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Why is it common practice to boil the solution near the equivalence point in the standardization of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) with acid?

Short Answer

Expert verified
Boiling helps expel CO2 gas for accurate titration.

Step by step solution

01

Introduction to the Problem

When performing a titration of sodium carbonate (\(\mathrm{Na}_{2} \mathrm{CO}_{3}\)) with a strong acid, a common practice is to boil the solution as the titration approaches the equivalence point. This step is critical for achieving accurate results due to specific chemical processes that occur under heated conditions.
02

Understanding the Chemistry

During the titration of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) with acid, \(\mathrm{CO}_{2}\) gas is released as one of the main products around the equivalence point:\[\mathrm{Na}_{2} \mathrm{CO}_{3} + 2\mathrm{H}^{+} \rightarrow 2\mathrm{Na}^{+} + \mathrm{H}_{2}\mathrm{O} + \mathrm{CO}_{2(g)}\]This carbon dioxide needs to be expelled efficiently from the solution so as not to be mistaken for other present carbonates or trapped in the solution.
03

Effect of Boiling

Boiling the solution aids in the complete removal of \(\mathrm{CO}_{2}\) gas. Heated solutions increase the rate of gas expulsion, reducing the chance of supersaturation or false end-point readings due to retained low-solubility carbon dioxide.
04

Improvement in Titration Accuracy

By boiling and therefore removing the \(\mathrm{CO}_{2}\), the solution ensures that reformation of \(\mathrm{NaHCO}_{3}\) is minimized, maintaining clearer end-point indication and improving the overall accuracy of the standardization process.

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Key Concepts

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

Understanding Sodium Carbonate and Its Role
Sodium carbonate, commonly known as soda ash or washing soda, is an important chemical compound with the formula \( \mathrm{Na}_{2} \mathrm{CO}_{3} \). It is a white, water-soluble salt that can form various hydrates and is commonly used in various industrial and household applications, such as glass making and detergent manufacture.

In titration experiments, sodium carbonate acts as a primary standard for acid-base titrations due to its high purity and stability. This makes it an excellent choice for standardizing acidic solutions because it provides reliable and consistent results. Understanding the chemistry and behavior of sodium carbonate is crucial for achieving accurate titration results.
What is the Equivalence Point?
The equivalence point is a key concept in titration, representing the moment when the number of moles of acid equals the number of moles of base in the solution. It is the "equal" point where neutralization occurs.

When titrating sodium carbonate with a strong acid, the equivalence point signifies the complete reaction between \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) and the acid, producing water and carbon dioxide. Detecting this point accurately is crucial, as it indicates that the titration process is complete, maximizing the precision of the measurement.

In the laboratory, visual indicators or pH meters can be used to identify this point, helping to determine when the reaction has reached its completion effectively.
How Carbon Dioxide is Removed
In the titration of sodium carbonate, carbon dioxide \( (\mathrm{CO}_{2}) \) is released as a gas around the equivalence point. Efficient removal of this carbon dioxide is essential to prevent errors in the titration process.

Boiling the solution is a common method used to remove \( \mathrm{CO}_{2} \), as heating the solution increases the gas's escape rate. This step helps maintain the accuracy of the titration by reducing the possibility of supersaturation or unintended reactions, such as the formation of sodium bicarbonate \((\mathrm{NaHCO}_{3})\) which can mislead the results.

By ensuring effective carbon dioxide expulsion, the titration process is more likely to achieve accurate and reliable results.
The Basics of Acid-Base Titration
Acid-base titration is a classic laboratory method used to determine the concentration of an unknown acid or base by reacting it with a solution of known concentration. This technique involves gradually adding the titrant to the analyte until the reaction reaches the equivalence point.

In the specific case of sodium carbonate titration, the reaction involves an acid reacting with \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) to produce water and \( \mathrm{CO}_{2} \). The precise determination of the endpoint is critical to accurately calculate the concentration of the unknown substance in the sample.

Understanding the purpose and procedures involved in acid-base titration ensures that outcomes are accurate, enhancing the reliability of experimental results across various chemical investigations.

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Most popular questions from this chapter

How would you prepare \(2.00 \mathrm{~L}\) of (a) \(0.10 \mathrm{M} \mathrm{KOH}\) from the solid? (b) \(0.010 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) from the solid? (c) \(0.150 \mathrm{M} \mathrm{HCl}\) from a reagent that has a density of \(1.0579 \mathrm{~g} / \mathrm{mL}\) and is \(11.50 \% \mathrm{HCl}(\mathrm{w} / \mathrm{w})\) ?

A \(50.00-\mathrm{mL}\) sample of a white dinner wine required \(24.57 \mathrm{~mL}\) of \(0.03291 \mathrm{M} \mathrm{NaOH}\) to achieve a phenolphthalein end point. Express the acidity of the wine in grams of tartaric acid \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} ; 150.09 \mathrm{~g} / \mathrm{mol}\right)\) per \(100 \mathrm{~mL}\). (Assume that two hydrogens of the acid are titrated.)

If \(1.000 \mathrm{~L}\) of \(0.1500 \mathrm{M} \mathrm{NaOH}\) was unprotected from the air after standardization and absorbed \(11.2 \mathrm{mmol}\) of \(\mathrm{CO}_{2}\), what is its new molar concentration when it is standardized against a standard solution of HCl using (a) phenolphthalein? (b) bromocresol green?

A 0.1401-g sample of a purified carbonate was dissolved in \(50.00 \mathrm{~mL}\) of \(0.1140 \mathrm{M} \mathrm{HCl}\) and boiled to eliminate \(\mathrm{CO}_{2}\). Back-titration of the excess \(\mathrm{HCl}\) required \(24.21 \mathrm{~mL}\) of \(0.09802 \mathrm{M} \mathrm{NaOH}\). Identify the carbonate.

A 0.9471-g sample containing dimethylphthalate, \(\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{COOCH}_{3}\right)_{2}(194.19 \mathrm{~g} / \mathrm{mol})\), and unreactive species was refluxed with \(50.00 \mathrm{~mL}\) of \(0.1215 \mathrm{M}\) \(\mathrm{NaOH}\) to hydrolyze the ester groups (this process is called saponification). $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{COOCH}_{3}\right)_{2}+2 \mathrm{OH}^{-} & \rightarrow \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COO})_{2}{ }^{2-} \\ &+2 \mathrm{CH}_{3} \mathrm{OH} \end{aligned} $$ After the reaction was complete, the excess \(\mathrm{NaOH}\) was back-titrated with \(24.27 \mathrm{~mL}\) of \(0.1644 \mathrm{M} \mathrm{HCl}\). Calculate the percentage of dimethylphthalate in the sample.

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