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Chapter 15 Extra Credit Name: ______________Key_______________ 1. Write the ionic equation for dissolution and the solubility product (K sp ) expression for each of the following slightly soluble ionic compounds: (a) PbCl 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) Ksp = [Pb 2+ ][ Cl ] 2 2. Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF 2 , Hg 2 Cl 2 , PbI 2 , or Sn(OH) 2 . CaF 2 Ksp = 4.0 x 10 −11 Hg 2 Cl 2 Ksp = 1.1 x 10 −18 PbI 2 Ksp = 1.4 x 10 −8 Sn(OH) 2 Ksp = 3 x 10 −27 Each of these break into a total of 3 ions (even mercury (I) chloride) so the Ksp’s can be directly compared, Sn(OH) 2 is less soluble than Hg 2 Cl 2 < CaF 2 < PbI 2 3. Calculate the molar solubility of each. Look up Ksp in the appendix. (a) PbI 2 Ksp = 1.4 x 10 −8 X = 0.0015 M (b) Ag 2 SO 4 Ksp = 1.2 x 10 −5 X = 0.014 M 4. Given the molar solubility, calculate K sp for each of the slightly soluble solids indicated: (a) AgBr: x = 5.7 × 10 –7 M, Ksp = x 2 Ksp = 3.2 x 10 −13 (b) PbF 2 : x = 2.1 × 10 –3 M, Ksp = x(2x) 2 Ksp = 3.7 x 10 −8 5. The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of solution. Calculate the solubility product (Ksp) for each. (a) BaSiF 6 , 0.026 g/100 mL (contains SiF 6 2− ) x = 0.00074M Ksp = 5.5 x 10 −7 (b) Ce(IO 3 ) 4 , 1.5 × 10 –2 g/100 mL MM Ce(IO 3 ) 4 = 840 g/mole X = 0.00018 M Ksp = [Ce 4+ ][IO 3 ] 4 = x(4x) 4 = 256 (x) 5 Ksp = 4.8 x 10 −17 5. Calculate the molar solubility of: (a) AgCl(s) in pure water. Ksp = 1.6 x 10 −10 x = 1.3 x 10 −5 M AgCl(s) in 0.010 M NaCl x = 1.6 x 10 −8 M How does the solubility change when a common ion is added? The solubility decreases.

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Page 1: Chapter 15 Extra credit key - blogs.nvcc.edublogs.nvcc.edu/alchm/files/2018/03/Chapter_15_Extra_credit_key.pdf · Chapter 15 Extra Credit Name: _____Key_____ 1. Write the ionic equation

Chapter15 ExtraCredit Name:______________Key_______________

1.Writetheionicequationfordissolutionandthesolubilityproduct(Ksp)expressionforeachofthefollowingslightlysolubleioniccompounds:

(a)PbCl2PbCl2(s)⇄Pb2+(aq)+2Cl−(aq)Ksp=[Pb2+][Cl−]2

2.Usesolubilityproductsandpredictwhichofthefollowingsaltsisthemostsoluble,intermsofmolesperliter,inpurewater:CaF2,Hg2Cl2,PbI2,orSn(OH)2.

CaF2 Ksp=4.0x10−11

Hg2Cl2 Ksp=1.1x10−18

PbI2 Ksp=1.4x10−8

Sn(OH)2 Ksp=3x10−27

Eachofthesebreakintoatotalof3ions(evenmercury(I)chloride)sotheKsp’scanbedirectlycompared,

Sn(OH)2islesssolublethanHg2Cl2<CaF2<PbI2

3.Calculatethemolarsolubilityofeach.LookupKspintheappendix.

(a)PbI2 Ksp=1.4x10−8 X=0.0015M

(b)Ag2SO4 Ksp=1.2x10−5 X=0.014M

4.Giventhemolarsolubility,calculateKspforeachoftheslightlysolublesolidsindicated:

(a)AgBr:x=5.7×10–7M,Ksp=x2 Ksp=3.2x10−13

(b)PbF2:x=2.1×10–3M,Ksp=x(2x)2Ksp=3.7x10−8

5.TheHandbookofChemistryandPhysicsgivessolubilitiesofthefollowingcompoundsingramsper100mLofsolution.Calculatethesolubilityproduct(Ksp)foreach.

(a)BaSiF6,0.026g/100mL(containsSiF62−)x=0.00074MKsp=5.5x10−7

(b)Ce(IO3)4,1.5×10–2g/100mLMMCe(IO3)4=840g/mole

X=0.00018MKsp=[Ce4+][IO3−]4=x(4x)4=256(x)5

Ksp=4.8x10−17

5.Calculatethemolarsolubilityof:

(a)AgCl(s)inpurewater.Ksp=1.6x10−10 x=1.3x10−5M

AgCl(s)in0.010MNaCl x=1.6x10−8M

Howdoesthesolubilitychangewhenacommonionisadded?Thesolubilitydecreases.

Page 2: Chapter 15 Extra credit key - blogs.nvcc.edublogs.nvcc.edu/alchm/files/2018/03/Chapter_15_Extra_credit_key.pdf · Chapter 15 Extra Credit Name: _____Key_____ 1. Write the ionic equation

(b)Calculatethemolarsolubilityof:CaF2(s)in0.00125MKFKsp=4.0x10−11=[Ca2+][F−]2=x(.00125+2x)2=x(.00125)2

x=2.6x10−6M

(c)Calculatethemolarsolubilityof:Ni(OH)2(s)inasolutionwithpHof12.00.Ksp=1.6x10−16

[OH−]=10−2Ksp=[Ni2+][OH−]2=(x)(.01+2x)2=x(.01)2x=1.6x10−12M

7.Willaprecipitateformgiventheconcentrationsindicated?(SeeappendixforKspvalues.)

(a)CaCO3:[Ca2+]=0.0020M,[CO32−]=0.010MKsp=8.7x10−9

Q=2.0x10−5Q>Ksp,soaprecipitatewillform.

(b)Mn(OH)2:[Mn2+]=1.0×10–4M,[OH–]=1.0×10–5M

Q=1.0x10−14Q<Ksp,aprecipitatewillnotform.

8.DrawtheLewisStructureforeachandLabeltheLewisAcidsandtheLewisBases(reactantsonly)(a)BF3+F−⟶ BF4−

LALB(b)Al(OH)3+OH−⟶ Al(OH)4−

LALB

DrawtheLewisStructureforeachandLabeltheLewisAcidsandtheLewisBases(reactantsonly)andpredicttheProducts.(a)NH3(g)+HCl(g)⟶

LB LA(b)NH4

++C2H5O− ⟶LALB

9.Avolumeof.080Lof2.0x10−3MBa(NO3)2(aq)isaddedto.020Lof5.0x10−3MLi2SO4(aq).Willaprecipitateform?

For:BaSO4:Ksp=2.3x10−8

Q=1.6x10−6.Aprecipitatewillform.