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?

  • 2
    $\begingroup$ 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 '14 at 15:49

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.


protected by orthocresol Aug 19 '18 at 16:57

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.