Which of the following is true when lead nitrate is treated with sodium hydroxide solution, and ppt. obtained is dissolved in excess of alkali ?

Options

A. White ppt. soluble in excess of alkali
B. White ppt, insoluble in excess of alkali
C. Green ppt. soluble in excess of alkali
D. Green ppt. insoluble in excess of alkali

Explanation

To solve the question regarding the reaction of lead nitrate with sodium hydroxide and the subsequent behavior of the precipitate, we will follow these steps:

### Step 1: Write the Chemical Reaction
When lead nitrate (Pb(NO3)2) is treated with sodium hydroxide (NaOH), a double displacement reaction occurs. The lead ions (Pb²⁺) react with hydroxide ions (OH⁻) to form lead hydroxide (Pb(OH)2), which is a white precipitate.

**Chemical Equation:**
\[ \text{Pb(NO}_3\text{)}_2 + 2 \text{NaOH} \rightarrow \text{Pb(OH)}_2 \downarrow + 2 \text{NaNO}_3 \]

### Step 2: Identify the Precipitate
The precipitate formed in this reaction is lead hydroxide (Pb(OH)2), which is white in color.

### Step 3: Dissolve the Precipitate in Excess Alkali
When excess sodium hydroxide is added to the white precipitate of lead hydroxide, it dissolves to form a complex ion known as sodium plumbite (Na2Pb(OH)4).

**Chemical Equation for Dissolution:**
\[ \text{Pb(OH)}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{Pb(OH)}_4 \]

### Step 4: Resulting Solution
The reaction results in a clear solution of sodium plumbite, indicating that the white precipitate has dissolved.

### Conclusion
The correct statement regarding the reaction is that the white precipitate (Pb(OH)2) is soluble in excess sodium hydroxide, leading to the formation of a clear solution.

### Final Answer
The correct option is: **A. A white precipitate, soluble in excess alkali.**

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