELECTROLYTE (WE) STRONG ELECTROLYTE (SE) WEAK ELECTROLYTE (WE)

2 Net ionic equations: Remember, in order to write a net ionic equation, you must write out a BALANCED molecular equation. Example: Sodium carbonate is mixed with bismuth (III) nitrate Translate: Na 2CO 3 + Bi(NO 3) 3 Write symbols for products: Na 2CO 3 + Bi(NO 3) 3 Na + NO 3 + Bi 3+ CO 3 2 Balance CHARGES of products: Na 2CO 3 + Bi(NO 3) 3 NaNO 3 + Bi 2(CO 3) 3 NO MORE TOUCHING THE SUBSCRIPTS!!!!!!!! Balance the equation: (ONLY CHANGE COEFICIENTS TO BALANCE EQUATION!!!) Initial: Na 2CO 3 + Bi(NO 3) 3 NaNO 3 + Bi 2(CO 3) 3 Multiply Na 2CO 3 by 3 so that the carbonates are equal on both sides. Trial 1: 3 Na 2CO 3 + Bi(NO 3) 3 NaNO 3 + Bi 2(CO 3) 3 Multiply NaNO 3 by 6 so that the nitrates are equal on both sides. Trial 2: 3 Na 2CO 3 + Bi(NO 3) 3 6 NaNO 3 + Bi 2(CO 3) 3 Multiply Bi(NO 3) 3 by 2 so that the bismuths are equal on both sides. Trial 3: 3 Na 2CO Bi(NO 3) 3 6 NaNO 3 + Bi 2(CO 3) 3 Reactants Products Initial Trial 1 Trial 2 Trial 3 Initial Trial 1 Trial 2 Trial 3 Na Na CO CO Bi Bi NO NO BALANCED!!!! Last step in writing the balanced molecular equation is to add in to the state subscripts (aq, l, g, s). Look at the solubility rules to determine if a salt is soluble. All acids are written as (aq) Add in state subscripts: So the balanced molecular equation is: The molecular equation is just your typical balanced chemical equation. It shows you possible reactions, but not necessarily what is really going on in your beaker. The main drawback to the molecular equation is that it shows all compound as single particles, regardless of state-of-matter, rather than how they are most likely to exist in water. The ionic equation (aka total ionic equation) on the other hand, is intended to show each compound in the way in which it is most likely to be found in water. Because strong electrolytes are ionized/dissociated ~100% in water, they are most likely to be encounter as ions, with little or none of the original particles from the molecular equation to be found. This means that any strong electrolyte is written as its constituent ions in the ionic equation. For example, MgCl 2(aq), a soluble salt and therefore a strong electrolyte, would be written as Mg 2+ (aq) and 2 Cl (aq) in the ionic equation. HClO 4(aq), a strong acid and therefore a strong electrolyte, would be written as H + (aq) and ClO 4 (aq). In contrast to strong electrolytes, weak electrolytes are ionized/dissociated less than ~1% in water. This means that less than 1 out of every 100 particles of a weak electrolyte are going to split into ions while more than 99 particles out of 100 are going to remain in the form of the original particle from the molecular equation. This means that are most likely going to encounter this compounds in particle form rather than ion form, so weak electrolytes are written in their original form in the ionic equation. For example, PbCl 2(s), an insoluble salt and therefore a weak electrolyte, would be written as PbCl 2(s) in the ionic equation. HCN (aq), a weak acid and therefore a weak electrolyte, would be written as HCN (aq). In both of these instances, the compound forms almost no ions, so the ions are not shown in the ionic equation. Non electrolytes are similar to weak electrolytes except that non electrolytes for NO ions in water. Because non electrolytes are ionized/dissociated 0% in water, they are found only in their original form from the molecular equation.

