Question

$PbC{l}_2$ is insoluble in cold water. Addition of $HCl$ increases its solubility due to:

• A. formation of soluble complex anions like $[PbC{l}_3{]}^{-}$
• B. oxidation of Pb(II) to Pb(IV)
• C. formation of $\left[Pb\right(H_{2}O{)}_6{]}^{2+}$
• D. formation of polymeric lead complexes

Answer 1

The correct option is A formation of soluble complex anions like $[PbC{l}_3{]}^{-}$

$PbC{l}_2\left(s\right)+C{l}^{\ominus }\to \left[PbC{l}_3{]}^{\ominus }\right(aq)$

$PbC{l}_2\left(s\right)+2C{l}^{\ominus }\to \left[PbC{l}_4{]}^{2-}\right(aq)$

The addition of chloride ions to a suspension of $PbC{l}_2$ gives rises to soluble complex ion. In these reactions, the additional chlorides break up the chloride bridges that comprise the polymeric framework of solid $PbC{l}_2\left(s\right)$