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Sagot :
Solution :
Compound Ksp
[tex]$PbF_2$[/tex] [tex]$3.3 \times 10^{-8}$[/tex]
[tex]$Ni(CN)_2$[/tex] [tex]$3 \times 10^{-23}$[/tex]
FeS [tex]$8 \times 10^{-19}$[/tex]
[tex]$CaSO_4$[/tex] [tex]$4.93 \times 10^{-5}$[/tex]
[tex]$Mg(OH)_2$[/tex] [tex]$5.61 \times 10^{-12}$[/tex]
Ksp of [tex]$Ni(CN)_2 << Ksp \text{ of}\ \ Mg(OH)_2$[/tex] and both compounds dissociate the same way. Hence [tex]$Mg(OH)_2$[/tex] is more soluble than [tex]$(B). \ Ni(CN)_2$[/tex]
[tex]$Mg(OH)_2$[/tex] is less soluble than [tex]$(A). \ \ PbF_2 \ ()Ksp \ PbF_2 > Ksp \ \text{ of } \ Mg(OH)_2$[/tex]
It is not possible to determine CD - [tex]$FeS \text{ or} \ CaSO_4$[/tex] is more or less soluble than [tex]$Mg(OH)_2$[/tex] as though they have a different Ksp values their molecular dissociation is also different and they may have a close solubility values.
[tex]$Ni(OH)_2$[/tex] can be directly compared with PbS, [tex]$AlPO_4, MnS$[/tex]
[tex]$\text{For } \ Ni(OH)_2$[/tex]
[tex]$AB_2(s) \rightarrow A^{2+} + 2B^{-}$[/tex]
[tex]$Ni(OH)_2(s) \rightarrow Ni^{2+} + 2OH^-$[/tex]
100
1-s s 2s
Ksp = [tex][A2+][B-]^2 = s \times (2s)^2 = 4s^3[/tex]
Hence they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.
For Silver Chloride
[tex]$AB(s) \rightarrow A^{x+}+B^{x-}$[/tex]
[tex]$AgCl(s) \rightarrow Ag^+ + Cl^-$[/tex]
1 0 0
1 - s s s
Ksp [tex]$=[A^{x+}][B^{x-}]=s \times s = s^2$[/tex]
Hence, they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.
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