Question

1A: Consider these compounds:

Answer 1

Solution :

Compound                      Ksp

$PbF_2$                               $3.3 \times 10^{-8}$

$Ni(CN)_2$                        $3 \times 10^{-23}$

FeS                                $8 \times 10^{-19}$

$CaSO_4$                           $4.93 \times 10^{-5}$

$Mg(OH)_2$                      $5.61 \times 10^{-12}$

Ksp of $Ni(CN)_2$ and both compounds dissociate the same way. Hence $Mg(OH)_2$ is more soluble than $(B). \ Ni(CN)_2$

$Mg(OH)_2$  is less soluble than $(A). \ \ PbF_2 \ ()Ksp \ PbF_2 > Ksp \ \text{ of } \ Mg(OH)_2$

It is not possible to determine CD - $FeS \text{ or} \ CaSO_4$  is more or less soluble than $Mg(OH)_2$  as though they have a different Ksp values their molecular dissociation is also different and they may have a close solubility values.

$Ni(OH)_2$  can be directly compared with PbS, $AlPO_4, MnS$

$\text{For } \ Ni(OH)_2$

$AB_2(s) \rightarrow A^{2+} + 2B^{-}$

$Ni(OH)_2(s) \rightarrow Ni^{2+} + 2OH^-$

100

1-s s 2s

Ksp = $[A2+][B-]^2 = s \times (2s)^2 = 4s^3$

Hence they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

For Silver Chloride

$AB(s) \rightarrow A^{x+}+B^{x-}$

$AgCl(s) \rightarrow Ag^+ + Cl^-$

1 0 0

1 - s s s

Ksp $=[A^{x+}][B^{x-}]=s \times s = s^2$

Hence, they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.