in the laboratory you are given the task of separating pb2+ and zn2+ ions in aqueous solution.
for each reagent listed below indicate if it can be used to separate the ions. type \y\ for yes or
\ for no. if the reagent can be used to separate the ions, give the formula of the precipitate. if it cannot, type
o\

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Accepted Answer

# Explanation:
## 1. For NaCl:
- When NaCl reacts with $Pb^{2 +}$ and $Zn^{2+}$ in aqueous solution:
    - The reaction with $Pb^{2+}$ is $Pb^{2+}+2Cl^{-}=PbCl_{2}\downarrow$ (since $PbCl_{2}$ is insoluble in cold water, soluble in hot water).
    - The reaction with $Zn^{2+}$ is $Zn^{2+}+2Cl^{-}=ZnCl_{2}$ ($ZnCl_{2}$ is soluble in water). So, it can be used to separate the ions.
## 2. For KOH:
- When KOH reacts with $Pb^{2 +}$ and $Zn^{2+}$ in aqueous solution:
    - The reaction with $Pb^{2+}$ is $Pb^{2+}+2OH^{-}=Pb(OH)_{2}\downarrow$.
    - The reaction with $Zn^{2+}$ is $Zn^{2+}+2OH^{-}=Zn(OH)_{2}\downarrow$ (and $Zn(OH)_{2}+2OH^{-}=ZnO_{2}^{2 -}+2H_{2}O$ in excess KOH, but initially both form precipitates, so it cannot be used to separate them simply).
## 3. For $Na_{2}SO_{4}$:
- When $Na_{2}SO_{4}$ reacts with $Pb^{2 +}$ and $Zn^{2+}$ in aqueous solution:
    - The reaction with $Pb^{2+}$ is $Pb^{2+}+SO_{4}^{2 -}=PbSO_{4}\downarrow$.
    - The reaction with $Zn^{2+}$ is $Zn^{2+}+SO_{4}^{2 -}=ZnSO_{4}$ ($ZnSO_{4}$ is soluble in water). So, it can be used to separate the ions.

# Answer:
1. Y, $PbCl_{2}$
2. N
3. Y, $PbSO_{4}$

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QUESTION IMAGE

Question

Explanation:

1. For NaCl:

2. For KOH:

3. For \(Na_{2}SO_{4}\):

Answer: