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Chapter 9: Problem 23

A 250.0 -mg sample of an organic weak acid is dissolved in an appropriate solvent and titrated with \(0.0556 \mathrm{M} \mathrm{NaOH}\), requiring \(32.58 \mathrm{~mL}\) to reach the end point. Determine the compound's equivalent weight.

Short Answer

Expert verified
The equivalent weight of the compound is approximately 138.07 g/mol.

Step by step solution

01

Calculate Moles of NaOH Used

First, we need to calculate the number of moles of NaOH used in the titration. The formula to find moles is given by: \[ \text{moles of NaOH} = M \times V \] where \( M = 0.0556 \; \text{M} \) is the concentration of NaOH, and \( V = 32.58 \; \text{mL} = 0.03258 \; \text{L} \) is the volume converted to liters. So: \[ \text{moles of NaOH} = 0.0556 \times 0.03258 = 0.001811048 \; \text{moles} \]
02

Relate Moles of NaOH to Moles of Acid

Assuming a 1:1 reaction between the weak acid and NaOH, the moles of NaOH will be equal to the moles of the weak acid titrated: \( 0.001811048 \; \text{moles of weak acid} \). This is because the equivalent point in titration occurs when the moles of acid equal the moles of base.
03

Find Equivalent Weight of the Acid

The equivalent weight of the acid is the mass of the acid per mole that will react with or supply one mole of H鈦� ions. Given that 250 mg of the sample was equivalent to \( 0.001811048 \; \text{moles} \), we calculate the equivalent weight using: \[ \text{Equivalent weight} = \frac{\text{mass of the acid in grams}}{\text{moles of acid}} = \frac{0.250}{0.001811048} \approx 138.07 \; \text{g/mol} \]

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Key Concepts

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

Titration
Titration is a laboratory technique used to determine the unknown concentration of a solution by reacting it with a solution of known concentration, called the titrant. In this exercise, titration helps us measure how much of a sodium hydroxide (\( \text{NaOH} \)) solution is needed to completely neutralize an organic weak acid. This point, known as the "end point" or "equivalence point," is reached when the number of moles of the acid equals the number of moles of the base.

Titration requires important equipment such as a burette, which allows for precise measurement of the titrant added. The process involves:
  • Adding the titrant gradually to the solution of unknown concentration.
  • Observing a color change or using an indicator to detect the end point.
  • Recording the volume of titrant used.
In our case, we know that it took 32.58 mL of 0.0556 M NaOH to reach the equivalence point. This known volume and concentration let us calculate the moles of NaOH used, setting the stage for determining the equivalent weight of the acid.
Moles of NaOH
The calculation of moles is fundamental in chemistry to express the amount of a substance. The formula used is: \[ \text{moles} = \text{concentration (M)} \times \text{volume (L)} \]

In this exercise, NaOH is the titrant with a concentration of 0.0556 M. We converted the volume used from milliliters to liters, giving 0.03258 L. Thus, the moles of NaOH are computed as:\[ \text{moles of NaOH} = 0.0556 \, \text{M} \times 0.03258 \, \text{L} = 0.001811048 \, \text{moles} \]

Through the principle of titration, we assume the reaction between NaOH and the weak organic acid is 1:1. Therefore, these moles equal those of the weak acid. This equivalence is crucial in calculating the equivalent weight as it provides a bridge between the measured mass of the acid and its reactive capacity.
Organic Weak Acid
Organic weak acids are a type of acid that partially dissociate in solution, meaning they do not fully release their hydrogen ions (\( \text{H}^+ \)). This partial dissociation characterizes them as weak acids, distinguishing them from strong acids that dissociate completely.

The calculation of an organic weak acid鈥檚 equivalent weight involves understanding its propensity to react with bases like NaOH, as demonstrated in titration. The equivalent weight is defined as the mass required to neutralize or provide one mole of \( \text{H}^+ \)
ions.

In the given exercise, a 250 mg sample of the organic weak acid was titrated with NaOH. The number of moles of acid determined from the moles of NaOH provides the insight to calculate the equivalent weight:
  • Find the moles of acid using titration data.
  • Equivalent weight formula uses mass (in grams) over moles of acid.
  • In our case, the calculation leads us to an equivalent weight of approximately 138.07 g/mol.
This value represents each mole of the weak acid in terms of its ability to react with reagents like NaOH, showing its reactive capacity in stoichiometry.

