The Sediment of Lead (II) Nitrate

The Sediment of Lead (II) Nitrate

Today, I’m going to do a common experiment about the sediment of Lead (II) Nitrate. This is a very quick demonstration showing that two solids can react together. White lead nitrate and white potassium iodide react to make yellow lead iodide.

I added 5 grams of each chemical into 95ml  of water so I could have 5 % of each.

I pour 10ml of potassium iodide solution into each test tube. And poured 3ml of lead (II) nitrate into the first test tube. Poured 6ml of lead (II)nitrate into the second test tube. And poured 9ml of lead (II) nitrate into the third test tube.

As you could see on the picture the more lead (II) nitrate I add to the potassium iodide, more sediment increases in the test tube.

The demonstration might have more impact if the test tubes are opaque and the yellow product can be poured out and shown to the unsuspecting audience. Have a white background available.

Point out that for a reaction to occur, particles of the reactants must meet. This is much easier in solution (where the particles are free to move) than in the solid state.

The reaction is:

Pb(NO₃)₂(s) + 2KI(s) →  2KNO₃(s) + PbI₂(s)

All of these compounds are white except lead iodide, which is yellow.

Lead ethanoate can be substituted for lead nitrate, but the reaction is much slower.

The experiment  Diffusion in liquids is a class practical using the same compounds but as solutions.

We must first convert from a word equation to a symbol equation:

$\text{Lead (II) Nitrate + Potassium Iodide}\to \text{Lead (II) Iodide + Potassium Nitrate}$

The lead (II) ion is represented as ${\text{Pb}}^{2+}$, whilst the nitrate ion is ${\text{NO}}_3^{-}$. To balance the charges, we require two nitrate ions per lead (II) ion, and so lead (II) nitrate is ${\text{Pb(NO}}_3{\text{)}}_2$ .

The potassium ion is ${\text{K}}^{+}$ and the iodide ion is ${\text{I}}^{-}$. The two charges balance in a $1:1$ratio, so potassium iodide is simply $\text{KI}$.

In lead (II) iodide, the charges balance in a $1:2$ ratio, so the formula is ${\text{PbI}}_2$.

Finally, in potassium nitrate, the charges balance in another $1:1$ ratio, giving a formula of ${\text{KNO}}_3$ .

The symbol equation is as follows:

${\text{Pb(NO}}_3{\text{)}}_2+\text{KI}\to {\text{PbI}}_2+{\text{KNO}}_3$

The most obvious change we must make, when balancing this equation, is to increase the number of nitrate ions on the right hand side of the equation. We can do this by placing a coefficient of $2$ before the potassium nitrate:

${\text{Pb(NO}}_3{\text{)}}_2+\text{KI}\to {\text{PbI}}_2+2{\text{KNO}}_3$

In doing this we have upset the balance of potassium ions on each side of the equation. Again, we can fix this: we must simply place another coefficient of $2$, this time before the potassium iodide:

${\text{Pb(NO}}_3{\text{)}}_2+2\text{KI}\to {\text{PbI}}_2+2{\text{KNO}}_3$

Checking over the equations once more, you will notice that we initially had 1 iodide ion on the right hand side, but 2 on the left. However, we already dealt with this in balancing our potassium ions. Now, our equation is balanced.

And that’s it! One last thing to add is that you may have noticed the irregularity in iodide ions rather than nitrate ions. In this case, you would have arrived at the same answer simply by working backwards.

Source: https://socratic.org/questions/how-do-you-write-the-the-reaction-of-lead-ii-nitrate-aq-with-sodium-iodide-aq-to