1 Salt Solubility Chapter 18. 2 Solubility product constant K sp K sp Unitless Unitless CaF 2(s) Ca...

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Salt SolubilitySalt Solubility

Chapter 18Chapter 18

Solubility product constantSolubility product constant

• KKspsp

• UnitlessUnitless

• CaFCaF2(s)2(s) Ca Ca2+2+(aq)(aq) + 2F + 2F--

(aq)(aq)

• KKspsp = [Ca = [Ca2+2+][F][F--]]22

• [Ca[Ca2+2+] = molar solubility (in M)] = molar solubility (in M)

• Get KGet Kspsp values from Appendix J, pages values from Appendix J, pages A-24-25A-24-25

We have problemsWe have problems

• KKspsp for BaSO for BaSO44 = 1.1 x 10 = 1.1 x 10-10-10 @ 25 @ 25°C°C– Calculate this salt’s solubility in water in Calculate this salt’s solubility in water in

g/L.g/L.

Solution Solution 2 2

4( ) ( ) 4 ( )

10 2 2 2sp 4

C +x +x

K 1.1 10 [ ][ ]

1.0 10

s aq aqBaSO Ba SO

Ba SO x

233 0.0024g golL mol L

Problem Problem

• If [CaIf [Ca2+2+] = 2.4 x 10] = 2.4 x 10-4-4 M, what is the M, what is the KKspsp value of CaF value of CaF22??

SolutionSolution

2+ -2(s) (aq) (aq)

2+ -sp

CaF Ca + 2F

I -- 0 0

C -- +x +2x

E -- x 2x

(2.4 10 ) (2 x 2.4 10 )

K = [Ca ][F ]

2 -4 -4 2 11[2.4 10 ][4.8 10 ] 5.5 10

Solubility and the Common Ion Solubility and the Common Ion EffectEffect

• Problem: How much AgCl (g/L) would Problem: How much AgCl (g/L) would dissolve in water given Kdissolve in water given Kspsp= 1.8 x 10= 1.8 x 10--

1010??

( ) ( ) ( )

-10 2sp

K =1.8 10 [ ][ ]

1.3 10

1.3 10 143 0.0019

s aq aqAgCl Ag Cl

Ag Cl x

g gM mol

Solubility and the Common Ion Solubility and the Common Ion EffectEffect

• If solid AgCl is placed in 1.00L of If solid AgCl is placed in 1.00L of 0.55M NaCl, what mass of AgCl (g/L) 0.55M NaCl, what mass of AgCl (g/L) would dissolve?would dissolve?

• Before solving this, would you expect Before solving this, would you expect it to be lesser or greater than in pure it to be lesser or greater than in pure water?water?– Think back to Le ChThink back to Le Châtelier’s Principle âtelier’s Principle

and the pH of buffersand the pH of buffers

( ) ( ) ( )

0 0.55

x x+0.55

K =1.8 10 [ ][ ] [ ][ 0.55] [ ][0.55]

(We can appro

s aq aqAgCl Ag Cl

Ag Cl x x x

-10 -8

ximate since K is very small.)

x = 3.3 10 [ ]

3.3 10 143 4.7 10

g gM mol L

Basic anions and salt Basic anions and salt solubilitysolubility• Do we recall what a conjugate Do we recall what a conjugate

acid/conjugate base is?acid/conjugate base is?

• If salt has conjugate base of weak acid, then If salt has conjugate base of weak acid, then salt more soluble than value given by Ksalt more soluble than value given by Kspsp

– Why?Why?

PbSPbS(s)(s) Pb Pb2+2+(aq)(aq) + S + S2-2-

(aq)(aq)

SS2-2-(aq)(aq) + H + H22OO(l)(l) HS HS--

(aq)(aq) + OH + OH--(aq)(aq); ;

Where net = KWhere net = Kspsp K Kbb (>K (>Kspsp))

Effect of pH on solubilityEffect of pH on solubility

• Mg(OH)Mg(OH)2 (s)2 (s) Mg Mg2+2+(aq)(aq) + 2OH + 2OH--

(aq)(aq)

• If one adds base, OHIf one adds base, OH--, which way will , which way will the equilibrium shift?the equilibrium shift?– Will more solid form or dissolve?Will more solid form or dissolve?

• If one adds acid, HIf one adds acid, H33OO++, which way will , which way will the equilibrium shift?the equilibrium shift?– Will more solid form or dissolve?Will more solid form or dissolve?

• MgMg33(PO(PO44))2 (s)2 (s) 3Mg 3Mg2+2+(aq)(aq) + 2PO + 2PO44

-3-3(aq)(aq)

• MgClMgCl2 (s)2 (s) Mg Mg2+2+(aq)(aq) + 2Cl + 2Cl--

(aq)(aq)

So what can we say about the So what can we say about the conjugate base and solubility?conjugate base and solubility?