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Most popular questions from this chapter

The amount of iron in a meteorite is determined by a redox titration using \(\mathrm{KMnO}_{4}\) as the titrant. A \(0.4185-\mathrm{g}\) sample is dissolved in acid and the liberated \(\mathrm{Fe}^{3+}\) quantitatively reduced to \(\mathrm{Fe}^{2+}\) using a Walden reductor. Titrating with \(0.02500 \mathrm{M} \mathrm{KMnO}_{4}\) requires \(41.27 \mathrm{~mL}\) to reach the end point. Determine the \(\% \mathrm{w} / \mathrm{w} \mathrm{Fe}_{2} \mathrm{O}_{3}\) in the sample of meteorite.

The amount of uranium in an ore is determined by an indirect redox titration. The analysis is accomplished by dissolving the ore in sulfuric acid and reducing \(\mathrm{UO}_{2}^{2+}\) to \(\mathrm{U}^{4+}\) with a Walden reductor. The solution is treated with an excess of \(\mathrm{Fe}^{3+}\), forming \(\mathrm{Fe}^{2+}\) and \(\mathrm{U}^{6+}\). The \(\mathrm{Fe}^{2+}\) is titrated with a standard solution of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) In a typical analysis a 0.315 -g sample of ore is passed through the Walden reductor and treated with \(50.00 \mathrm{~mL}\) of \(0.0125 \mathrm{M} \mathrm{Fe}^{3+}\). Back titrating with 0.00987 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) requires \(10.52 \mathrm{~mL}\). What is the \(\% \mathrm{w} / \mathrm{w} \mathrm{U}\) in the sample?

The concentration of \(\mathrm{CO}\) in air is determined by passing a known volume of air through a tube that contains \(\mathrm{I}_{2} \mathrm{O}_{5}\), forming \(\mathrm{CO}_{2}\) and \(\mathrm{I}_{2}\). The \(\mathrm{I}_{2}\) is removed from the tube by distilling it into a solution that contains an excess of \(\mathrm{KI}\), producing \(\mathrm{I}_{3}^{-}\). The \(\mathrm{I}_{3}^{-}\) is titrated with a standard solution of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\). In a typical analysis a 4.79 - \(\mathrm{L}\) sample of air is sampled as described here, requiring \(7.17 \mathrm{~mL}\) of \(0.00329 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) to reach the end point. If the air has a density of \(1.23 \times 10^{-3} \mathrm{~g} / \mathrm{mL},\) determine the parts per million CO in the air.

Calculate or sketch titration curves for the following acid-base titrations. a. \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(0.0500 \mathrm{M} \mathrm{HCl}\) b. \(50.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{HCOOH}\) with \(0.100 \mathrm{M} \mathrm{NaOH}\) c. \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NH}_{3}\) with \(0.100 \mathrm{M} \mathrm{HCl}\) d. \(50.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M}\) ethylenediamine with \(0.100 \mathrm{M} \mathrm{HCl}\) e. \(50.0 \mathrm{~mL}\) of \(0.0400 \mathrm{M}\) citric acid with \(0.120 \mathrm{M} \mathrm{NaOH}\) f. \(50.0 \mathrm{~mL}\) of \(0.0400 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) with \(0.120 \mathrm{M} \mathrm{NaOH}\)

After removing the membranes from an eggshell, the shell is dried and its mass recorded as \(5.613 \mathrm{~g} .\) The eggshell is transferred to a \(250-\mathrm{mL}\) beaker and dissolved in \(25 \mathrm{~mL}\) of \(6 \mathrm{M}\) HCl. After filtering, the solution that contains the dissolved eggshell is diluted to \(250 \mathrm{~mL}\) in a volumetric flask. A \(10.00-\mathrm{mL}\) aliquot is placed in a \(125-\mathrm{mL}\) Erlenmeyer flask and buffered to a \(\mathrm{pH}\) of 10 . Titrating with \(0.04988 \mathrm{M}\) EDTA requires \(44.11 \mathrm{~mL}\) to reach the end point. Determine the amount of calcium in the eggshell as \(\% \mathrm{w} / \mathrm{w} \mathrm{CaCO}_{3}\).
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