# Lead nitrate solution and potassium iodide solution when react with each other what precipitate is formed?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

53
Jun 19, 2017

#### Explanation:

Assuming that you're dealing with lead(II) nitrate, "Pb"("NO"_3)_2, you can say that this reaction will produce aqueous potassium nitrate and lead(II) iodide, an insoluble solid that precipitates out of solution.

The balanced chemical equation that describes this double replacement reaction looks like this

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

The complete ionic equation looks like this

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

The net ionic equation, which doesn't include 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))^(-))))

will look like this

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

You can thus say that the reaction will produce lead(II) iodide, a yellow insoluble solid that will precipitate out of the solution.

• 16 minutes ago
• 20 minutes ago
• 25 minutes ago
• 26 minutes ago
• A minute ago
• 8 minutes ago
• 13 minutes ago
• 14 minutes ago
• 15 minutes ago
• 15 minutes ago
• 16 minutes ago
• 20 minutes ago
• 25 minutes ago
• 26 minutes ago