Question

How many of the following reactions will form solid precipitates?

Cu(NO₃)₂ + 2HCl (aq) ⟶ CuCl₂ (?) + 2HNO₃ (?)
AgNO₃ (aq) + HCl (aq) ⟶ AgCl (?) + HNO₃ (?)
Fe(OH)₃ (s) + 3HCl (aq) ⟶ FeCl₃ (?) + 3 H₂O (?)
Cu(NO₃)₂ (aq) + 2 NaOH (aq) ⟶ Cu(OH)₂ (?) + 2NaNO₃ (?)
Cu(OH)₂ (s) + 4 NH₄OH (aq) ⟶ Cu(NH₃)₄(OH)₂ (?) + 4 H₂O (?)
a) 1
b) 2
c) 3
d) 4

Answer 1

Solid precipitates will form in 2 of the given reactions: when silver nitrate reacts with hydrochloric acid to form silver chloride, and when copper(II) nitrate reacts with sodium hydroxide to form copper(II) hydroxide.

Step-by-step explanation:

To determine how many of the listed reactions will form solid precipitates, let's analyze each one individually based on the solubility rules of ionic compounds in water.

Cu(NO₃)₂ + 2HCl (aq) → CuCl₂ (?) + 2HNO₃ (?) - Copper(II) chloride is soluble in water; no precipitate forms.

AgNO₃ (aq) + HCl (aq) → AgCl (?) + HNO₃ (?) - Silver chloride is insoluble and will precipitate.

Fe(OH)₃ (s) + 3HCl (aq) → FeCl₃ (?) + 3 H₂O (?) - There is no precipitate because Fe(OH)₃ is already a solid and FeCl₃ is soluble.

Cu(NO₃)₂ (aq) + 2 NaOH (aq) → Cu(OH)₂ (?) + 2NaNO₃ (?) - Copper(II) hydroxide is insoluble, hence it will form a precipitate.

Cu(OH)₂ (s) + 4 NH₄OH (aq) → Cu(NH₃)₃4(OH)₂ (?) + 4 H₂O (?) - The complex ion [Cu(NH₃)₃4]^(2+) is soluble in water, so no precipitate will form.

Therefore, solid precipitates will form in 2 of the given reactions: the second and the fourth, making the answer (b) 2.