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Chapter 6: Problem 68

Bubbling \(\mathrm{CO}_{2}\) through which of the following will produce a white precipitate : (1) \(\mathrm{NaAlO}_{2}\) (2) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) (3) \(\mathrm{NaOH}\) (4) \(\mathrm{Mg} \mathrm{SO}_{4}\)

Short Answer

Expert verified
Bubbling \(\mathrm{CO}_{2}\) through \(\mathrm{NaAlO}_{2}\) produces a white precipitate.

Step by step solution

01

Analyzing \(\mathrm{CO}_{2}\) with \(\mathrm{NaAlO}_{2}\).

When \(\mathrm{CO}_{2}\) is passed through \(\mathrm{NaAlO}_{2}\) (sodium aluminate) solution, \(\mathrm{Al(OH)}_{3}\) (aluminum hydroxide) precipitate is formed, which is white in color. The reaction is: \[ \mathrm{2NaAlO_{2} + 3CO_{2} + 3H_{2}O \rightarrow 2Al(OH)_{3} + 3NaHCO_{3}} \].
02

Analyzing \(\mathrm{CO}_{2}\) with \(\mathrm{Na}_{2} \mathrm{CO}_{3}\).

When \(\mathrm{CO}_{2}\) is passed through \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) (sodium carbonate) solution, \(\mathrm{NaHCO}_{3}\) (sodium bicarbonate) is formed, which is soluble in water. Therefore, no precipitate is formed. The reaction is: \[ \mathrm{Na_{2}CO_{3} + CO_{2} + H_{2}O \rightarrow 2NaHCO_{3}} \].
03

Analyzing \(\mathrm{CO}_{2}\) with \(\mathrm{NaOH}\).

When \(\mathrm{CO}_{2}\) is passed through \(\mathrm{NaOH}\) (sodium hydroxide) solution, \(\mathrm{Na_{2}CO_{3}}\) or \(\mathrm{NaHCO_{3}}\) is formed depending on the concentration. Both compounds are soluble in water and thus, no precipitate is formed. The reactions are: \[ \mathrm{CO_{2} + 2NaOH \rightarrow Na_{2}CO_{3} + H_{2}O} \] and \[ \mathrm{CO_{2} + NaOH \rightarrow NaHCO_{3}} \].
04

Analyzing \(\mathrm{CO}_{2}\) with \(\mathrm{MgSO_{4}}\)

When \(\mathrm{CO}_{2}\) is passed through \(\mathrm{MgSO_{4}}\) (magnesium sulfate) solution, no precipitation occurs. There is no reaction between them that leads to the formation of a precipitate.
05

Conclusion: Identifying the correct solution

Based on the reactions analyzed above, \(\mathrm{CO}_{2}\) passing through \(\mathrm{NaAlO}_{2}\) is the only reaction that produces a white precipitate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

sodium aluminate reaction with CO2
When \({CO}_{2}\) interacts with \(\text{NaAlO}_{2}\) (sodium aluminate), a white precipitate forms. This is because \({CO}_{2}\) reacts with \(\text{NaAlO}_{2}\) in the presence of water to produce aluminum hydroxide \((\text{Al(OH)}_{3})\), which is a white solid.
The reaction can be written as: \[ \mathrm{2NaAlO_{2} + 3CO_{2} + 3H_{2}O \rightarrow 2Al(OH)_{3} + 3NaHCO_{3}} \]
Summary of process:
  • Aluminum hydroxide \(\text{Al(OH)}_{3}\) is insoluble and appears as a white precipitate.
  • New products formed are aluminum hydroxide and sodium bicarbonate \(\text{NaHCO}_{3}\), both white in color.
This means when \(\text{CO}_{2}\) is bubbled through a sodium aluminate solution, the clear reaction product is a white precipitate, making this reaction unique.
sodium carbonate reaction with CO2
When you bubble \({CO}_{2}\) through \(\text{Na}_{2} \text{CO}_{3}\) (sodium carbonate) solution, no precipitate forms. Instead, sodium bicarbonate \(\text{NaHCO}_{3}\) is produced. Sodium bicarbonate is soluble in water.
The reaction involved is: \[ \text{Na}_{2} \text{CO}_{3} + \text{CO}_{2} + \text{H}_{2} \text{O} \rightarrow 2\text{NaHCO}_{3} \]
Key points about the reaction:
  • Products: Sodium bicarbonate \(\text{NaHCO}_{3}\) forms which is a soluble compound
  • No physical change in the solution as sodium bicarbonate dissolves completely in water.
Therefore, the reaction of \({CO}_{2}\) with sodium carbonate does not produce a precipitate, making the solution clear.
sodium hydroxide reaction with CO2
When \({CO}_{2}\) comes into contact with \(\text{NaOH}\) (sodium hydroxide), different compounds form based on the concentration of sodium hydroxide. Both sodium carbonate \(\text{Na}_{2} \text{CO}_{3}\) and sodium bicarbonate \(\text{NaHCO}_{3}\) can be produced, and both are soluble in water.
The reactions sometimes differ:
  • For sodium carbonate: \[ \text{CO}_{2} + 2\text{NaOH} \rightarrow \text{Na}_{2} \text{CO}_{3} + \text{H}_{2} \text{O} \]
  • For sodium bicarbonate: \[ \text{CO}_{2} + \text{NaOH} \rightarrow \text{NaHCO}_{3} \]
Important things to know:
  • No precipitate formation as both sodium carbonate \(\text{Na}_{2} \text{CO}_{3}\) and sodium bicarbonate \(\text{NaHCO}_{3}\) dissolve in water.
  • Solutions remain clear.
This means passing \({CO}_{2}\) through sodium hydroxide will not produce any white precipitate either.
magnesium sulfate reaction with CO2
If you bubble \({CO}_{2}\) through \(\text{MgSO}_{4}\) (magnesium sulfate), no white precipitate forms. There's no significant interaction between carbon dioxide and magnesium sulfate in solution,
No reaction occurs as shown:
  • Magnesium sulfate remains stable
  • No visible changes or new product formation
  • Solution remains clear
This clearly indicates that among the given compounds, \(\text{NaAlO}_{2}\) is the only one that reacts with \({CO}_{2}\) to produce a white precipitate.

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