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Chapter 9: Problem 11

Using a ladder diagram, explain why the presence of dissolved \(\mathrm{CO}_{2}\) leads to a determinate error for the standardization of \(\mathrm{NaOH}\) if the end point's \(\mathrm{pH}\) is between \(6-10\), but no determinate error if the end point's \(\mathrm{pH}\) is less than 6.

Short Answer

Expert verified
At pH 6-10, dissolved CO2 creates bicarbonate that reacts with NaOH, causing error. At pH < 6, CO2 remains mainly as gas or weak acid, causing no error.

Step by step solution

01

Understand the impact of dissolved CO2

Dissolved CO2 in water forms carbonic acid (H2CO3), which partially dissociates to bicarbonate (HCO3^-) and carbonate ions (CO3^2-). This can lower the pH of the solution.
02

Consider the standardization process at pH 6-10

In this pH range, bicarbonate ions (HCO3^-) are present in the solution. These ions react with NaOH, causing additional consumption of base, which results in erroneous determination.
03

Analyze the situation when pH is less than 6

At pH < 6, the environment is acidic enough that CO2 primarily remains as dissolved gas or weak carbonic acid. This state does not significantly interfere with the NaOH titration process.
04

Use a ladder diagram for visualization

A ladder diagram helps visualize species distribution across the pH scale. It shows CO2/H2CO3 is predominant at pH < 6, HCO3^- between pH 6-10, and CO3^2- above pH 10.

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Key Concepts

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

Ladder Diagrams
Ladder diagrams are visual tools used to help understand the pH-dependent behavior of chemical species in a solution. They offer a simplified view of how different forms of a substance exist at various pH levels. For example, when dealing with dissolved carbon dioxide
  • Below pH 6, carbonic acid (\(\mathrm{H_2CO_3}\)) dominates.
  • The pH range of 6-10 is where the bicarbonate ion (\(\mathrm{HCO_3^-}\)) is the prevalent species.
  • Above pH 10, the carbonate ion (\(\mathrm{CO_3^{2-}}\)) takes over.
Using a ladder diagram in acid-base titrations helps visualize how the presence of specific species can affect chemical reactions. In the specific case of the \(\mathrm{NaOH}\) titration, it's critical to know which species dominate because they can react with the base, influencing the amount needed to reach the endpoint. This is particularly relevant in the pH range where bicarbonate ions are prevalent.
Determinate Error
Determinate error in titrations refers to a systematic error that consistently skews results in a specific direction. During an acid-base titration, such errors can occur if substances other than the acid or base you are titrating affect the endpoint determination. For the standardization of \(\mathrm{NaOH}\) against an acid, such as when dissolved \(\mathrm{CO_2}\) is present, determinate error can occur because
  • The bicarbonate ions (\(\mathrm{HCO_3^-}\)) react with \(\mathrm{NaOH}\) in the pH range of 6-10, using up more \(\mathrm{NaOH}\) than is needed to neutralize the acid alone.
  • This results in the erroneous calculation of the base's concentration.
If the pH is less than 6, \(\mathrm{CO_2}\) primarily remains in dissolved form and does not contribute significantly to the reaction, thus not causing a determinate error in this case.
Carbonic Acid
Carbonic acid (\(\mathrm{H_2CO_3}\)) is a weak acid formed when carbon dioxide (\(\mathrm{CO_2}\)) is dissolved in water. It is an important factor in the carbon cycle and plays a role in atmospheric chemistry. When \(\mathrm{CO_2}\) dissolves in water, equilibrium is established between carbon dioxide, carbonic acid, and hydrogen carbonate. In the context of an acid-base titration, dissolved \(\mathrm{CO_2}\) forming carbonic acid can cause complications due to:
  • Its ability to lower the pH, contributing to an acidic environment.
  • The presence of \(\mathrm{H_2CO_3}\) and its subsequent dissociation into bicarbonate ions, which react with bases like \(\mathrm{NaOH}\).
Understanding carbonic acid's role is crucial to predicting and preventing titration errors especially when the pH is not significantly low, such as pH 6-10, where bicarbonate forms predominate.
Bicarbonate Ion
The bicarbonate ion (\(\mathrm{HCO_3^-}\)) is an intermediate form in the equilibrium between carbon dioxide and carbonate ion in aqueous solutions. It appears predominantly when the pH ranges between 6 and 10. In acid-base titrations, especially when standardizing \(\mathrm{NaOH}\), bicarbonate ions are especially important because:
  • They can react with \(\mathrm{NaOH}\) to form carbonate ions, consuming additional base.
  • This additional consumption leads to errors in determining the exact concentration of \(\mathrm{NaOH}\).
To minimize these issues, it's critical to understand how these ions behave across different pH levels. Recognizing when bicarbonate ions are prevalent helps in anticipating and mitigating determinate errors in titrations.

