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Solubility Equilibria

Solubility Product Constant

Ksp

for saturated solutions at equilibrium

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Comparingvalues.

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Solubility Product (Ksp) = [products]x/[reactants]y but.....

reactants are in solid form, so Ksp=[products]x

i.e. A2B3(s) 2A3+ + 3B2– Ksp=[A3+]2 [B2–]3

Given: AgBr(s) Ag+ + Br–

In a saturated solution @25oC, the [Ag+] = [Br– ]= 5.7 x 10–7 M. Determine the Ksp value.

-132-7sp 10x 3.3 10x 7.5Br AgK

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Problem: A saturated solution of silver chromate was to found contain 0.022 g/L of Ag2CrO4. Find Ksp

Eq. Expression: Ag2CrO4 (s) 2Ag+ + CrO42–

Ksp = [Ag+]2[CrO42–]

So we must find theconcentrations of each

ion and then solvefor Ksp.

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Problem: A saturated solution of silver chromate was to contain 0.022 g/L of Ag2CrO4. Find Ksp

Eq. Expression: Ag2CrO4 (s) 2Ag+ + CrO42–

Ksp = [Ag+]2[CrO42–]

Ag+: L

Agmol

0.022 g Ag2CrO4

L g332

mol

42CrO Ag1

Ag2

1.33 x 10–4

CrO4–2: 0.022g Ag2CrO4

L L

CrO mol

-24

g 332

mol4

-24

AgCrO1

CrO 16.63 x 10–5

-12-52-4sp 10x 1.16 10x 63.610x 33.1K

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Problems working from Ksp values.

Given: Ksp for MgF2 is 6.4 x 10–9 @ 25 oC

Find: solubility in mol/L and in g/L

MgF2(s) Mg2+ + 2F– Ksp = [Mg2+][F–]2

I.C.E.

N/A 0 0N/A +x +2xN/A +x +2x

Ksp= [x][2x]2 = 4x3

6.4 x 10–9 = 4x3

223-3

-9

MgF Mg 10x 1.2 4

10x 4.6 x

now for g/L:L

MgF mol10x 2.1 2-3

L

MgF g 2

mol

g 62.3 7.3 x 10–2

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The common ion effect “Le Chatelier”overhead fig 17.16

What is the effect of adding NaF?

CaF2(s) Ca2+ + 2F-

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Solubility and pH

CaF2(s) Ca2+ + 2F–

Add H+ (i.e. HCl)

2F– + H+ HF

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Solubility and pH

Mg(OH)2(s) Mg2+ + 2OH–

Adding NaOH?

Adding HCl?

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The common ion effect “Le Chatelier”

Why is AgCl less soluble in sea water than in fresh water?

AgCl(s) Ag+ + Cl–

Seawater containsNaCl

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Problem: The solubility of AgCl in pure water is 1.3 x 10–5 M.What is its solubility in seawater where the [Cl–] = 0.55 M? (Ksp of AgCl = 1.8 x 10–10)

AgCl(s) Ag+ + Cl–

I.C.E.

N/A 0 0.55N/A +x +xN/A +x 0.55 + x

Ksp= [Ag+][Cl–]

Ksp= [x][0.55 + x] try dropping this x

Ksp = 0.55x

1.8 x 10–10 = 0.55x

x = 3.3 x 10–10 = [Ag+]=[AgCl]

“AgCl is much less soluble in seawater”

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more Common ion effect:

a. What is the solubility of CaF2 in 0.010 M Ca(NO3)2?Ksp(CaF2) = 3.9 x 10–11

CaF2(s) Ca2+ + 2F–

[Ca2+] [F–]

I.C.E.

0.010 0 +x +2x0.010 + x 2x

Ksp= [0.010 + x][2x]2 [0.010][2x]2 = 0.010(4x2)

3.9 x 10–11 = 0.010(4x2)

x = 3.1 x 10–5 M Ca2+ from CaF2 so = M of CaF2

Now YOU determine the solubility of CaF2 in 0.010 M NaF.

Ksp=[Ca2+][F-]2

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Answer: 3.9 x 10–7 M Ca2+

CaF2(s) Ca2+ + 2F–

0 0.010 +x 2x x 0.010 + 2x

Ksp = [x][0.010 + 2x]2

3.9 x 10-11 =x(0.010)2

x(0.010)2

x = 3.9 x 10-7

What does x tell us

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Reaction Quotient (Q): will a ppt. occur?

Use Q (also called ion product) and compare to Ksp

Q < Ksp reaction goes

Q = KspEquilibrium

Q > Kspreaction goes

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Problem: A solution is 1.5 x 10–6 M in Ni2+. Na2CO3 is added to make the solution 6.0 x 10–4 M in CO3

2–. Ksp(NiCO3) = 6.6 x 10–9.Will NiCO3 ppt?

