Ch.18 - Aqueous Ionic Equilibrium

Tro - Chemistry: A Molecular Approach 6th Edition

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Tro 6th Edition

Ch.18 - Aqueous Ionic Equilibrium

Problem 83

Chapter 18, Problem 83

A 0.229-g sample of an unknown monoprotic acid is titrated with 0.112 M NaOH. The resulting titration curve is shown here. Determine the molar mass and pKa of the acid.

Verified step by step guidance

Identify the equivalence point on the titration curve, which is the point where the pH rapidly changes. This occurs around 25 mL of NaOH added.

Calculate the moles of NaOH added at the equivalence point using the formula: moles of NaOH = Molarity of NaOH * Volume of NaOH (in liters).

Since the acid is monoprotic, the moles of NaOH at the equivalence point are equal to the moles of the unknown acid.

Determine the molar mass of the unknown acid using the formula: Molar mass = mass of the acid sample / moles of the acid.

To find the pKa of the acid, locate the half-equivalence point on the titration curve, which is at half the volume of the equivalence point (around 12.5 mL). The pH at this volume is equal to the pKa of the acid.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Monoprotic Acid

A monoprotic acid is an acid that can donate only one proton (H⁺) per molecule during a chemical reaction. This characteristic simplifies the titration process, as the equivalence point occurs when the amount of base added is stoichiometrically equivalent to the amount of acid present. Understanding the behavior of monoprotic acids is crucial for determining their molar mass and pKa from titration data.

Titration Curve

A titration curve is a graphical representation of the pH of a solution as a function of the volume of titrant added. For a monoprotic acid titrated with a strong base like NaOH, the curve typically shows a gradual increase in pH, followed by a steep rise at the equivalence point, and then levels off. Analyzing the shape of the curve helps identify the equivalence point and calculate the pKa of the acid.

Molar Mass and pKa Calculation

The molar mass of the acid can be calculated using the mass of the acid sample and the volume of NaOH required to reach the equivalence point, applying stoichiometry. The pKa, which indicates the strength of the acid, can be determined from the pH at the half-equivalence point, where half of the acid has been neutralized. These calculations are essential for characterizing the unknown acid in the titration experiment.

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Molar Mass Calculation Example

Related Practice

Textbook Question

Methyl red has a pKa of 5.0 and is red in its acid form and yellow in its basic form. If several drops of this indicator are placed in a 25.0-mL sample of 0.100 M HCl, what color will the solution appear? If 0.100 M NaOH is slowly added to the HCl sample, in what pH range will the indicator change color?

Textbook Question

Consider the titration curves (labeled a and b) for two weak acids, both titrated with 0.100 M NaOH.

(i) Which acid solution is more concentrated?

Textbook Question

Consider the titration curves (labeled a and b) for two weak bases, both titrated with 0.100 M HCl. (a)

(b)

(ii) Which base has the larger Kb?

Textbook Question

Consider the titration curves (labeled a and b) for two weak acids, both titrated with 0.100 M NaOH.

(ii) Which acid has the larger Ka?

Textbook Question

A 20.0-mL sample of 0.115 M sulfurous acid (H₂SO₃) solution is titrated with 0.1014 M KOH. At what added volume of base solution does each equivalence point occur?