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Most popular questions from this chapter

A quantitative analysis for aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, K_{\mathrm{b}}=3.94 \times 10^{-10}\right)\) is carried out by an acid-base titration using glacial acetic acid as the solvent and \(\mathrm{HClO}_{4}\) as the titrant. A known volume of sample that contains \(3-4 \mathrm{mmol}\) of aniline is transferred to a \(250-\mathrm{mL}\) Erlenmeyer flask and diluted to approximately \(75 \mathrm{~mL}\) with glacial acetic acid. Two drops of a methyl violet indicator are added, and the solution is titrated with previously standardized \(0.1000 \mathrm{M} \mathrm{HClO}_{4}\) (prepared in glacial acetic acid using anhydrous \(\mathrm{HClO}_{4}\) ) until the end point is reached. Results are reported as parts per million aniline. (a) Explain why this titration is conducted using glacial acetic acid as the solvent instead of using water. (b) One problem with using glacial acetic acid as solvent is its relatively high coefficient of thermal expansion of \(0.11 \% /{ }^{\circ} \mathrm{C}\). For example, \(100.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) occupies \(100.22 \mathrm{~mL}\) at \(27^{\circ} \mathrm{C}\). What is the effect on the reported concentration of aniline if the standardization of \(\mathrm{HClO}_{4}\) is conducted at a temperature that is lower than that for the analysis of the unknown? (c) The procedure calls for a sample that contains \(3-4\) mmoles of aniline. Why is this requirement necessary?

Calculate or sketch the titration curve for a \(50.0 \mathrm{~mL}\) solution of a \(0.100 \mathrm{M}\) monoprotic weak acid \(\left(\mathrm{p} K_{\mathrm{a}}=8.0\right)\) with \(0.1 \mathrm{M}\) strong base in a nonaqueous solvent with \(K_{\mathrm{s}}=10^{-20}\). You may assume that the change in solvent does not affect the weak acid's \(\mathrm{p} K_{\mathrm{a}}\). Compare your titration curve to the titration curve when water is the solvent.

An acid-base titration can be used to determine an analyte's equivalent weight, but it can not be used to determine its formula weight. Explain why.

Before the introduction of EDTA most complexation titrations used \(\mathrm{Ag}^{+}\) or \(\mathrm{CN}^{-}\) as the titrant. The analysis for \(\mathrm{Cd}^{2+},\) for example, was accomplished indirectly by adding an excess of \(\mathrm{KCN}\) to form \(\mathrm{Cd}(\mathrm{CN})_{4}^{2-}\), and back titrating the excess \(\mathrm{CN}^{-}\) with \(\mathrm{Ag}^{+},\) forming \(\mathrm{Ag}(\mathrm{CN})_{2}^{-} .\) In one such analysis a \(0.3000-\mathrm{g}\) sample of an ore is dissolved and treated with \(20.00 \mathrm{~mL}\) of \(0.5000 \mathrm{M} \mathrm{KCN}\). The excess \(\mathrm{CN}^{-}\) requires \(13.98 \mathrm{~mL}\) of \(0.1518 \mathrm{M} \mathrm{AgNO}_{3}\) to reach the end point. Determine the \(\% \mathrm{w} / \mathrm{w}\) Cd in the ore.

The concentration of \(\mathrm{CO}_{2}\) in air is determined by an indirect acid-base titration. A sample of air is bubbled through a solution that contains an excess of \(\mathrm{Ba}(\mathrm{OH})_{2},\) precipitating \(\mathrm{BaCO}_{3} .\) The excess \(\mathrm{Ba}(\mathrm{OH})_{2}\) is back titrated with HCl. In a typical analysis a 3.5-L sample of air is bubbled through \(50.00 \mathrm{~mL}\) of \(0.0200 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2} .\) Back titrating with \(0.0316 \mathrm{M} \mathrm{HCl}\) requires \(38.58 \mathrm{~mL}\) to reach the end point. Determine the \(\mathrm{ppm} \mathrm{CO}_{2}\) in the sample of air given that the density of \(\mathrm{CO}_{2}\) at the temperature of the sample is \(1.98 \mathrm{~g} / \mathrm{L}\).
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