The reaction quotient, QThe reaction quotient, Q

• Q tells us:Q tells us:– Whether it’s at Whether it’s at

equilibriumequilibrium• If not, which way it’ll If not, which way it’ll

shiftshift

• CaFCaF2(s)2(s) Ca Ca2+2+(aq)(aq)+ 2F+ 2F--

(aq)(aq)

• KKspsp = [Ca = [Ca2+2+][F][F--]]22

• Q = [CaQ = [Ca2+2+][F][F--]]22

• If Q = KIf Q = Kspsp, then @ , then @ eq.eq. & soln is saturated& soln is saturated

• If Q < KIf Q < Kspsp, then will , then will shift to right shift to right (dissolve more) & (dissolve more) & soln is unsaturatedsoln is unsaturated

• If Q > KIf Q > Kspsp, then will , then will shift to left (not shift to left (not dissolve anymore, dissolve anymore, ppt out) & soln is ppt out) & soln is supersaturated (will supersaturated (will ppt out)ppt out)

Check out the heating pads & Check out the heating pads & videovideo

• http://genchem.chem.wisc.edu/demohttp://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/11solutinstrations/Gen_Chem_Pages/11solutionspage/crystallization_from_super.honspage/crystallization_from_super.htmtm

ProblemProblem

• AgCl placed in water. After a certain AgCl placed in water. After a certain amount of time, concentration= 1.2 x amount of time, concentration= 1.2 x 1010-5-5 M. M.

• Has the system reached eq.? Has the system reached eq.?

• If not, will more solid dissolve? If not, will more solid dissolve?

• How much more? How much more?

• (K(Kspsp = 1.8 x 10 = 1.8 x 10-10-10))

5 5 10

[ ][ ] [1.2 10 ][1.2 10 ] 1.4 10

1.8 10 1.4 10

Thus, the system is not yet at equilibrium and will proceed to the right,

dissolving more solid.

At equilibrium, 1.3 10 mol of each i

Q Ag Cl

on are present.

Therefore, 1.3 10 1.2 10 0.1 10 moles required to dissolve to reach eq.

0.1 10 143 1 10 of additional AgCl will dissolve

mol mol mol

gmol gmol

Selective precipitationSelective precipitation

• Will a precipitate form, given the ion Will a precipitate form, given the ion concentrations?concentrations?

• What concentrations of ions are What concentrations of ions are needed to bring about precipitation?needed to bring about precipitation?

• Basically, compare Q to KBasically, compare Q to Kspsp

Problem Problem

• If [MgIf [Mg2+2+] = 1.5x10] = 1.5x10-6-6 M, and enough M, and enough hydroxide ions are added to make hydroxide ions are added to make the solution 1.0 x 10the solution 1.0 x 10-4 -4 M of hydroxide M of hydroxide ions, will Mg(OH)ions, will Mg(OH)22 precipitate? precipitate?

• If not, will it occur if the If not, will it occur if the concentration of hydroxide ions is concentration of hydroxide ions is increased to 1.0 x 10increased to 1.0 x 10-2-2 M? M?

• KKspsp = 5.6 x 10 = 5.6 x 10-12-12

2+ -2(s) (aq) (aq)

6 4 2 14

6 -2 2 10

Mg(OH) Mg +2OH

[1.5 10 ][1.0 10 ] 1.5 10

(1.5 10 ) (5.6 10 )

Unsaturated solution and precipitation doesn't occur yet.

When [OH ]=1.0 10 :

[1.5 10 ][1.0 10 ] 1.5 10

10 12(1.5 10 ) (5.6 10 )

Precipitation will occur.

Complex ion equilibriaComplex ion equilibria

• Complex ionComplex ion = a central = a central metal ion (emetal ion (e-- acceptor = acceptor = Lewis acid) bound to 1 or Lewis acid) bound to 1 or more more ligandsligands

• Ligand Ligand = neutral = neutral molecule or ion that acts molecule or ion that acts as Lewis base (eas Lewis base (e-- donor) donor) with the central metal with the central metal ionion

• KKff values on page A-26, values on page A-26, Appendix KAppendix K

+ -(s) 3(aq) 3 2 (aq) (aq)AgCl + 2NH [Ag(NH ) ] Cl

Example Example

• What is KWhat is Knet net for the reaction below?for the reaction below?

+ -(s) 3(aq) 3 2 (aq) (aq)AgCl + 2NH [Ag(NH ) ] Cl

+ -(s) 3(aq) 3 2 (aq) (aq)

+ - -10(s) (aq) (aq) sp

+ + 7(aq) 3(aq) 3 2 (aq) f

+ - ++ -3 2 3 2

sp f2 + 23 3

AgCl + 2NH [Ag(NH ) ] Cl

AgCl Ag Cl ; K 1.8 10

Ag 2NH [Ag(NH ) ] ; K = 1.6 10

[Ag(NH ) ] [Cl ] [Ag(NH ) ][Ag ][Cl ] K K (1.

[NH ] [Ag ][NH ]netK

-10 7 38 10 ) (1.6 10 ) 2.9 10

Problem Problem

• What is the net rxn and value of KWhat is the net rxn and value of Knetnet for dissolving AgBr in a solution for dissolving AgBr in a solution containing Scontaining S22OO33

2-2-??

• KKspsp = 5.4 x 10 = 5.4 x 10-13-13

• KKff = 2.0 x 10 = 2.0 x 101313

+ - -13(s) (aq) (aq) sp

+ 2- 3- 13(aq) 2 3 (aq) 2 3 2 (aq) f

2- 3- -(s) 2 3 (aq) 2 3 2 (aq) (aq)

-13 13sp f

AgBr Ag Br ; K 5.4 10

Ag 2S O [Ag(S O ) ] ; K = 2.0 10

net rxn: AgBr + 2S O [Ag(S O ) ] Br

K K (5.4 10 ) (2.0 10 ) 10.netK

1 Salt Solubility Chapter 18. 2 Solubility product constant K sp K sp Unitless Unitless CaF 2(s) Ca...

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