Solubility Lesson 3 Calculating Ksp

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# Solubility Lesson 3 Calculating Ksp - PowerPoint PPT Presentation

Solubility Lesson 3 Calculating Ksp. The Molar Solubility is the molarity required to saturate or fill the solution at any given temperature. 1. The solubility ( s ) of BaCO 3 is 5.1 x 10 -5 M @ 25 0 C. Calculate the solubility product or Ksp .

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Solubility

Lesson 3

Calculating

Ksp

The Molar Solubilityis the molarity required to saturate or fill the solution at any given temperature.

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

The Molar Solubilityis the molarity required to saturate of fill the solution at any given temperature.

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

Ba2+

CO32-

BaCO3(s)

BaCO3(s)⇌ Ba2+ + CO32-

s s s

Ksp = [Ba2+][CO32-]

Ksp = [s][s]

Ksp = s2

Ksp = (5.1 x 10-5)2

Ksp = 2.6 x 10-9

Ksp

Solubility Product

Saturated solutions- at equilibrium

No Units

Increasing Temperatureincreasesthe Ksp

dissociation equation

PbBr2(s)⇌ Pb2+ + 2Br-

solubility

s s 2s

equilibrium expression

Ksp = [Pb2+][Br-]2

substitute & solve

Ksp = [s][2s]2

Ksp = 4s3

Ksp = 4(0.012)3

Ksp = 6.9 x 10-6

Note that the Br- is doubledand then squared!

3.If0.00243 g of Fe2(CO3)3 is required to saturate 100.0 mL of solution. What is the solubility product?

Fe2(CO3)3⇌ 2Fe3+ + 3CO32-

s 2s 3s

Ksp = [Fe3+]2[CO32-]3

s =

0.00243 g

x 1 mole

291.6 g

Ksp = [2s]2[3s]3

0.100 L

Ksp = 108s5

=8.333 x 10-5 M

Ksp = 108(8.333 x 10-5)5

Ksp = 4.34 x 10-19

4. A 200.0 mL sample of a saturated solution of Mg(OH)2 weighs 222.1210 g. When the beaker containing the solution is evaporated to dryness it weighs 22.1213 g. The mass of the empty beaker is 22.1200 g. Calculate the Ksp.

Mass of Beaker + Mg(OH) 22.1213 g

note sig figs- 4th decimal

-Mass of Beaker -22.1200 g

Mass of Mg(OH)20.0013 g

Mg(OH)2⇌ Mg2+ + 2OH-

s s 2s

0.0013 g

x 1 mole

Ksp = [Mg2+][OH-]2

s =

58.3 g

= [s][2s]2 = 4s3

0.2000 L

=4(1.1149 x 10-4)3

= 1.1149 x 10-4 M

= 5.5 x 10-12

5. 40.00 mL of a saturatedBa(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

Titration 2HCl + 1Ba(OH)2

0.02910 L

0.0400 L

0.300 M

? M

x 1mole Ba(OH)2

0.02910 L HCl

x 0.300 moles

2 moles HCl

1 L

[Ba(OH)2] =

0.0400 L

s= 0.1091 M

Ksp

Ba(OH)2⇌ Ba2+ + 2OH-

s s 2s

Ksp = [Ba2+][OH-]2

= [s][2s]2

= 4s3

= 4(0.1091)3

= 5.20 x 10-3