When a solution of lead(II) nitrate is mixed with a solution of potassium iodide, a yellow precipitate is formed.
(i) Name the yellow precipitate formed.
(ii) Write the balanced chemical equation for this reaction with state symbols.
(iii) What type of reaction is this?
Generated by claude-sonnet-4-6 · 2026-06-26 00:54 · grounding rag
Model Answer
(i) The yellow precipitate formed is lead(II) iodide (PbI₂).
(ii) Balanced chemical equation with state symbols:
$$\text{Pb(NO}_3)_2\text{(aq)} + 2\text{KI(aq)} \rightarrow \text{PbI}_2\text{(s)} + 2\text{KNO}_3\text{(aq)}$$
(iii) This is a double displacement reaction (also called a precipitation reaction), because the ions of the two reactants exchange places and an insoluble precipitate (PbI₂) is formed.
Source: Chapter 1, Section 1.2.4 Double Displacement Reaction
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Explanation
- (i) Just naming PbI₂ with its correct IUPAC name earns the mark.
- (ii) Examiners check: correct formulae, coefficient 2 before KI and KNO₃, and all four state symbols — (aq), (aq), (s) for the precipitate, (aq). Missing state symbols or unbalanced equation loses marks.
- (iii) Both terms ("double displacement" and "precipitation") are acceptable; mentioning both is safer. The key reasoning is ion exchange + precipitate formation. The textbook explicitly identifies this lead iodide reaction as a double displacement/precipitation reaction.