# Identifying 5 unknown substances

I need help identifying the following 5 substances; G, Y, T, L and J

G is a white solid which is insoluble in water. Consider the following reactions carried out on solid G.

Step 1:
Excess dilute nitric acid is added to a small amount of G in a test-tube. A colourless and odourless gas Y is given off and it is noticed that Y extinguishes a lighted splint.
The solution remaining in the test tube is divided into two different test tubes, sample A and sample B.
Step 2:
A few drops of potassium iodide solution are added to a sample A and a bright yellow precipitate L is observed.
Step 3:
Sodium hydroxide solution is added drop wise to sample B. A dense white precipitate, T, is immediately observed. Precipitate T dissolves when excess sodium hydroxide solution is added to it.
Step 4:
When some G is heated in a hard glass tube, gas Y which is colourless and odourless is given off. The solid J remaining in the hard glass tube changes colour on cooling from red-brown to yellow.

These are my solutions:

• G = $$\ce{PbCO3}$$ because
• Y = $$\ce{CO2}$$ - colourless and odourless gas
• L = $$\ce{PbI2}$$
• T = $$\ce{Pb(OH)2}$$
• J = $$\ce{PbO2}$$

Are they correct or not?

• Your solutions sound good so far. But why do you think product I is $\ce{PbO2}$ and not, for example, $\ce{PbO}$? How would you explain the color change during cooling of I? – Jannis Andreska Apr 10 '14 at 15:49

If you don't have any idea:

• Some of the test are apparently useful to identify the cation, others tackle the anion.
• Is it likely that the flame-extinguishing gas comes from the cation?
• If it is not the cation, it must be the anion from which the gas is formed.
• Think in the main property of diluted nitric acid. Is it an oxidant in diluted form, or just something else?

I almost completely agree with the solution given in the question, except for the last step.

The three known lead oxides are

1. $\ce{PbO2}$: dark brown to black, strong oxidant
2. $\ce{Pb3O4}$: red-orange, used as a rust-proof primer (in German: Bleimenninge)
3. $\ce{PbO}$: (normally) yellow

When dark $\ce{PbO2}$ is heated, it splits off oxygen to form red $\ce{Pb3O4}$, when it's heated up more, more oxygen is released and yellow $\ce{PbO}$ is formed. Properly done in a quartz tube the colours of the German flag can be produced this way.

In the puzzle, we are not starting with $\ce{PbO2}$ but with $\ce{PbCO3}$.

When $\ce{PbCO3}$(= G) is heated, $\ce{CO2}$ (= Y) is released. The questions mentions a hard glass tube, which suggests that the tube has to withstand high temperatures.

Let's assume that the reaction simply is $\ce{PbCO3 ->[\Delta] PbO + CO2 ^}$

A possible explanation for the colour change would be that at high temperatures, red $\ce{\alpha-PbO}$ is initially formed, which converts to yellow $\ce{\beta-PbO}$ upon cooling.