Solubility and Simultaneous Equilibria
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1 Chemistry Matter and Its Changes 4/e Brady Senese Chapter 19: Solubility and Simultaneous Equilibria Reference: Chapter 15 in Chemistry 6/e by Steven S. Zumdahl & Susan A. Zumdahl 19.1 An insoluble salt is in equilibrium with the solution around it
2 For solids dissolving to form aqueous solutions Ag CrO 4 ( sp Ag (aq) CrO 4 (aq) sp solubility product constant The is the product of the molar concentrations of the dissolved ions of the solute. effect ( ): The shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction. AgCl( Ag (aq) Cl (aq) adding NaCl shifts equilibrium position sp can be used to determine if a precipitation will occur: - If the ion product sp, the solution is supersaturated and a precipitate will form. - If the ion product sp, the solution is saturated and no precipitate will form. - If the ion product sp, the solution is unsaturated and no precipitate will form. Solubility ( vs. solubility product ( sp ) sp is s is (at a given temperature) (especially with a common ion present) 4
3 Precipitation of bismuth sulfide (Bi S ). 5 sp can be calculated by determining the molar solubility or the number of moles of salt dissolved in one liter of the solution. Example: The molar solubility of PbF in a 0. M Pb(NO ) solution at 5 o C is.1-4 mol L -1. What is sp for PbF? SOLUTION: I C E PbF ( 0. but x is themolar solubilityso that sp Pb x (0. x)(x).9 0. x 8 F 0 x x sp (0..1 [ Pb [ F (0. x)(x) 4 )(.1 (the value from Table19.1is.6 4 ) 8 ) 6
4 Molar solubility can be calculated from sp. Example: What is the molar solubility of PbI in 0. M NaI? SOLUTION: PbI Pb I I 0 0. C x x E x 0. x sp [ Pb [ I (from Table 19.1) [ I sp 0. x 0. thus x(0.) or x M ( x << 0. so assumption is valid) Example 19.1 ~
5 19. Solubility equilibria of metal oxides and sulfides involve reactions with water 9 Some metal oxides dissolve in water by reacting with H O. : Most water-insoluble metal oxides dissolve in acid. : Water-insoluble metal oxides can form in basic solution. : Metal sulfides are similar to metal oxides. Some sulfides dissolve by reacting with water : Na S(H ONa (aq)hs (aq)oh (aq) Some metal ions are so reactive that they react with H S directly. These active ions include Cu, Pb, and Ni. Cu H S( aq) CuS( H :
6 The solubility of CuS in water is describe by: CuS( Cu S But the S CuS( H O reactswith water so that Cu HS theion product is now[ Cu [ HS OH [ OH and sp [ Cu [ HS [ OH Both OH - and HS - would react (to form H O and H S) in acidic solutions a modification to the ion product. CuS( H O H Cu H S( aq) H O removing water CuS( H [ Cu [ H [ H S H S( aq) which is described by theacid solubilityproduct, spa Cu spa 11 The metal sulfides can be collected into two groups: The acid-insoluble sulfides, which dissolve in neither acidic nor basic solution. The acid-soluble or base-insoluble sulfides that dissolve in acidic, but not basic, solutions. 1
7 Fig : The separation of Cu and Hg from Ni and Mn using H S Metal ions can be separated by selective precipitation 14
8 Selective precipitation means causing one metal ion to precipitate while holding the other in solution. Selective precipitation by control of ph applies to any system where the anion comes from a weak acid Consider the solubility of metal carbonates The amount of [CO - depends on [H : H CO HCO HCO CO H H a1 a Thus, the [CO - : as ph increases (because [H ) as ph decreases (because [H ) Example Example 19.9 Example: Over what range of ph is it possible to selectively precipitate Cu and Ni as their sulfides from a solution initially 0.0M in both ions? ( [H S0.1M in a saturated solution.) ANALYSIS: Cu is an acid-insoluble sulfide ( spa 6-16 ) and Ni is a base-insuble sulfide ( spa 4 1 ). Thus NiS will dissolve in acidic solution while CuS will not. The ph limits for the precipitation of both sulfides in saturates H S must be calculated. 16
9 SOLUTION: Consider CuSfirst CuS( H The Cu [ H this[ H (aq) will stay in solution only if [ Cu [ H spa H (0.0)(0.1) 16 6 could never be reached so Cu no matter how acidic the solution is. Cu S S( aq) [ H spa is above 6 1 M [ Cu [ H [ H precipitates S 17 Now consider Ni NiS( H The Ni [ H or a ph >. (aq) precipitate from solution when [ H [ Ni [ H spa Ni S H S( aq) (0.0)(0.1) 1 4 is below 0.005M [ Ni [ H [ H Thus, if we maintain the ph of the solution of 0.0 M Cu and 0.0 M Ni at ph., as we make the solution saturated in H S, virtually all the Cu will precipitate as CuS, but all the Ni will stay in solution. spa S 18
10 Fig. 15.1: A schematic diagram of the classic method for separating the common cations by selective precipitation Complex ions participate in equilibria in aqueous solutions 0
11 Complex Ion ( ): A charged species consisting of a metal ion (Lewis acid surrounded by ligands (Lewis base. : Cu 4H O Cu( H O) Coordination compounds ( ): Compounds that contain complex ions are called and the complex itself is sometimes called a coordination complex. Coordination Number ( ): Number of ligands attached to a metal ion. (Most common are ) 4 1 Formation (Stability) Constants: The equilibrium constants characterizing the stepwise addition of ligands to metal ions. The inverse of the formation constant is called the instability constant, inst. Cu( NH form inst ) 4 [ Cu( NH) [ Cu [ NH 1 form Cu 4 4 [ Cu [ NH [ Cu( NH ) 4NH
12 The solubility of a slightly soluble salt increases when one of its ions can be changed into soluble complex ion. (left) Aqueous ammonia is added to silver chloride (white). (right) Silver chloride, insoluble in water, dissolves to form Ag(NH ) (aq) and Cl - (aq). AgCl ( NH (aq) Ag(NH ) (aq) Cl - (aq) H (aq)ag(nh ) (aq)cl - (aq) NH4 (aq)agcl( Example: Calculate the solubility of silver chloride in 0. M NH. SOLUTION: AgCl( [ Ag [ Cl NH AgCl( NH sp Ag form c Cl [ Ag( NH ) [ Ag [ NH sp Ag form 1.8 [ Ag( NH ) [ NH Ag( NH 1.6 Ag( NH 7 ), ) [ Cl combining Cl.9 4
13 AgCl( NH c [ Ag( NH) [ Cl [ NH x 0. x c Ag( NH ) x (0. x) Cl I C x x x E 0. -x x x or x M The solubility of AgCl in pure water is sp 1. thus, the solubility is a factor of times larger in 0.M NH 5 1. Example M 5 Fig. 15.1: The separation of the Group I ions in the classic scheme of qualitative analysis. 6
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