NiCO3 Ni2+ + CO32–

Ksp = [Ni2+][CO32–]

Q = [Ni2+][CO32–]

Q = [1.5 x 10–6][6.0 x 10–4] = 9.0 x 10–10

Q < Ksp no ppt.

We must compare Q to Ksp.

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Problem: 0.50 L of 1.0 x 10–5 M Pb(OAc)2 is combined with 0.50 L of 1.0 x 10–3 M K2CrO4.

a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–14

Pb(OAc)2(aq) + K2CrO4(aq) PbCrO4(s) + 2KOAc(aq)

then: PbCrO4(s) Pb2+ + CrO42– Ksp= [Pb2+][CrO4

2–]

[Pb2+]:L

Pb mol

2

0.50 L

L

Pb(OAc) mol10x 0.1 2-5

2

2

Pb(OAc) 1

Pb 1

1 L5.0 x 10–6

[CrO42-]:

L

CrO mol

-24

0.50 L

L

CrOK mol10x 1.0 42-3

42

-24

CrOK 1

CrO 15.0 x 10-4

Q = [5.0 x 10-6][5.0 x 10-4] = 2.5 x 10-9

compare to Ksp: Q > Ksp so a ppt. will occur

1 L

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b. find the Eq. conc. of Pb2+ remaining in solution after the PbCrO4 precipitates.

Since [Pb2+] = 5.0 x 10-6 and [CrO42-] = 5.0 x 10-4 and there is a

1:1 stoichiometry, Pb2+ is the limiting reactant.

PbCrO4(s) Pb2+ + CrO42–

I. (after ppt.)

C.E.

0 5.0 x 10-4 - 5.0 x 10-6 = 5.0 x 10-4

+x +x x 5.0 x 10–4 + x

Ksp = [x][5.0 x 10–4 + x] Try dropping the “+ x” term.

Ksp(PbCrO4) = 1.8 x 10–14

1.8 x 10–14 = x(5.0 x 10-4)x = [Pb2+] = 3.6 x 10–11

This is the concentration of Pb2+ remaining in solution.

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Complex ion formation:

Ag+(aq) + NH3(aq) Ag(NH3)+(aq)

AgCl(s) Ag+ + Cl–

Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2+(aq)

Ag+(aq) + 2NH3(aq) {Ag(NH3)2}+(aq)

formation or stability constant:

72

3

23f 10x 1.7

NH Ag

)NH(AgK

Ksp= 1.8 x 10–10

For Cu2+: Cu2+ + 4NH3 [Cu(NH3)4]2+(aq)

K1 x K2 x K3 x K4 = Kf = 6.8 x 1012

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Solubility and complex ions:

Problem: How many moles of NH3 must be added to dissolve 0.050 mol of AgCl in 1.0 L of H2O? (KspAgCl = 1.8 x 10–10 ; Kf[Ag(NH3)2]+ = 1.6 x 107)

AgCl(s) Ag+ + Cl–

Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)

sum of RXNS: AgCl(s) + 2NH3 Ag(NH3)2+(aq) + Cl–

fsp2

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23overall K x K

NH

Cl )NH(Ag K

= 2.9 x 10–3

Now use Koverall to solve the problem:

Koverall= 2.9 x 10–3 =

NH

Cl )NH(Ag 2

3

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2 3NH

0.050 0.050

[NH3]eq = 0.93 but ..... How much NH3 must we add?

[NH3]total= 0.93 + (2 x 0.050) = 1.03 M2 ammonia’s for each Ag+

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Fractional Precipitation: “ppting” one ion at a time. Compounds must have different Ksp values (i.e. different solubilities)

Example: Ksp CdS = 3.9 x 10–29 and KspNiS = 3.0 x 10–21

? Which will ppt. first? least soluble

Problem: A solution is 0.020 M in both Cd2+ and Ni2+. Just before NiS begins to ppt., what conc. of Cd2+ will be left in solution?

Approach: Find conc. of S2– ion when Ni2+ just begins to ppt. since Cd2+ will already be ppting. Then use this S2– conc. to find Cd2+.

NiS(s) Ni2+ + S2– Ksp= 3.0 x 10–21 = [0.020][S2–]

[S2–] = 1.5 x 10–19 M when Ni2+ just begins to ppt.

So: CdS(s) Cd2+ + S2– Ksp= 3.9 x 10–29 = [Cd2+][1.5 x 10–19]

[Cd2+] = 2.6 x 10–10 M when NiS starts to ppt.