Question

If the anode compartment contains [so2−4]= 1.30 m in equilibrium with pbso4(s, what is the ksp of pbso4?

Answer 1

Answer:

The solubility product lead sulfate at anode is 1.69.

Explanation:

$PbSO_4\rightleftharpoons Pb^{2+}+SO_4^{2-}$

                    S      S

Concentration of sulfate ion at anode:

$[SO_4^{2-}]=S=1.30 M$

1 mole of lead sulfate dissociates into 1 mol of lead ion and 1 mole of sulfate ion.So.

$SO_4^{2-}=[Pb^{2+}]=S$

The solubility product of the lead sulfate is:

$K_{sp}=S\times S=S^2=(1.30 M)^2=1.69$

The solubility product lead sulfate at anode is 1.69.

Answer 2

The anode compartment contains [SO2-4] with a concentration of 1.30 m, this is in equilibrium with PbSO4 (s). The dissociation of PbSO4 is:

PbSO4 + H2O ---> Pb 2+ + SO4 2-

The Ksp of PbSO4:

Ksp = [Pb2+] [SO4-2]