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

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

However, upon the addition of additional base with stirring, part of the precipitate dissolves. The undissolved precipitate contains $\mathrm{Cd}$ but no Zn. Describe what happened in terms of net ionic equations.

Question

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

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

However, upon the addition of additional base with stirring, part of the precipitate dissolves. The undissolved precipitate contains $\mathrm{Cd}$ but no Zn. Describe what happened in terms of net ionic equations.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the initial reaction**: When $6.0 \, \text{M} \, \text{NaOH}$ is added to a solution containing equimolar concentrations of $\text{Cd(NO}_3\text{)}_2$ and $\text{Zn(NO}_3\text{)}_2$, both $\text{Cd(OH)}_2$ and $\text{Zn(OH)}_2$ precipitate out initially. The net ionic equations for these reactions are: $$ \text{Cd}^{2+} + 2\text{OH}^- \rightarrow \text{Cd(OH)}_2 $$ $$ \text{Zn}^{2+} + 2\text{OH}^- \rightarrow \text{Zn(OH)}_2 $$ ‖ ‖ 2 ⋮ **Addition of more base**: Upon the addition of more $\text{NaOH}$, $\text{Zn(OH)}_2$ dissolves because it forms a soluble complex ion, $\text{Zn(OH)}_4^{2-}$. The net ionic equation for this reaction is: $$ \text{Zn(OH)}_2 + 2\text{OH}^- \rightarrow \text{Zn(OH)}_4^{2-} $$ However, $\text{Cd(OH)}_2$ does not form a similar soluble complex and remains undissolved. ‖ ‖ 3 ⋮ **Final precipitate**: The undissolved precipitate contains only $\text{Cd(OH)}_2$ because $\text{Zn(OH)}_2$ has dissolved in the excess $\text{NaOH}$. ‖ ∻Answer∻ ⚹ The undissolved precipitate contains $\text{Cd(OH)}_2$ but no $\text{Zn(OH)}_2$. ⚹ ∻Key Concept∻ ⚹ Solubility of hydroxides in excess base ⚹ ∻Explanation∻ ⚹ $\text{Zn(OH)}_2$ dissolves in excess $\text{NaOH}$ to form $\text{Zn(OH)}_4^{2-}$, while $\text{Cd(OH)}_2$ does not form a soluble complex and remains as a precipitate. ⚹◊What is the balanced net ionic equation for the reaction between $Cd^{2+}$ and $OH^-$ ions?⍭ Generate me a similar question◊

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