CHEM 101/105 Aqueous Solution Chemistry Lect-06 Ionic solutes are involved in three types of reactions in aqueous solutions: precipitation reactions - sparingly soluble ionic substances produced when two solutions are mixed acid/base reactions - involves hydrogen and hydroxide ions (the constituent ions of water) oxidation/reduction reactions - involves gain and loss (i.e., exchange) of electrons between reactants Precipitation reactions in aqueous solutions So, what happens when two solutions with ionic solutes are mixed? ANS: A precipitate may, or may not form. Must decide whether some combination of ions present in the mixed solutions could form a sparingly soluble substance, i.e., whether or not a precipitate could form. Let's DEMO this for solutions of nickel(ii) nitrate, barium chloride, sodium carbonate, copper(ii) sulfate, and potassium hydroxide. In each case: (a) Start by listing the ions present in the separated solutions. (b) Test whether any combination of ions forms an insoluble substance upon mixing. (c) If a precipitate forms then write a net ionic equation to represent the process. 1. When a solution of sodium carbonate is mixed with one of barium chloride does a precipitate form? Do the test. If yes, then either one or both of the ionic salts sodium chloride, and/or barium carbonate is insoluble. A ppt does form, and it is BaCO 3. Write the net ionic equation for this precipitation reaction: Ba 2+ + CO 3 2- = BaCO 3(s) Note that sodium ions and chloride ions do not appear in the net ionic equation, because there is no net change in their condition; they begin as aquated Na 1+ and Cl 1- ions, and remain as aquated Na 1+ and Cl 1- ions - both before and after the precipitation reaction. So they are not included in the net ionic equation. They are called "spectator" ions, they go along for the ride but do not participate in precipitation formation. 2. Mix solutions of nickel(ii) nitrate and potassium hydroxide. Does a precipitate form? Do the test. If yes, then some combination of the ions present in solution must have formed an insoluble ionic substance, i.e., a precipitate. Lets analyze these solutions, decide what precipitated, and write a net ionic balanced equation for the precipitation reaction. 3. Mix solutions of copper(ii) sulfate and potassium hydroxide. Does a precipitate form? Do the test. If yes, then some combination of the ions present in solution must have formed an insoluble ionic substance, i.e., a precipitate. Analyze starting solutions and decide what precipitated. Write a net ionic balanced equation for the precipitation reaction. Results of such tests lead to Rules of Solubility, used to predict whether or not an insoluble product will be formed. Solubility Rules are presented in terms of combinations of ions forming insoluble substances. Combinations of the ions appearing in the Precipitation Table below will form precipitates when indicated cation and anion are both present: anions NO 3 1 - Cl 1 - SO 4 2 - OH 1 - CO 3 2 - PO 4 3 - S 2 - cations Gp I & NH 4 1 + Gp II Cations Ba 2 +, Sr 2 +, Pb 2 + ALL ALL ALL T.M. & Other Cations Ag 1 +, Hg 2 2 +, Pb 2 + ALL ALL ALL ALL
Solubility rules are grouped and listed according to anions as follows: If the anion component is a nitrate anion (NO 3-1 ), then combination with any cation will NOT precipitate. This is worded as, "all NITRATES are soluble". Most salts containing chloride anion (Cl -1 ) are soluble - except when the cation is Ag +1, Hg 2 +2, or Pb +2. This is abbreviated to, "all chlorides are soluble, except AgCl, Hg 2 Cl 2, and PbCl 2 ". Most salts containing sulfate anion (SO 4-2 ) are soluble, except BaSO 4, SrSO 4, and PbSO 4. Most salts containing carbonate anion (CO -2 3 ), or hydroxide anion (OH -1 ) are insoluble, except GROUP I and NH 1+ 4 cations, and as noted. Most salts containing sulfide anions (S -2 ) are