# What is Pb(IV) + NiCl_2 ->?

Jul 26, 2017

${\text{Pb"^(4+)(aq) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + "Cl}}_{2} \left(g\right)$

This is a "simplified" double-replacement reaction (i.e. we ignore some anion coupled with the ${\text{Pb}}^{4 +}$), so we initially get---noting that $I V = 4$:

${\text{Pb"^(4+)(aq) + "NiCl"_2(aq) -> "PbCl"_4(s) + "Ni}}^{2 +} \left(a q\right)$

The ${\text{NiCl}}_{2} \left(a q\right)$ is rather deliquescent (i.e. hygroscopic), and sops up surrounding moisture to form an aqueous solution.

${\text{PbCl}}_{4} \left(s\right)$ is going to be written as a solid for now, but it decomposes in aqueous solution into ${\text{PbCl}}_{2} \left(s\right)$ and ${\text{Cl}}_{2} \left(g\right)$. ${\text{Ni}}^{2 +}$ is evidently going to form, but to balance the charge, we must rebalance the mass.

${\text{Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> "PbCl"_4(s) + 2"Ni}}^{2 +} \left(a q\right)$

And now, we incorporate the decomposition.

${\text{Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> cancel("PbCl"_4(s)) + 2"Ni}}^{2 +} \left(a q\right)$

cancel("PbCl"_4(s)) stackrel("H"_2"O"" ")(->) "PbCl"_2(s) + "Cl"_2(g)

$\text{-----------------------------------------------------------------}$

${\text{Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> "PbCl"_2(s) + 2"Ni"^(2+)(aq) + "Cl}}_{2} \left(g\right)$

We expect the nickel(II) chloride to dissociate, and thus we are not done:

${\text{Pb"^(4+)(aq) + cancel(2"Ni"^(2+)(aq)) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + cancel(2"Ni"^(2+)(aq)) + "Cl}}_{2} \left(g\right)$

This gives us:

$\textcolor{b l u e}{{\text{Pb"^(4+)(aq) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + "Cl}}_{2} \left(g\right)}$