202-NYB-05 Chapter 16 - Lecture

3/27/23




Chapter 16
202-NYB-05




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include
wedont
thereactant
solubility product K id
Dissolving a solid in water:
CaF2(s) H 2O Ca2+(aq) + 2F-(aq)

Ca2+(aq) + 2F-(aq) CaF2(s)

CaF2(s) Ca2+(aq) + 2F-(aq) equilibrium

Solubility Product Ksp = [Ca2+][F-]2



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, 3/27/23




Ksp punit cuzconstant
Solubility:
is an equilibrium position. Units usually mol/L. Will
change if a common ion is present.
Solubility product Ksp

µ is a constant at any one temperature.

ZnCO 3 has a solubility of 1.4x10-5 mol/L at 25˚C, what
is its Ksp value?
ZnCO 3(s) Zn2+(aq) + CO 32-(aq)

Ksp = [Zn2+][CO 32-] = (1.4x10-5)2 = 2.0x10-10
o o

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I x xx
x x
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Ksp x




What is the Ksp of Sr3(PO 4)2 if it has a solubility of
2.5x10-7 mol/L at 25˚C?
Sr3(PO 4)2(s) 3Sr2+(aq) + 2PO 43-(aq)
[Initial] 0M 0M
[Change] -2.5x10 -7 M + 3 x 2.5x10 -7M + 2 x 2.5x10 -7 M
[Equil.] 7.5x10-7 M 5.0x10-7 M



Ksp = [Sr2+]3[PO 43-]2
= (7.5x10-7)3(5.0x10-7)2
= 1.1x10-31


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35dropex
10sedropinnine
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it Kspiscloseto
Interneglecttheinitialconcentration
The Ksp of Cu(OH)2 is 1.6x10-19 at 25˚C, what is its
solubility at 25˚C?
Nd
Cu(OH)2(s) Cu2+(aq) + 2OH -(aq)
[Initial] 0M 1x 10-7 M come
[Change] -x M +x M +2x M
[Equil.] xM 1x10-7 + 2x M
largerthankwthen drop
it Kspisway thechangeinthis
Ksp = [Cu2+][OH-]2 = 1.6x10-19
= x(1x10-7 + 2x)2 (assume 2x is greater than 1x10-7)
gy
=4x3 = 1.6x10-19 drop a ixio
x = 3.4x10-7 mol/L


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