# What halide salts are insoluble?

Aug 27, 2017

Here's what I got.

#### Explanation:

Start by writing the chemical formulas of the reactants and of the products.

• "lead(II) nitrate " -> " Pb"("NO"_3)_2
• $\text{potassium iodide " -> " KI}$
• ${\text{lead(II) iodide " -> " PbI}}_{2}$
• ${\text{potassium nitrate " -> " KNO}}_{3}$

Now, your reaction takes place in aqueous solutions. Three of the four chemical species involved in the reaction will be in the aqueous state, $\text{(aq)}$, and one, potassium iodide, will be in the solid state, $\text{(s)}$.

The unbalanced chemical equation looks like this

${\text{Pb"("NO"_ 3)_ (2(aq)) + "KI"_ ((aq)) -> "PbI"_ (2(s)) darr + "KNO}}_{3 \left(a q\right)}$

To balance this chemical equation, multiply the potassium iodide by $2$ and the potassium nitrate by $2$.

You will end up with

${\text{Pb"("NO"_ 3)_ (2(aq)) + 2"KI"_ ((aq)) -> "PbI"_ (2(s)) darr + 2"KNO}}_{3 \left(a q\right)}$

If you want, you can write the complete ionic equation by breaking up the soluble compounds, i.e. the ones that are in the aqueous state, into their respective cations and anions

${\text{Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"K"_ ((aq))^(+) + 2"I"_ ((aq))^(-) -> "PbI"_ (2(s)) darr + 2"K"_ ((aq))^(+) + 2"NO}}_{3 \left(a q\right)}^{-}$

If you eliminate the spectator ions, i.e. the ions that are present on both sides of the equation

"Pb"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(2"K"_ ((aq))^(+)))) + 2"I"_ ((aq))^(-) -> "PbI"_ (2(s)) darr + color(red)(cancel(color(black)(2"K"_ ((aq))^(+)))) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))

you will get the net ionic equation

${\text{Pb"_ ((aq))^(2+) + 2"I"_ ((aq))^(-) -> "PbI}}_{2 \left(s\right)} \downarrow$

Lead(II) iodide is a yellow insoluble solid that precipitates out of the solution. Aug 27, 2017

This illustrates a general phenomenon of aqueous solubility......

#### Explanation:

All NITRATES are soluble; and ALL halides are soluble EXCEPT for those of $P {b}^{2 +}$, $H {g}_{2}^{2 +}$, $\text{mercurous ion}$, and $A {g}^{+}$ in aqueous solution. This is an experimental phenomenon that simply must be known at A level and higher.

And we must able to reproduce the reactions.....

$P {b}^{2 +} + 2 {X}^{-} \rightarrow P b {X}_{2} \left(s\right) \downarrow$

$H {g}_{2}^{2 +} + 2 {X}^{-} \rightarrow H {g}_{2} {X}_{2} \left(s\right) \downarrow$

$A {g}^{+} + {X}^{-} \rightarrow A g X \left(s\right) \downarrow$

In all cases both charge and mass are conserved, as they must be if we are to represent chemical reality.

In the case of the silver halides, in particular, this reaction gives silver chloride as a very curdy white solid from aqueous solution; silver bromide as a cream solid; and silver iodide as a yellow solid.