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What would you observe if you add excess dil. NaOH solution to $ZnC{{l}_{2}}$ solution?
A. A white ppt
B. A white ppt which later dissolves
C. A green ppt
D. A green ppt which later dissolves

Last updated date: 24th Jun 2024
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Hint: Sodium hydroxide is basic in nature. It reacts with metal chlorides and forms metal hydroxides as the products. We can detect the formed metal hydroxide in the form of precipitate. By adding an excess amount of sodium hydroxide the formed metal hydroxide precipitate is going to soluble.

Complete step by step solution:
- In the question it is given what we are going to observe when excess dil. NaOH solution is added to $ZnC{{l}_{2}}$ solution.
- The molecular formula of sodium hydroxide is NaOH.
- We know that dilute sodium hydroxide reacts with zinc chloride and forms zinc hydroxide in the form of a white precipitate.
- The chemical reaction of sodium hydroxide with zinc chloride is as follows.
\[2NaOH+ZnC{{l}_{2}}\to \underset{White\text{ }ppt}{\mathop{Zn{{(OH)}_{2}}}}\,\downarrow +2NaCl\]

- In the above reaction two moles of sodium hydroxide reacts with one moles of zinc chloride and forms one mole of zinc hydroxide (white ppt) and two moles of sodium chloride as the products.
- But in the question it is given that an excess amount of dilute sodium hydroxide is going to react with zinc chloride.
- Means when we are going to add more sodium hydroxide we can observe the following chemical reaction.
\[2NaOH+ZnC{{l}_{2}}\to \underset{White\text{ }ppt}{\mathop{Zn{{(OH)}_{2}}}}\,\downarrow +2NaCl\xrightarrow{2NaOH}\underset{sodium\text{ }zincate}{\mathop{N{{a}_{2}}Zn{{O}_{2}}}}\,+2{{H}_{2}}O\]

- Means the formed zinc hydroxide is going to convert into sodium zincate which dissolves in water.
- Therefore we observed the formation of a white ppt and later it dissolves when excess sodium hydroxide is added to zinc chloride.
So, the correct answer is “Option B”.