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Solubility Product Constant 6-5. Ksp. is a variation on the equilibrium constant for a solute-solution equilibrium. remember that the solubility equilibrium is based on a saturated solution. Solids are not written in the equilibrium expression. CaF 2 (s) Ca 2+ (aq) + 2F - (aq).

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slide2
is a variation on the equilibrium constant for a solute-solution equilibrium.
  • remember that the solubility equilibrium is based on a saturated solution.
  • Solids are not written in the equilibrium expression.
slide3

CaF2 (s) Ca 2+ (aq) + 2F -(aq)

Ksp = [Ca 2+] [F -]2

Where Ksp is the solubility product constant, or simply the solubility product.

slide5
Given the solubility, calculate the Ksp of the substance.

1).Write the chemical equation for the substance dissolving and dissociating.

2) Write the Ksp expression.

3) Insert the concentration of each ion and multiply out.

slide6

Calculating the solubility product:

CuBr has a measured solubility of 2.0 10-4 M at 25 C. Calculate the Ksp value.

CuBr (s) Cu + (aq) + Br -(aq)

Ksp = [Cu +] [Br -]

= (2.0 10-4 )(2.0 10-4 )

= 4.0 10-8

slide7

x

EX: Calculate the Ksp for Bi2 S3 , which has a solubility of 1.36 x 10-15 mol/L at 25°C.

Bi2S3 (s) 2 Bi3+ (aq) + 3 S2- (aq)

2(1.36 x 10-15 ) 3(1.36 x 10-15)

2.72 x 10-15 4.08 x 10-15

x

3x

2x

slide8
Ksp = [Bi 3+]2 [S 2-]3

= (2.72 x 10-15)2 (4.08 x 10-15)3

= 5.02 x 10-73

slide9
Determine the Ksp of calcium fluoride (CaF2 ), given that its molar solubility is 2.14 x 10-2 M.
  • CaF2 (s) → Ca2+ (aq) + 2 F¯ (aq)
  • The Ksp expression is:

Ksp = [Ca2+ ] [F¯]2

2x

x

slide10
Ksp = [Ca2+ ] [F¯]2
  • Ksp = (2.14 x 10-2 ) (4.28 x 10-2 )2
  • Ksp = 3.92 x 10-5
convert g l to mol l
Convert g/l to mol/l

The solubility of Fe(OH)2 is found to be 1.4 x 10-3 g/l. What is the Ksp value?

Change the g/l to mol/l.

Molar mass of Fe(OH)2 is 89.87g/mol

Solubility is = 1.56 x 10-5 mol/l

Fe(OH)2(s) → Fe2+(aq) + 2OH-(aq)

x x 2x

Ksp = [Fe2+ ][OH- ]2

=(1.56 x 10-5 ) (2[1.56 x 10-5 ]) 2

=1.5 x 10-14

slide12
Calculate the Ksp of

silver bromide: ( AgBr ), if the

solubility is 1.3 x 10 -4 g / L .

slide13
We can also use a known value of Ksp to find out the solubility of a substance.
  • Write out the chemical equation for the substance
  • Write the Ksp expression
  • Solve for x
slide14
If the Ksp is 1.8 x 10 -8 of silver chloride, find the solubility.
  • write Ksp expression for silver chloride.
  • AgCl (s) ------> Ag+ (aq) + Cl- (aq)
  • x x
slide15
Ksp = [Ag + (aq) ] [Cl - (aq)]
  • 1.8 x 10 -8 = x2
  • 1.3 x 10-4 = x
slide16
The Ksp value for Cu(IO3)2 is

1.4 x 10-7 at 25°C. Calculate its solubility at this temperature.

Cu(IO3)2 (s) Cu2+ (aq) + 2 IO31- (aq)

x

x

2x

slide17
Ksp = [Cu 2+] [IO31-]2
  • 1.47 x 10-7 = (x) (2x)2
  • 1.47 x 10-7 = 4x3
  • 1.47 x 10-7 = 4x3

4 4

3.3 x 10-3 = x

Solubility of Cu(IO3)2

slide18
If Ksp of calcium hydroxide is 5.5 x 10-6, find the solubility.

Ca(OH)2 (s)→ Ca+2 (aq) + 2OH-(aq)

x=0.011 M