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Solubility Review

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Solubility

Review

1.Solubility is a measure of the maximum amount of solid that will dissolve in a volume of water.

Units: g/Lmol/Lg/100mL

Or even: mL/Lfor CO2(g) in water

Ksp

Solubility Product

Saturated solutions

No Units

Only Changes with Temperature

No ICE!

Step #1Write a dissociation equation for the low solubility salt

2.The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp.

Dissociation net ionicBaCO3(s)⇌ Ba2+ + CO32-

s s s

Ksp = [Ba2+][CO32-]

Ksp = s2

Ksp = (5.1 x 10-5)2

Ksp = 2.6 x 10-9

Adding BaCO3does notincrease [Ba2+] or [CO32-] but it does increase the rate of dissolving and crystallization.

Formation Net Ionic: Ba2+ + CO32-BaCO3(s)

Ba2+

CO32-

BaCO3(s)

2.The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp.

BaCO3(s)⇌ Ba2+ + CO32-

s s s

Ksp = [Ba2+][CO32-]

Ksp = s2

Ksp = (5.1 x 10-5)2

Ksp = 2.6 x 10-9

Adding BaCO3does notincrease [Ba2+] or [CO32-] but it does increase the rate of dissolving and crystallization.

Formation Net Ionic:Ba2+ + CO32-BaCO3(s)

Ba2+

CO32-

BaCO3(s)

3.PbCl2(s)⇌Pb2++2Cl-

Ksp = 4s3

4.200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur?

PbCl2(s)⇌Pb2++2Cl-

2000.10 M3000.20 M500500

0.040 M0.12 M

TIP=[Pb2+][Cl-]2

TIP=[0.040][0.12] 2

=5.8 x 10-4

Ksp=1.2 x 10-5 TIP > Ksp ppt forms

5.Changing the Solubility of a Salt

E+Ca(OH)2(s)⇄ Ca2+ + 2OH-

Substance AddedEffect on Molar SolubilityEffect on Ksp

Ca(NO3)2Decreasenone

Na2SO4Increasenone

NaOHDecreasenone

HCl Increasenone

NaClnonenone

Ca(OH)2nonenone

Increase TemperatureIncreaseIncrease

Decrease TemperatureDecreaseDecrease

6.Mg(OH)2 will have the greatest solubility in:

Mg(OH)2(s) ⇌ Mg2+ +2OH-

A.NaOHOH- lowers solubility

B.Mg(NO3)2Mg2+ lowers solubility

C.H2ONo effect solubility

D.AgNO3Ag+ increases solubility by reacting with OH-

7.Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

AgCl(s)⇄ Ag+ + Cl-

0.20 M

Ksp=[Ag+][Cl-]

1.8 x 10-10 = [0.20][Cl-]

[Cl-] =9.0 x 10-10 M

BaCl2(s)⇄ Ba2+ + 2Cl-

4.5 x 10-10 M9.0 x 10-10 M

0.50 L x 4.5 x 10-10molex 208.3 g = 4.7 x 10-8 g

1 L mole

8.Ionic Solutions:NaClHClNH4NO3

Molecular Solutions:C12H22O11

9.What is the answer?

A.NaCl

B.CaCl2

C.AgCl

D.AgBr

10.Ionic Solutions:NaClHClNH4NO3

Molecular Solutions:C12H22O11

11.What is the answer?

A.NaClHigh

B.CaCl2High

C.AgClLow1.8 x 10-12

D.AgBrLow5.4 x 10-13