General Chemistry II Chapter 20

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1 1 General Chemistry II Chapter 0 Ionic Equilibria: Principle There are many compounds that appear to be insoluble in aqueous solution (nonelectrolytes). That is, when we add a certain compound to water it appears to remain in a solid state. Although we call such compounds insoluble, they do, in fact, ionize to some very small degree. Insoluble compounds will dissolve in water BUT are only very slightly soluble. Ballpark: Compounds that dissolve in water at a concentration of 0.00 moles per liter or greater are considered soluble. Those compounds that do not dissolve at concentrations of at least 0.00 M are considered insoluble compounds. The material covered in this portion of the chapter will require that you are able to recognize the name and formulas of many common compounds (namely salts). If you do not have this knowledge, I suggest you review: Fundamentals (F-) Solubility Product Constant (K ) The equilibrium constant for slightly soluble compounds is called the solubility product constant and abbreviated as K. The K of a compound is the product of the concentrations of the ions in a saturated (aqueous) solution. A saturated aqueous solution is defined as the mass of a compound the will dissolve in 1.0 liter of water. For insoluble compounds the mass of the dissolved compound is very small; therefore, there are only a very small number of dissociated ions in the aqueous solution. In general: A a B b aa bb - K = [A ] a [B ] b Where A represents the cation/s and B represents the anion/s that is formed in the aqueous solution. The small case letters represent the number of cations and anions that are in the balanced formula unit. 5 6

2 7 Often, the generic symbols for metals (cations in salts) are represented with M and the X for nonmetals (anions in salts). M X M X M X M X E.g. The solubility product of lead (II) sulfide: The equilibrium equation for lead (II) sulfide is written as follows: PbS (s) Pb S Because the activity of a solid is one, we write the equilibrium constant as: K = [Pb ][S ] 8 Since we are multiplying the concentrations of the resulting ions, we express the equilibrium constant as the solubility product. K = [Pb ][S ] is the solubility product expression for lead (II) sulfide. The solubility product of magnesium hydroxide: Mg(OH) (s) K = [Mg ][OH ] Mg OH 9 10 The solubility product of zinc phohate: Zn (PO ) (s) K = [Zn ] [PO ] Zn PO Determination of the Solubility Product Determine the K for zinc phohate if 1.18 x 10 grams of zinc phohate dissolved to make.0 liters of a saturated solution. A saturated aqueous solution is defined as the mass of a compound that will dissolve in 1.0 liter of water. 11 1

3 1 Determination of the Solubility Product In order to determine the K : We need to know molar concentration of the dissolved zinc phohate From this quantity we can determined the molar concentration of each ion formed in the solution. M of Zn PO solubility Determination of the Solubility Product ( ) 7 7 ( M) ( M) ( ) ( ) ( ) ( ) gzn PO 1 molzn PO =.0 L 86.11gZn PO 7 = M () ( ) Zn PO s Zn aq PO [ Zn ] =.5 10 M [ PO ] =.0 10 M Determine the K K = [ Zn ] [ PO ] = 7 7 (.5 x10 ) (.0 x10 ) = 8. x 10 Determination of molar solubility Molar solubility is the number of moles of a compound that will dissolve to give a liter of a saturated aqueous solution. i.e., the moles of a compound that will dissolve in a liter of water. As a rule, K is only calculated to significant figures Ex 1: Lead (II) sulfide has a K of 8. x (a) What is the molar solubility of lead (II) sulfide? (b) What mass of lead (II) sulfide is contained in the saturated solution? PbS s Pb aq S aq ( ) ( ) ( ) a). Let x equal the concentration of lead (II) ions and sulfide ions. K = [ Pb ][ S ] [ Pb ] = x [ S ] = x ( )( ) x 8 = x x = x 1.9 x 10 = x = [ Pb ] = [ S ] 17 18

4 PbS () s Pb ( aq) S ( aq) 1.9 x 10 M 1.9 x10 M 1.9 x10 M The molar solubility of the PbS (s) is.9 x This means that only.9 x 10 1 moles of PbS (s) dissolves in one liter of water. Another way of looking at it is that when.9 x 10 1 moles of PbS (s) dissolve in a liter of water, the aqueous solution is saturated. b) The mass of PbS in the saturated solution: (The mass that dissolved.) 9.7 gpbs.9 x10 1 moles PbS x 6.9 x10 1 = g PbS mol PbS 19 0 Often it is necessary to understand what is given or what is being asked. Recognition depends on a clear understanding of the definitions introduced in this unit. The solubility of a compound is the amount of the compound that dissolves in a ecified volume of solution. Solubility is expressed as either grams/liter or grams/100. ml. The molar solubility of a compound is the number of moles that dissolve to give one liter of saturated solution. The solubility product expression for a compound is the product of the concentration of its constituent ions, each raised to the power that correonds to the number of ions in one formula unit of the compound. 1 Ex. Lead (II) iodide has a K of 8.7 x What is the molar solubility of lead (II) iodide? PbI s Pb aq I aq () ( ) ( ) x x K = [ Pb ][ I ] [ Pb ] = x [ I ] = x ( )( ) x = x x = x x = x 1. x 10 = x= [ Pb ] = [ PbI]

