Triethylamine, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N},\) has a \(\mathrm{p} K_{b}\) value of \(2.99 .\) Is triethylamine a stronger base than ammonia, \(\mathrm{NH}_{3} ?\)

Short Answer

Expert verified
Yes, triethylamine is a stronger base than ammonia, as its \(\mathrm{p}K_b\) value (2.99) is lower than ammonia's \(\mathrm{p}K_b\) value (4.75).

Step by step solution

01

Compare \(\mathrm{p}K_b\) Values

Compare the \(\mathrm{p}K_b\) values of triethylamine and ammonia: Triethylamine: \(\mathrm{p}K_b = 2.99\) Ammonia: \(\mathrm{p}K_b = 4.75\)
02

Determine Which Base is Stronger

Since the \(\mathrm{p}K_b\) of triethylamine (2.99) is lower than the \(\mathrm{p}K_b\) of ammonia (4.75), triethylamine is a stronger base than ammonia.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding pKb Values
When it comes to analyzing the strength of bases in chemistry, one essential tool is understanding pKb values. The pKb value is a numeric scale used to specify the strength of a base. It is closely related to the pKa value, which is used for acids. The lower the pKb value, the stronger the base.

But why is this the case? The pKb value is logarithmically linked to the base dissociation constant (Kb), which provides a measure of how readily a base forms its conjugate acid by accepting a proton in water. Mathematically, it is expressed as pKb = -log(Kb). The negative logarithm means that as Kb increases (indicating a stronger base), the resulting pKb decreases. Therefore, bases with lower pKb values are considered stronger because they are more effective at accepting protons.
Base Strength Determination
To determine the relative strength of a base, comparing pKb values is a convenient method. As seen in the exercise, triethylamine has a pKb of 2.99, while ammonia has a pKb of 4.75. Since triethylamine's value is lower, it indicates a stronger base. However, understanding the concept in depth requires recognizing that the base dissociation constant (Kb) is the true descriptor of base strength and the pKb value is simply a more approachable representation of this constant.

In the aqueous solutions where base strength is typically measured, a stronger base will be more likely to accept a proton, thus shifting the equilibrium towards the conjugate acid. This results in a higher Kb and consequently, a lower pKb. It’s crucial to remember that this measurement is relative and depends on the solvent and the specific base being compared.
Amine Chemistry
Amines are a key class of compounds in organic chemistry, characterized by the presence of a nitrogen atom with a lone pair that can accept a proton, making them bases. Generally speaking, the basicity of an amine is influenced by its alkyl substituents – the groups attached to the nitrogen.

Triethylamine, for instance, has three ethyl groups connected to its nitrogen, while ammonia lacks these alkyl substituents. The presence of alkyl groups in amines such as triethylamine tends to increase the electron density available to the nitrogen, enabling it to more readily accept a proton. This effect is known as alkyl group inductive effect, where the electron-donating nature of the alkyl groups enhances the base strength of the amine. Consequently, triethylamine, with its increased electron density due to the ethyl groups, is a stronger base compared to ammonia, which solely relies on its lone pair of electrons for basicity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the molar concentration of \(\mathrm{OH}^{-}\) ions in a \(0.075 \mathrm{M}\) solution of ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} ; K_{b}=6.4 \times 10^{-4}\right) .\) Calcu- late the \(\mathrm{pH}\) of this solution.

Which of the following solutions has the higher pH? (a) a \(0.1 M\) solution of a strong acid or a \(0.1 M\) solution of a weak acid, (b) a \(0.1 \mathrm{M}\) solution of an acid with \(K_{a}=2 \times 10^{-3}\) or one with \(K_{a}=8 \times 10^{-6},(\mathrm{c}) \mathrm{a} 0.1 \mathrm{M}\) solution of a base with \(\mathrm{p} K_{b}=4.5\) or one with \(\mathrm{p} K_{b}=6.5\).

(a) Write a chemical equation that illustrates the autoionization of water. (b) Write the expression for the ion-product constant for water, \(K_{w}\). Why is \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) absent from this expression? (c) A solution is described as basic. What does this statement mean?

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} \cdot\) Calculate \(\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature. $$ \begin{array}{llll} \hline \text { pH } & \text { pOH } & \text { [H }^{+} \text {] } & \text { [OH }^{-} \text {] } & \text { Acidic or basic? } \\ \hline 5.25 & & & \\ & 2.02 & & \\ & & 4.4 \times 10^{-10} M & & \\ & & & 8.5 \times 10^{-2} M & \\ \hline \end{array} $$

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free