3 The short version of the above is, when writing the ionic equation, strong electrolytes are written as ions, while weak and non electrolytes are written as molecules/formula units. To write the ionic equation, we start from the molecular equation (from above) and determine into which electrolyte categories each of the substances falls. Na 2CO 3(aq) is a soluble salt (rule 1) and therefor a STRONG ELECTROLYTE Bi(NO 3) 3(aq) is a soluble salt (rule 2) and therefor a STRONG ELECTROLYTE NaNO 3(aq) is a soluble salt (rule 1) and therefor a STRONG ELECTROLYTE Bi 2(CO 3) 3(s) is an insoluble salt (rule 7) and therefor a WEAK ELECTROLYTE * Now that we know what type of electrolyte each compound is, we can write the ionic equation. This is done by breaking up ALL STRONG electrolytes into ions. You break up ONLY the strong electrolytes; weak electrolytes and non-electrolytes are left alone. Make sure not to break up polyatomic ions; i.e. leave NO 3 - (aq) alone instead of splitting it into nitrogen and oxygen. Also, make sure you have the correct number of each ion present. So, the ionic equation looks like this: 6 Na + (aq) + 3 CO 3 2 (aq) + 2 Bi 3+ (aq) + 6 NO 3 (aq) 6 Na + (aq) + 6 NO 3 (aq) + Bi 2(CO 3) 3(s) That is it for the ionic equation. Each of the compounds is represented in the form in which it is most likely to be found in water. Now, remember what it means to be reaction. A reaction is the result of a chemical change which means that in order to have reacted, the substance must change chemically. In other words, in order to be part of the reaction, the substance must be different on the reactant and product sides of the equation. If the particle is identical on both sides of the reaction, then it was not part of the reaction and is then referred to as a spectator. The final step is to write the net ionic equation. The net ionic equation only shows those particles that participated (not to be confused with precipitated) in the reaction. To write the net ionic equation, the ionic equation is rewritten excluding all spectators. The easiest way to do this is to neatly cross out anything that is IDENTICAL on both sides of the ionic equation, like so: 6 Na + (aq) + 3 CO 3 2 (aq) + 2 Bi 3+ (aq) + 6 NO 3 (aq) 6 Na + (aq) + 6 NO 3 (aq) + Bi 2(CO 3) 3(s) The ions that are crossed out are all spectators. Rewrite anything that is not crossed out. This is the NET IONIC EQUATION!! Examples: 3 CO 3 2 (aq) + 2 Bi 3+ (aq) Bi 2(CO 3) 3(s) Write the net ionic equation of the reaction between silver sulfate and calcium chloride yielding silver chloride and calcium sulfate. Balance the equation: Ag 2SO 4 + CaCl 2 2AgCl + CaSO 4 Add in state subscripts: Cross out spectators: Ag 2SO 4(aq) + CaCl 2(aq) 2AgCl (s) + CaSO 4(aq) Ag2SO4(aq) + CaCl2(aq) 2 AgCl(s) + CaSO4(aq) 2 Ag + (aq) + SO4 2 (aq) + Ca 2+ (aq) + 2 Cl (aq) 2 AgCl(s) + Ca 2+ (aq) + SO4 2 (aq) 2 Ag + (aq) + SO 4 2 (aq) + Ca 2+ (aq) + 2 Cl (aq) 2 AgCl (s) + Ca 2+ (aq) + SO 4 2 (aq) Ag + (aq) + Cl (aq) AgCl(s) (note: the remaining coefficients reduced to the lowest common denominator) * WE is a small lie that will be corrected in 2 nd semester gen. chem. This lie does NOT change the process however.

