III.2 Calculating Solubility and Ion Concentrations Solubility ***This is a re-visitation to Chemistry 11: translating grams/l to moles/l (M) and back again. Grams moles (M) L L Since Solubility is a measure of the MASS of solute present in a saturated solution; solubility is a measure of the Grams/L. Molar Solubility (not surprisingly) is a measure of MOLES of solute present in a saturated solution: molar solubility is measured in Moles/L or M. Exercises - Now!!! 1. Given a 250 ml saturated solution containing 18.6 g of CaCl 2. Find solubility and molar solubility. 2. Given the molar solubility of PbI 2 of 1.37 X 10-3, find solubility. 1
3. How much PbI 2 would there be in 250 ml of saturated solution? Hebden p.77-80 Exercises Concentrations of Ions Dissociation of Ions Solutes dissociate into ions having the same proportions in solution as they did in solid form. Therefore, if 1 mole of Na 3 PO 4 is dissolved and diluted to a final volume of 1 L then: Na 3 PO 4 (s) 3 Na + (aq) + PO 4 3- (aq) Moles of particles (in 1 L) 1 3 + 1 Molar concentration (M) 1 3 + 1 NB. What does mean? Suppose 34.0 g of Na 3 PO 4 is dissolved in water to make 750 ml of solution. What is the [Na + ]? What is the [PO 4 3- ]? 2
Dilution Problems Remember: M 1 V 1 = M 2 V 2 Suppose 5.0 ml of 0.20M Cl - is added to 15.0 ml of 0.035 M Br -. What is [Br - ]? What is [Cl - ]? Hebden p.81 Exercises III.3 Predicting the Solubility of Salts Definitions Insoluble - strictly speaking does not exist. Actually, negligibly soluble degree of solubility is disregarded. Low solubility - slightly soluble but of sufficient degree that it cannot be ignored. < 0.1M. Soluble - solubility of 0.1 M or greater. Introducing The Table Hebden p.82 top of page - p. 83 examples Aside: Cu + (insoluble) 3
Cu 2+ (soluble) Hebden p. 83 Exercises A few generalizations from The Table Compounds containing alkali metals, H +, NH + 4, or NO - 3 are soluble alkali metals, H +, NH + 4, or NO - 3 do not precipitate from solution. if you need to get a specific anion (- ion) into solution; use a Na + compound to do it - if you need to get a specific cation (+ ion) into solution; use a NO 3 compound to do it. Hebden p. 84 Exercises III.4 Three Types of Equations formula equations - shows all reactants and products by their chemical formulae. 2AgNO 3 (aq) + Na 2 CO 3 (aq) Ag 2 CO 3 + 2NaNO 3 (aq) complete ionic equations - shows all soluble ionic species broken into their respective ions. 2Ag + (aq) + 2NO 3 - (aq) + 2Na + (aq) + CO 3 - (aq) Ag 2 CO 3 (s) + 2Na + (aq) + 2NO 3 - (aq) spectator ions (do not take part in reaction) 4
net ionic equations - shows only species that are actively involved in the reaction (i.e. sans spectator ions). 2Ag + (aq) + CO - 3 (aq) Ag 2 CO 3 (s) Rules for Writing Equations 1) identify ion combinations which will form precipitates (using The Table ) 2) if a precipitate forms, complete and balance a formula equation showing formation of a precipitate (i.e. (s) ) 3) write and balance the complete ionic equation 4) write and balance the net ionic equation. (Simplify if needed) ***You can short-cut to Net Ionic Equation by 1) identify ion combinations which will form precipitates (using The Table ) 2) write and balance the net ionic equation. (Simplify if needed) Hebden p. 87 Exercises III.5 Separating Mixtures of Ions by Precipitation Methods Hebden! 5
Hebden! Hebden! Hebden p.90 Exercises III.6 Solubility Product (Ksp) First, a bit of Review Since Solubility is a measure of the MASS of solute present in a saturated solution; solubility is a measure of the Grams/L. Molar Solubility (not surprisingly) is a measure of MOLES of solute present in a saturated solution: molar solubility is measured in Moles/L or M. Solubility Product (Ksp) is the value obtained when the concentrations of of the ions in a saturated solution are multiplied together. Equilibrium Equation: CaF 2 Ca 2+ (aq) + 2F - (aq) Solubility Product Expression: Ksp= [Ca 2+ ][F - ] 2 6