insoluble, except main GROUPS I and II cations, and ammonium cation (NH 4 +1 ). Apply these "solubility rules" to write balanced net ionic chemical equations for each of the examples below. In addition, solve any associated stoichiometric questions. a. Will a precipitate form when solutions of aluminum sulfate and sodium chloride are mixed? b. Will a precipitate form when solutions of ammonium chloride and lead(ii) nitrate are mixed? (How many grams precipitate could be formed from reaction of 42.78 ml of 0.125 M NH 4 Cl?) c. Will a precipitate form when solutions of uranium(iii) nitrate and sodium carbonate are mixed? (How many grams precipitate can be formed from reaction of 29.31 ml of 0.0951 M Na 2 CO 3, with 32.13 ml of 0.0544 M U(NO 3) 3? FW of U 2 (CO 3 ) 3 = 656 amu ) d. It is desired to remove iron(ii) ions from a water supply. What anion(s) might be used? Solution Stoichiometry - Ground Rules 1. Matter is conserved in chemical reactions. (Law of Conservation of Mass) 2. Standard Mass Substance = (atomic mass )(# atoms ) 3. Molarity( M ) is a unit of solution concentration; M moles solute = = Vol. of solution in Liter units r r moles Liters mass substance 4. where moles solute ( actually any solid) = standard mass substance 5. and, moles solute ( in a solution) = ( Molarity)( Vol. in Liters) 6. For all chemical reactions: moles A = moles B [ Conversion Factor( s)... ]
A. Starting Information: Concentrations expressed in Terms of... But what's really present in the solution is... reactant Al 2 ( SO 4 ) 3 2 solution A Al 3+ 2 ( aq) + 3 SO4 ( aq) and 1+ reactant NaCl Na + Cl solution B 1 Net Ionic equation: Will any combination of above ions produce a precipitate? A check of the Solubility Rules shows that aluminum chloride, AlCl 3, would be soluble, and also that sodium sulfate, Na 2 SO 4, would be soluble as well. So no combination of the above ions in a solution will produce an insoluble substance, so no precipitate forms when these two solutions are mixed. moles A = moles B [ C.F.] No quantitative information is supplied by the problem so no stoichimoetric calculations are required.
B. Starting Information: Concentrations expressed in Terms of... 0.125 M reactant NH 4 Cl NH + Cl solution A and 42.78 ml? M But what's really present in the solution is... 1+ 1 4 2+ 1 + 3 reactant Pb(NO 3 ) 2 Pb 2 NO solution B? Vol. Net Ionic equation: Lead(II) cations and chloride anions will Will any combination of above ions produce a precipitate? 2+ 1 2 (s) form insoluble lead(ii) chloride: Pb + 2 Cl PbCl moles A = moles B [ C.F.] moles PbCl 2 = moles NH 4 Cl [ C.F. ] (? g Std. mass ) = ( 0.125 M)(0.04278L) 1? mass = 0.743 grams ppt. 1 mole Cl 1 mole PbCl 1 mole NH Cl 2 4 2 mole Cl 2 +
C. Starting Information: Concentrations expressed in Terms of... 0.0544 M reactant U(NO 3 ) 3 U + 3 NO solution A and 32.13 ml 0.0951 M reactant Na 2 CO 3 2 solution B 29.31 ml But what's really present in the solution is... 3+ 1 3 1+ 2 + 3 Na CO Will any combination of above ions produce a precipitate? Net Ionic equation: Uranium(III) cations and carbonate anions 3+ 2 3 2 3 3 (s) will form insoluble uranium(iii) carbonate: 2 U + 3 CO U ( CO ) moles A = moles B [ C.F.] Recognize that this is a limiting reagent type of-a-problem because actual amounts of BOTH reactants are given. So, two calculations are needed - one for each reactant - to see which forms the lesser amount of product.
Calc. I moles ppt = moles U(NO 3 ) 3 [ C.F. ] = ( 0.0544 M)(0.03213 L) = 8.739 E 4 3+ 1 mole U 1 mole U ( NO ) 3 3 1 mole U CO ( ) 3+ 2 mole U 2 3 3 Calc. II moles ppt = moles Na 2 CO 3 [ C.F. ] = ( 0.0951 M)(0.02931 L) = 9.291 E 4 2-1 mole CO3 1 mole Na CO 2 3 1 mole U (CO ) 2-3 mole CO3 2 3 3 So, the limiting reagent is uranium(iii) nitrate. Now convert from moles ppt to mass ppt.? g ppt = 8.739 E 4 moles ppt. 656 grams 1 mole U (CO ) 2 3 3 = 0.573 grams of ppt.