5 5 Determination of ion concentration Ex. Aluminum hydroxide has a K of 1.9 x 10. What is the concentration of the aluminum ion? Al OH s Al aq OH aq ( ) ( ) ( ) ( ) x x Let x equal the concentration of aluminum ions and hydroxide ions. K = [ Al ][ OH] [ Al ] = x [ OH ] = x 1.9 x10 = ( x ) x x 7x = = 7x =.9 x10 9 = [ Al ] 6 We will do the following example in class: Magnesium hydroxide is a slightly soluble substance. If the ph of a saturated solution of magnesium hydroxide is 10.9, calculate the K for magnesium hydroxide. The common ion effect The solubility of a compound will decrease if the solution contains an ion that is the same as one of the ions produced by the insoluble compound. 7 8 E.g. What is the molar solubility of lead (II) fluoride in 0.10 M NaF? The K for PbF is.7 x % () ( ) NaF s Na aq F 0.10 M 0.10 M 0.10 M PbF s Pb aq F aq X 0.10 x () ( ) ( ) Let x equal the concentration of lead (II) ions and fluoride ions. K [ ][ ] = Pb F [ Pb ] = x [ F] = 0.10M ( the amount of F from PbF is very small and can be ignored).7 x ( x)( ) = = x(0.01).7 x10 6 = x = [ Pb ] = [ PbF ] Very small 9 0

6 1 The Reaction Quotient in Precipitation Reactions Q < K The concentration of ions have not reached saturation The forward reaction is favored No ppt is formed More solid will dissolve if there is any available The Reaction Quotient in Precipitation Reactions Q = K The concentration of ions have just reached saturation. Neither the forward reaction nor the reverse reaction is favored. The solid solution The Reaction Quotient in Precipitation Reactions Q > K The concentration of ions are beyond saturation The reverse reaction is favored A ppt is formed No more solid will be made Predicting the formation of a precipitate. Ex: Will a ppt. form when 1.00 g of silver nitrate is added to 50.0 ml of M NaCl? If so, would you expect the ppt. to be visible? AgNO () s Ag ( aq) NO ( aq) NaCl ( aq) Na ( aq) Cl ( aq) AgNO () s NaCl ( aq) AgCl() s NaNO ( aq) [ Cl] = 0.050M 1.00g AgNO 1.00 mol AgNO [ ] = 1.00 mol Ag = Ag x x 0.050L g AgNO 1.00 mol AgNO = 0.118M K ( ) AgCl = x Q = [ Ag ][ Cl] = (0.118)(0.050) = 5.9 x10 Q > K ; a ppt will form >>> , Since x x the ppt should be visible. Sometimes we would like to extract a single kind of ion from a solution that contains several different ions. One way to accomplish this is to exploit the differences in the solubility product of the different ions. This procedure usually must be carried out carefully because on many occasions the ecies used to ppt. out one ion will also ppt. one of the ions that you do not want to remove. 5 6

7 7 Example: Solid lead (II) nitrate is slowly added to a sol n that is M each in sodium hydroxide, potassium carbonate, and sodium sulfate. In what order will the solid lead (II) nitrate ppt. the different ions? a) Pb(NO ) (s) Pb NO What do we know? Concentration of ions. K [OH] = M Pb(OH) =.8 x [CO ] = M PbCO = 1.5 x 10 1 [SO ] = M PbSO = 1.8 x What is the [Pb ] needed to ppt. each ion? Let x equal [Pb ]. Pb(OH) K = [Pb ][OH ].8 x = x (0.015) 1. x 10 1 = x = [Pb ] PbCO K = [Pb ][CO ] 1.5 x 10 1 = x (0.015) 1.0 x = x = [Pb ] PbSO K = [Pb ][SO ] 1.8 x 10 8 = x (0.015) 1. x 10 6 = x = [Pb ] 9 0 Which anion did the lead II ion precipitate first, second then third? with OH 1. x 10 1 = [Pb ] with CO 1.0 x = [Pb ] with SO 1. x 10 6 = [Pb ] Order of precipitation: 1st to rd Simultaneous Equilibra Simultaneous Equilibra Involving Slightly Soluble Compounds Many weak acids and weak bases can form insoluble compounds. Consider the following: Mg(NO ) Mg NO NH OH NH Mg(NO ) NH Mg(OH) (s) NO NH Pb(OH), PbCO, PbSO 1

8 Simultaneous Equilibra Adding the salt of ammonia to the solution (ammonia and magnesium nitrate) will decrease the amount of magnesium ion that will precipitate. Why? Remember that a weak base plus the salt of a weak base will create a buffer solution. Simultaneous Equilibra How does the buffer solution prevent the precipitation of magnesium ions? NH Cl NH Cl Add the ammonium ion to ammonia: NH NH H O NH NH OH The salt of the weak base creates hydronium ions that react with the hydroxide ions to produce water. The [OH ] is decreased. Simultaneous Equilibra Ex: If a solution is made with M Mg(NO ), M NH, and M NH NO, will the magnesium hydroxide precipitate? What is the ph of the solution? a) First we will determine the [OH ]. The solution contains both a weak base and the salt of a weak base so that we have a buffered solution. poh = pk log [ salt] K ( NH ) = 1.80 x105 b [ base] b =.7 log =.69 [ OH ] = Will this = concentration.0 x 10 5 of hydroxide ion ppt. the magnesium ion? K = [ Mg ][ OH] = 1.5x10 Q ( ) 11 = x =.6 x10 Q > K, Mg( OH ) will ppt. 5 6 b. To determine the ph: From the K it is determined that the amount of Mg that precipitates is very, very small compared to the [Mg ] originally in the solution. We can assume that the [Mg ] is equal to 0.090M after the rxn has occurred. K = [ Mg ][ OH] 1.5 x10 11 = (0.090)[ OH] 1.7 x10 10 = [ OH] 1. x10 5 = [ OH] poh =.89 ph = = Solubility Product We will do the following examples in class. 1) Calculate the solubility of AgCN in a solution that is buffered at [H O ] = M, with [HCN] = 0.01 M. ) How many moles of Cr(OH) will dissolve in 1.00 L of a solution with a ph of 5.00? ) What is the ph of a saturated solution of Mg(OH)? 8

9 End of Chapter 0 9

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