Question

pper(ii) hydroxide and sodium phosphate undergo a double replacement reaction. predict the formulae of the two products of the reaction.
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cu₂po₄ and naoh
cu₃po₄ and na(oh)₂
cu₂na and po₄oh
cuna₂ and po₄(oh)₃
cu₃(po₄)₂ and naoh

Explanation:

Step1: Identify reactants' formulas

Copper(II) hydroxide: $\ce{Cu(OH)2}$ (Cu has +2 charge, $\ce{OH^-}$ has -1 charge, so 1 Cu²⁺ and 2 $\ce{OH^-}$). Sodium phosphate: $\ce{Na3PO4}$ (Na⁺ has +1 charge, $\ce{PO4^{3-}}$ has -3 charge, so 3 Na⁺ and 1 $\ce{PO4^{3-}}$).

Step2: Apply double - replacement (swap cations/anions)

In double - replacement, cations ($\ce{Cu^{2+}}$ and $\ce{Na^+}$) and anions ($\ce{OH^-}$ and $\ce{PO4^{3-}}$) swap. So:

• For the first product: $\ce{Cu^{2+}}$ combines with $\ce{PO4^{3-}}$. To balance charges: 3 $\ce{Cu^{2+}}$ (total +6) and 2 $\ce{PO4^{3-}}$ (total -6) → $\ce{Cu3(PO4)2}$.
• For the second product: $\ce{Na^+}$ combines with $\ce{OH^-}$. $\ce{Na^+}$ (+1) and $\ce{OH^-}$ (-1) → $\ce{NaOH}$.

Answer:

E. $\ce{Cu3(PO4)2}$ and $\ce{NaOH}$ (assuming the last option is E, as per the options: the last option is $\ce{Cu3(PO4)2}$ and $\ce{NaOH}$)