Question

The Ksp of Fe(OH)2 is 1.8 x 10–15. At what pH will a 0.50 M Fe2+ solution begin to show precipitation of Fe(OH)2?

Rank the molar solubility of the following substances in an increasing order.

Drag and drop to order

1

A

BaF2 (Ksp=2.4 x 10−5)

2

B

CaF2 (Ksp=4.0 x 10−11)

3

C

PbF2 (Ksp=4 x 10−8)

Answer 1

Fe(OH)_{2 (s)} Fe²⁺ (aq) + 2OH^- (aq)

K_sp=[Fe²⁺] [OH^- ]² = 1.8*10^-15 and Ka = [Fe²⁺] [OH^- ]²

So , we can use Ksp vale as Ka value.

Now P^H = 1/2 (p^Ka - logC) = 1/2 [ -log(1.8*10^-15) - log 0.5 ] = 7.52

You can see that BaF₂ , CaF₂ , PbF₂ all dissociate in same way.

MF₂(s) ⇌ M²⁺(aq) + 2F¯(aq)

K_sp = [M²⁺] [F¯]²

If molar solubility is x then, K_sp = (x_i) (x_i)² = x³ x_i =

So, The molecule having Ksp greater will have greater molar solubility.

Hence order is : CaF₂ < PbF₂ < BaF₂