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Sia

青州's Question
Chemistry
Posted 2 months ago

Question: When 6.0MNaH6.0 \mathrm{M} \mathrm{NaH} solution is added to a solution containing equimolar concentrations of Cd(NO\mathrm{C} d(\mathrm{NO}, and Zn(NO)\mathrm{Zn}(\mathrm{NO}), the precipitate that first forms contains both Zn\mathrm{Zn} and Cd\mathrm{Cd}. However,...

When 6.0MNaH6.0 \mathrm{M} \mathrm{NaH} solution is added to a solution containing equimolar concentrations of Cd(NO\mathrm{Cd}(\mathrm{NO}, and Zn(NO)\mathrm{Zn}(\mathrm{NO}), the precipitate that first forms contains both Zn\mathrm{Zn} and Cd\mathrm{Cd}.

However, upon the addition of additional base with stirring, part of the precipitate dissolves. The undissolved precipitate contains Cd\mathrm{Cd} but no Zn. Describe what happened in terms of net ionic equations.
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Answer from Sia
Posted 2 months ago
Solution
1
Identify the initial reaction: When 6.0MNaOH6.0 \, \text{M} \, \text{NaOH} is added to a solution containing equimolar concentrations of Cd(NO3)2\text{Cd(NO}_3\text{)}_2 and Zn(NO3)2\text{Zn(NO}_3\text{)}_2, both Cd(OH)2\text{Cd(OH)}_2 and Zn(OH)2\text{Zn(OH)}_2 precipitate out initially. The net ionic equations for these reactions are: Cd2++2OHCd(OH)2 \text{Cd}^{2+} + 2\text{OH}^- \rightarrow \text{Cd(OH)}_2 Zn2++2OHZn(OH)2 \text{Zn}^{2+} + 2\text{OH}^- \rightarrow \text{Zn(OH)}_2
2
Addition of more base: Upon the addition of more NaOH\text{NaOH}, Zn(OH)2\text{Zn(OH)}_2 dissolves because it forms a soluble complex ion, Zn(OH)42\text{Zn(OH)}_4^{2-}. The net ionic equation for this reaction is: Zn(OH)2+2OHZn(OH)42 \text{Zn(OH)}_2 + 2\text{OH}^- \rightarrow \text{Zn(OH)}_4^{2-} However, Cd(OH)2\text{Cd(OH)}_2 does not form a similar soluble complex and remains undissolved
3
Final precipitate: The undissolved precipitate contains only Cd(OH)2\text{Cd(OH)}_2 because Zn(OH)2\text{Zn(OH)}_2 has dissolved in the excess NaOH\text{NaOH}
Answer
The undissolved precipitate contains Cd(OH)2\text{Cd(OH)}_2 but no Zn(OH)2\text{Zn(OH)}_2.
Key Concept
Solubility of hydroxides in excess base
Explanation
Zn(OH)2\text{Zn(OH)}_2 dissolves in excess NaOH\text{NaOH} to form Zn(OH)42\text{Zn(OH)}_4^{2-}, while Cd(OH)2\text{Cd(OH)}_2 does not form a soluble complex and remains as a precipitate.

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