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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?

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    $\begingroup$ 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? $\endgroup$
    – Jannis Andreska
    Apr 10, 2014 at 15:49

1 Answer 1

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If you don't have any idea:

  • Start with the hypothesis that your starting material is the salt of some metal.
  • 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.

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