Testing for Lead ion | Qualitative Analysis of Pb²⁺| Lead(II)

Lead is an amphoteric metal and its oxides and hydroxides are also amphoteric compounds. In qualitative analysis of inorganic chemistry, there are tests to identify Pb²⁺ ion from other cations. Lead ion forms soluble compounds and insoluble compounds in water and some of them have colours. PbCO₃, PbSO₄ are insoluble white precipitates

In this tutorial, we will learn followings of lead ion.

1. Tests for Pb⁺² ion under qualitative analysis reactions and physical observations
2. Problems of identifying Lead(II) ion from other cations

Lead as a metal

Lead is located at p block. Lead does not react with water. But reacts with dilute acids and emit hydrogen gas. Lead is an amphoteric element.

Tests for Lead +2 ion - Qualitative Analysis

Following tests are done to identify Zn²⁺ ion and they are explained in detail in this tutorial.

• Gold dust experiment
• With aqueous NaOH
• With sulfuric acid or aqueous sulfate solution
• Dilute HCl and Pb²⁺ reaction
• Addition of K₂CrO₄

Gold dust experiment | With aqueous KI solution

Treatment of Pb²⁺ aqueous solution with aqueous KI solution, will give a dark yellow precipitate (PbI₂).

Pb²⁺_(aq) + KI_(aq) → PbI_2(s) + K⁺

This dark yellow precipitate dissolve on boiling and form colourless solution. This mixture recrystallizes upon cooling. This experiment is known as gold dust experiment.

Addition of aqueous NaOH

Addition of aqueous NaOH solution to aqueous Pb₂₊ ion solution, will give an insoluble hydroxide which is a white precipitate. This white precipitate is Lead(II) hydroxide ( Pb(OH)₂ ).

Pb²⁺_(aq) + NaOH_(aq) → Pb(OH)_2(s) + Na⁺_(aq)

This white precipitate is soluble in excess NaOH solution and give a colourless solution. Here, Lead(II) hydroxide will convert to complex ion, [Pb(OH)₄]^2-

Pb(OH)_2(aq) + NaOH_(aq) → Na₂[Pb(OH)₄]_(aq)

Addition of sulfuric acid or aqueous sulfate ion solution

Add dilute sulfuric acid or aqueous sulfate ion solution to Pb²⁺ ion solution. It will give Lead(II) sulfate (PbSO₄) which is deposited as a white precipitate in the water.

Dilute sulfuric acid completely dissociates and release sulfate ion in the water.

Pb²⁺_(aq) + H₂SO_4(aq) → PbSO_4(s) + H⁺_(aq)

Pb²⁺_(aq) + Na₂SO_4(aq) → PbSO_4(s) + Na⁺_(aq)

PbSO₄ is not soluble in hot water.

Addition of dilute HCl to aqueous Pb²⁺ ion solution

When dilute HCl s added to aqueous Pb²⁺ solution, a white colour precipitate is given. This white precipitate is Lead(II) chloride (PbCl₂).

Pb²⁺_(aq) + HCl_(aq) → PbCl_2(s) + H⁺_(aq)

PbCl₂ is insoluble in NaOH. But, this precipitate is soluble in hot water. On cooling, again separate out as needles.

Addition of potassium dichromate (K₂CrO₄)

Addition of potassium dichromate solution into aqueous Pb²⁺ solution, forms Lead(II) chromate (PbCrO₄) which is a yellow colour precipitate.

Pb²⁺_(aq) +K₂CrO₄(aq) → PbCrO_4(s) + K⁺_(aq)

Lead(II) chromate is insoluble in acetic acid or aqueous ammonium hydroxide solution. But it is soluble in NaOH or HNO₃.

Lead(II) chromate turns orange on heating.

Summary of testing Lead(II) ion

Precipitates of lead(II) ion and colours

• Lead(II) carbonate - white
• Lead(II) sulfate - white
• Lead(II) chloride - white
• Lead(II) iodide - dark yellow

Questions asked by students

How do you identify silver chloride and lead chloride?

First of all, I would like to tell it is good to study testing for silver ion occurrence.

Ok. Lets find out how we identify these two compounds.

Silver chloride and lead chloride are white precipitates. So we cannot identify them by dissolving them in water or observing colour. Now, add AgCl and PbCl₂ to water separately. You can see two white precipitates are deposited at bottom of the solutions.

Now add an excess of aqueous ammonia solution to both solutions which contains precipitates. You can see one precipitate dissolve and form colourless solution. But other white precipitate remains unchanged.

Silver chloride precipitate dissolves in excess ammonia solution and form colourless solution.