The pH at which aniline is expected to exist predominantly in protonated form is to be determined. The pH at which it can be expected to exist predominantly in uncharged form is to be determined. The pH at which an equal mixture of the two forms is expected is to be determined. Concept introduction: Aniline is a weak base whose protonated form has a pKa of about 4.6. In an aqueous solution, a weak base is protonated by water and exists in equilibrium with the protonated form and the relative concentrations of the two species at equilibrium are determined by the pH of the solution. At pH values above the pKa, the protonated form is favored, while below the pKa, the deprotonated form is favored. The ratio equals 1 at a pH equal to pKa. A difference of 1 unit between the pH and pKa corresponds to a difference in concentrations by a factor of 10. Therefore, if the pH of the solution is 1 unit lower than the pKa, the concentration of the protonated form is 10 times greater than the concentration of the deprotonated form. Conversely, when the pH is greater than pKa by 1 unit, the concentration of the deprotonated form is greater than the concentration of the protonated form by a factor of 10. In other words, the protonated form is predominant when the solution pH is lower than the pKa by one unit or more. The deprotonated form is predominant when the pH is higher than the pKa by one unit or more. When the concentrations of the two forms are known or the ratio of concentrations is known, the pH can be calculated using the Henderson-Hasselbalch equation.

Question

Want to see the full answer?

Answer 1

Question

Chapter 6, Problem 6.44P

Interpretation Introduction

Interpretation:

The pH at which aniline is expected to exist predominantly in protonated form is to be determined. The pH at which it can be expected to exist predominantly in uncharged form is to be determined. The pH at which an equal mixture of the two forms is expected is to be determined.

Concept introduction:

Aniline is a weak base whose protonated form has a pKa of about 4.6. In an aqueous solution, a weak base is protonated by water and exists in equilibrium with the protonated form and the relative concentrations of the two species at equilibrium are determined by the pH of the solution. At pH values above the pKa, the protonated form is favored, while below the pKa, the deprotonated form is favored. The ratio equals 1 at a pH equal to pKa.

A difference of 1 unit between the pH and pKa corresponds to a difference in concentrations by a factor of 10. Therefore, if the pH of the solution is 1 unit lower than the pKa, the concentration of the protonated form is 10 times greater than the concentration of the deprotonated form. Conversely, when the pH is greater than pKa by 1 unit, the concentration of the deprotonated form is greater than the concentration of the protonated form by a factor of 10. In other words, the protonated form is predominant when the solution pH is lower than the pKa by one unit or more. The deprotonated form is predominant when the pH is higher than the pKa by one unit or more.

When the concentrations of the two forms are known or the ratio of concentrations is known, the pH can be calculated using the Henderson-Hasselbalch equation.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Rank the labeled protons (Ha-Hd) in order of increasing acidity, starting with the least acidic. НОН НЬ OHd Онс

Can the target compound at right be efficiently synthesized in good yield from the unsubstituted benzene at left? ? starting material target If so, draw a synthesis below. If no synthesis using reagents ALEKS recognizes is possible, check the box under the drawing area. Be sure you follow the standard ALEKS rules for submitting syntheses. + More... Note for advanced students: you may assume that you are using a large excess of benzene as your starting material. C :0 T Add/Remove step G

The following equations represent the formation of compound MX. What is the AH for the electron affinity of X (g)? X₂ (g) → 2X (g) M (s) → M (g) M (g) M (g) + e- AH = 60 kJ/mol AH = 22 kJ/mol X (g) + e-X (g) M* (g) +X (g) → MX (s) AH = 118 kJ/mol AH = ? AH = -190 kJ/mol AH = -100 kJ/mol a) -80 kJ b) -30 kJ c) -20 kJ d) 20 kJ e) 156 kJ