4 Okay, so what about the states on molecular compounds? For most of these, I will have to tell you what the state is. If it is a molecular compound, whether it is (aq), (g), (l), or (s), it does not change or break up EVER (molecular compounds are nonelectrolytes)!!. You should know that CO2 is a gas and water is a liquid unless it is otherwise stated. Remember, strong acids are strong electrolytes and will therefore ionize (i.e. HCl (aq) H + (aq) + Cl (aq)). If the acid that is present is NOT one of these 7 acids, then it is a WEAK acid (weak electrolyte) and DOES NOT EVER BREAK UP IN THE N.I.E.!!!!! (i.e. HCO 2H (aq) stays HCO 2H (aq)) Write the N.I.E. for the reaction between calcium hydroxide and hypochlorous acid forming calcium hypochlorite and water. Cross out spectators: Ca(OH)2 (aq) + 2 HClO(aq) Ca(ClO)2 (aq) + 2 HOH(l) SE WE SE NE Ca 2+ (aq) + 2 OH (aq) + 2 HClO(aq) Ca 2+ (aq) + 2 ClO (aq) + 2 HOH(l) (HClO is a weak acid and weak electrolyte, so it does NOT break up) Ca 2+ (aq) + 2 OH (aq) + 2 HClO (aq) Ca 2+ (aq) + 2 ClO (aq) + 2 HOH(l) OH (aq) + HClO(aq) ClO (aq) + HOH(l) Write the N.I.E. for the reaction between barium hydroxide and nitric acid which gives barium nitrate and water. Ba(OH) 2 (aq) + 2 HNO 3 (aq) Ba(NO 3) 2 (aq) + 2 HOH (l) SE SE SE NE Ba 2+ (aq) + 2 OH (aq) + 2 H NO 3 (aq) Ba 2+ (aq) + 2 NO 3 (aq) + 2 HOH (l) (HNO 3 is a strong acid) OH (aq) + H + (aq) HOH (l) Write the N.I.E. for the reaction between ammonium sulfate and potassium iodide giving potassium sulfate and ammonium iodide. (NH 4) 2SO 4(aq) + 2 KI (aq) K 2SO 4(aq) + 2 NH 4I (aq) 2 NH 4 + (aq) + SO 4 2 (aq) + 2 K + (aq) + 2 I (aq) 2 K + (aq) + SO 4 2 (aq) + 2 NH 4 + (aq) + 2 I (aq) NO REACTION (when everything is a strong electrolyte, everything is a spectator and there is no reaction) Write the net ionic equation for each of the following reactions (assume all are in water solution): 1) aluminum hydroxide and sulfuric acid react 2) ammonium sulfide and lithium oxide 3) potassium hydroxide and selenic acid react 4) hydrobromic acid and silver nitrate 5) nitric acid and calcium hydroxide react 6) rubidium chromate and chromium (III) acetate are combined 7) ammonium phosphate and manganese (IV) sulfate 8) cesium arsenate and potassium sulfite 9) magnesium oxide and hydroiodic acid 10) lithium sulfate and barium nitrate react 11) sodium hydroxide and benzoic acid (HC 7H 5O 2) react to give sodium benzoate (NaC 7H 5O 2) and water 12) solid magnesium metal and acetic acid react to form magnesium acetate and hydrogen gas 13) iron (III) carbonate and chloric acid to form iron (III) chlorate, carbon dioxide, and water 14) sodium fluoride and calcium nitrate 15) titanium (III) iodide and mercury (I) acetate 16) bismuth (V) chlorate and aluminum nitrate react 17) solid calcium bicarbonate and hydrobromic acid calcium bromide, water, and carbon dioxide 18) dihydrogen monosulfide gas and cadmium chloride 19) lithium oxalate and calcium perchlorate 20) strontium hydroxide and hydrochloric acid react 21) oxalic acid and barium hydroxide react

5 22) ammonium carbonate + copper(ii) nitrate 23) lead(ii) hydroxide + hydrochloric acid 24) barium carbonate + hydrochloric acid 25) aluminum + tin(iv) sulfate aluminum sulfate + tin 26) potassium + gold(iii)chloride potassium chloride + gold 27) ammonium phosphate + sodium nitrate 28) silver nitrate + potassium iodide 29) magnesium sulfate + lithium bromide 30) strontium chloride + sodium carbonate 31) nitric acid + calcium carbonate 32) zinc + barium bromide barium + zinc bromide 33) ammonium chromate + barium chloride 34) copper(ii) sulfate + sodium sulfide 35) iron (II) chloride + ammonium phosphate 36) potassium oxalate + calcium nitrate 37) silver nitrate + sodium carbonate 38) sodium hydroxide + copper(ii) nitrate