WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Chemistry of the P-Block
Group 4 oxides and chlorides, and lead in solution: distinguishing lead from silver ions
- 3 things to know
- A2 Unit 3
Overview
Eleven recall questions on group 4 oxides and chlorides, and lead in solution: distinguishing lead from silver ions.
What you need to know
Answer each one before you reveal it.
A student adds sodium hydroxide solution slowly to a lead(II) salt solution and reports no precipitate at all. Suggest what went wrong.
Both the chloride and the iodide of lead(II) are insoluble. Why does the colour of the precipitate differ so markedly?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Describe what happens when the yellow precipitate formed from lead(II) ions and iodide ions is heated with water and then allowed to cool slowly.
Model answer
The precipitate dissolves in the hot water to give a colourless solution, and on slow cooling lead(II) iodide crystallises out as glittering golden-yellow plates. The effect is often called a shower of gold.
- 2
Explain how you could use sodium hydroxide solution and potassium iodide solution together to confirm the presence of lead(II) ions in a colourless solution.
Model answer
Sodium hydroxide gives a white precipitate that dissolves in excess to a colourless solution, showing an amphoteric hydroxide. Potassium iodide then gives a bright yellow precipitate, which among the common cations is characteristic of lead(II). The two results together confirm Pb²⁺.
- 3
Lead(II) chloride dissolves when concentrated hydrochloric acid is added. Explain why, and name the species formed.
Model answer
A high concentration of chloride ions converts the insoluble chloride into a soluble complex ion, PbCl₂(s) + 2Cl⁻(aq) → [PbCl₄]²⁻(aq). The species formed is the tetrachloroplumbate(II) ion, and the solution becomes colourless and clear.
- 4
Why should chloride ions be avoided when preparing a solution of a lead(II) salt?
Model answer
Lead(II) chloride is only sparingly soluble in cold water, so chloride ions precipitate the lead out of solution as a white solid. A nitrate, which is soluble, should be used instead.
- 5
Silver ions also give a white precipitate with chloride ions and a yellow precipitate with iodide ions. How could you distinguish silver(I) from lead(II) using aqueous ammonia?
Model answer
Silver chloride dissolves in dilute aqueous ammonia to give a colourless solution of the diamminesilver(I) ion, whereas lead(II) chloride does not dissolve in ammonia. The precipitate that dissolves therefore contains silver.
- 6
A student adds sodium hydroxide solution slowly to a lead(II) salt solution and reports no precipitate at all. Suggest what went wrong.
Model answer
Probably too much alkali was added in total before looking, or the mixture was not shaken, so the lead(II) hydroxide redissolved in the excess to give the colourless plumbate(II) ion. The alkali must be added dropwise with shaking, watching after each drop for the white precipitate.
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