SolubilityReview

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

Units: g/L mol/L g/100mL

Or even: mL/L for CO2(g) in water

KspSolubility ProductSaturated solutionsNo UnitsOnly Changes with Temperature No ICE!

Step #1 Write 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 ionic 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 BaCO3 does not increase [Ba2+] or [CO32-] but it does

increase the rate of dissolving and crystallization.Formation Net Ionic: Ba2+ + CO3

2- 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 BaCO3 does not increase [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-

200 0.10 M 300 0.20 M500 500

0.040 M 0.12 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 Added Effect on Molar Solubility Effect on Ksp Ca(NO3)2 Decrease none

Na2SO4 Increase none

NaOH Decrease none

HCl Increase none

NaCl none none

Ca(OH)2 none none

Increase Temperature Increase Increase

Decrease Temperature Decrease Decrease

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

Mg(OH)2(s) ⇌ Mg2+ + 2OH-

A. NaOH OH- lowers solubility

B. Mg(NO3)2 Mg2+ lowers solubility

C. H2O No effect solubility

D. AgNO3 Ag+ 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 M 9.0 x 10-10 M

0.50 L x 4.5 x 10-10 mole x 208.3 g = 4.7 x 10-8 g1 L mole

8. Ionic Solutions: NaCl HCl NH4NO3

Molecular Solutions: C12H22O11

9. What is the answer?

A. NaCl

B. CaCl2

C. AgCl

D. AgBr

10. Ionic Solutions: NaCl HCl NH4NO3

Molecular Solutions: C12H22O11

11. What is the answer?

A. NaCl High

B. CaCl2 High

C. AgCl Low 1.8 x 10-12

D. AgBr Low 5.4 x 10-13