Since the value of Ksp depends on the product of ion concentratios, the larger the Ksp, the more soluble the salt. Small Ksp s indicate a less soluble salt. Note the following cases: Case [Ca 2+ ] [F - ] [Ca 2+ ][F - ] 2 1 4.46 X 10-3 1.81 X 10-4 1.46 X 10-10 2 1.21 X 10-3 3.47 X 10-4 1.46 X 10-10 3 3.32 X 10-4 6.64 X 10-4 1.46 X 10-10 4 9.65 X10-5 1.23 X 10-3 1.46 X 10-10 decreasing increasing constant Does this make sense? It should! Okay, solve some problems! Write out the equilibrium equation showing the dissolving salt Write out the solubility product expression. THINK about the information. Go for it! Hebden p. 94 Examples esp. #3 Hebden p. 95 Exercises III.7 Predicting Whether a Precipitate Will Form ***This is similar to predicting a shift in equilibrium, only it is applied to salts in solution where: The Trial Ion Product is designated Q and it s size in relation to the actual Ksp value determines whether the concentration is big enough to form a precipitate or not. 7
Given: AgCl(s) Ag + (aq) + Cl - (aq) Ksp (from Table ) = [Ag + ][Cl - ] = 1.8 X10-10 Suppose we mixed two solutions together; one containing a known [Ag + ] and the other a known [Cl - ]. The question could be asked whether a precipitate would be formed or not. By calculating Q and comparing it to the standard Ksp value we can predict the answer. If Q < Ksp solution is not saturated and no precipitate will form. If Q = Ksp solution is barely saturated If Q > Ksp solution is more than saturated and additional Ag + and Cl - will combine to form a precipitate. Problem: Will a precipitate form if you mix 5.0 ml of 6.0 X 10-5 M Ag + with 10.0 ml of 4.2 X 10-6 M Cl -? ***Careful what is the final volume? What is the final M? How do you find it? (M 1 V 1 = M 2 V 2 ) Hebden p. 97 Example 2. 8
Will a precipitate form if you add 25.0 ml of 4.50 X10-3 M Pb(NO 3 ) 2 to 35.0 ml of 2.80 X10-3 M MgI 2? Hebden p. 98-99 Exercises 9
III.8 Solubility And Titrations Titration a process in which two volumes of solutions are reacted as follows: Solution A: Known volume but unknown concentration Solution B: Measured volume of known concentration The purpose of the reaction is to determine the concentration of Solution A. Knowing exactly how much of Solution B it takes to react completely with A allows us to determine how much solute was present (in A). In order to determine exactly when all A is reacted we must use an indicator (eg CrO 4 2- ). Demo titration In revisiting our favorite reaction from Chemistry 11: Pb(NO 3 ) 2 (aq) + 2KI(aq) PbI 2 (s) + 2KNO 3 (aq) The following net Ionic Equation can be derived Pb 2+ (aq) + 2I - (aq) PbI2(s) Suppose we titrate 25.0 ml of KI (with unknown concentration) with 0.010 M Pb(NO 3 ) 2 and use a chromate indicator to determine endpoint by the reaction: Pb 2+ (aq) + CrO 2-4 (aq) PbCrO 4 (s) (yellow) (red-orange) What would happen? 10
Hebden p. 99 101 Read and rationalize way through examples III.9 Removing Pollution and Hardness From Water via Precipitation Reactions 11
Pollution Removal Harmful ions (eg positive metallic ions) can be precipitated out of water by addition of other ions (eg negative ions). If the combination of these two ions has a low Ksp, they will readily couple and drop out of the water. See Hebden p. 102 Example Hard Water Water is made hard by the occurrence of Ca 2+ or Mg 2+ ions. {Aside - These enter the water by dissolution of mineral deposits by either slightly acidic rain-water (carbon dioxide in the air dissolves in rain water to give carbonic acid) or the full-blown acid rain (air borne sulfide and sulphate pollutants dissolves in rain water to produce sulfuric acid)} Hard water comes in two flavors: Permanently hard No HCO 3 - is present Can only be removed if precipitated with Na 2 CO 3 (aka washing soda) Temporarily hard HCO 3 - is present Mg 2+ (aq) +CO 3 2- (aq) MgCO 3 (s) Can be precipitated out by Na 2 CO 3 or by heating. Mg 2+ (aq) +CO 3 2- (aq) MgCO 3 (s) or Mg 2+ (aq) + 2HCO 3 - + HEAT MgCO 3 (s) + CO 2 (g) +H 2 0(l) Hebden p. 104 Exercises 12
III.10 The Common Ion Effect, Altering the Solubility of A Salt and Le Chatelier s Principle. Increasing and decreasing the solubility of the salt AgCl is an application of LeChatelier s Principle; however, manipulation of the SOLID salt s solubility is attained through manipulation of the ion. AgCl(s) Ag+(aq) + Cl-(Aq) Ag + (NO 3 - ) (Na+)Cl - To decrease solubility means a shift to the left. Increasing one of the product ions by addition of a common ion does this. Add AgNO 3 or NaCl for example. AgCl(s) Ag+(aq) + Cl-(Aq) + + S 2- Pb 2+ Increasing solubility means a shift to the right. Precipitating out one of the product ions does this. To do this one must consult the Solubility Table and use a compound the ion of which will readily form the desired precipitate and yet not contain any of the ions present initially. Hebden p.108 Exercises 13