Question

Identify the conjugate acids and bases in the following pairs of substances: (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+} /\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\) (b) \(^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2} \mathrm{COOH} /^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}\) \((\mathrm{c})^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-} / \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}\) \((d)^{-} O O C-C H_{2}-C O O H /^{-} O O C-C H_{2}-C O O^{-}\) (e) \(^{-} \mathrm{OOC}-\mathrm{CH}_{2}-\mathrm{COOH} / \mathrm{HOOC}-\mathrm{CH}_{2}-\mathrm{COOH}\)

Short answer

(a) Base, Conjugate Acid. (b) Acid, Conjugate Base. (c) Acid, Conjugate Base. (d) Acid, Conjugate Base. (e) Conjugate Acid, Base.

Step-by-step solution

Identify Conjugate Acid-Base Pair for (a)

The first pair is \(\text{(\text{CH}_{3})_{3}\text{NH}^{+}}/\text{(\text{CH}_{3})_{3}\text{N}}\). To identify the conjugate acid-base pair, recall that a conjugate acid is formed when a base gains a proton (\text{H}^{+}), and a conjugate base is formed when an acid loses a proton. Here, \(\text{(\text{CH}_{3})_{3}\text{N}}\) can gain a proton to form \(\text{(\text{CH}_{3})_{3}\text{NH}^{+}}\), so \(\text{(\text{CH}_{3})_{3}\text{N}}\) is the base and \(\text{(\text{CH}_{3})_{3}\text{NH}^{+}}\) is the conjugate acid.

Identify Conjugate Acid-Base Pair for (b)

The second pair is \(\text{H}_{3}\text{N}-\text{CH}_{2}\text{COOH} / \text{H}_{3}\text{N}-\text{CH}_{2}\text{COO}^{-}\). \(\text{H}_{3}\text{N}-\text{CH}_{2}\text{COOH}\) can lose a proton to form \(\text{H}_{3}\text{N}-\text{CH}_{2}\text{COO}^{-}\), so \(\text{H}_{3}\text{N}-\text{CH}_{2}\text{COOH}\) is the acid and \(\text{H}_{3}\text{N}-\text{CH}_{2}\text{COO}^{-}\) is the conjugate base.

Identify Conjugate Acid-Base Pair for (c)

The third pair is \(\text{H}_{3}\text{N}-\text{CH}_{2}\text{COO}^{-} / \text{H}_{2}\text{N}-\text{CH}_{2}\text{COO}^{-}\). \(\text{H}_{3}\text{N}-\text{CH}_{2}\text{COO}^{-}\) can lose a proton to form \(\text{H}_{2}\text{N}-\text{CH}_{2}\text{COO}^{-}\), hence \(\text{H}_{3}\text{N}-\text{CH}_{2}\text{COO}^{-}\) is the acid and \(\text{H}_{2}\text{N}-\text{CH}_{2}\text{COO}^{-}\) is the conjugate base.

Identify Conjugate Acid-Base Pair for (d)

The fourth pair is \(\text{}^{-} \text{OOC-CH}_{2}-\text{COOH} / ^{-} \text{OOC-CH}_{2}-\text{COO}^{-}\). \(\text{}^{-} \text{OOC-CH}_{2}-\text{COOH}\) can lose a proton to form \(\text{}^{-} \text{OOC-CH}_{2}-\text{COO}^{-}\), making \(\text{}^{-} \text{OOC-CH}_{2}-\text{COOH}\) the acid and \(\text{}^{-} \text{OOC-CH}_{2}-\text{COO}^{-}\) the conjugate base.

Identify Conjugate Acid-Base Pair for (e)

The fifth pair is \(\text{}^{-} \text{OOC-CH}_{2}-\text{COOH} / \text{HOOC-CH}_{2}-\text{COOH}\). \( \text{HOOC-CH}_{2}-\text{COOH}\) can gain a proton to form \(\text{}^{-} \text{OOC-CH}_{2}-\text{COOH}\), so here, \(\text{HOOC-CH}_{2}-\text{COOH}\) is the base and \(\text{}^{-} \text{OOC-CH}_{2}-\text{COOH}\) is the conjugate acid.

Key concepts

acid-base pairs

In the realm of chemistry, acids and bases are like dance partners that always come in pairs. When an acid donates a proton (H+), it transforms into its partner, a conjugate base. Conversely, when a base accepts a proton, it becomes a conjugate acid. This relationship is key to understanding many chemical reactions.
Imagine a scenario where a substance, let's call it HA, donates a proton to become A-. Here, HA is the acid, and A- is its conjugate base. On the flip side, if a substance, say B, accepts a proton, it transforms into BH+, making B the base and BH+ the conjugate acid.
For instance, in the pair (CH3)3N and (CH3)3NH+, (CH3)3N can accept a proton to form (CH3)3NH+. Thus, (CH3)3N is the base, and (CH3)3NH+ is the conjugate acid.
Each acid-base reaction involves two acid-base pairs. Understanding these pairs is essential for identifying how reactions proceed.

conjugate acid

A conjugate acid forms when a base gains a proton. This process is crucial because it's the basis of many chemical behaviors and reactions. Let's look at a specific example: Consider the pair (H3N-CH2-COOH / H3N-CH2-COO-).
In this scenario, H3N-CH2-COO- is capable of accepting a proton to turn into H3N-CH2-COOH. So, H3N-CH2-COO- acts as a base, and upon gaining the proton, it turns into its conjugate acid, H3N-CH2-COOH.
Recognizing conjugate acids in various pairs can help predict reaction outcomes, making it easier to understand complex chemical processes. Always remember, conjugate acids are the result of proton gain by bases!

conjugate base

When an acid loses a proton, it becomes a conjugate base. This concept is pivotal because it demonstrates how substances can change roles during reactions. Let's analyze the pair H3N-CH2-COOH / H3N-CH2-COO-.
Here, H3N-CH2-COOH can lose a proton, transforming into H3N-CH2-COO-. Hence, H3N-CH2-COOH is the acid, and after losing a proton, it becomes its conjugate base, H3N-CH2-COO-.
The pattern is consistent across different pairs, making it a reliable method for identifying substances in acid-base reactions. Knowing how to identify conjugate bases can help you understand the direction and products of chemical reactions.

proton transfer

Proton transfer is the heart of acid-base reactions. It involves the movement of a proton (H+) from one molecule to another, driving the transformation of acids to bases and vice versa. Imagine you have a pair like H2O (water) and OH- (hydroxide). When H2O loses a proton, it turns into OH-, and when OH- gains a proton, it turns back into H2O.
Consider the pair HOOC-CH2-COOH and -OOC-CH2-COOH. When HOOC-CH2-COOH (the acid) loses a proton, it becomes -OOC-CH2-COOH, its conjugate base. In reverse, if -OOC-CH2-COOH gains a proton, it transforms back into HOOC-CH2-COOH, becoming the conjugate acid.
Understanding proton transfer is crucial, as it helps explain how molecules interact and change during reactions. This knowledge allows better prediction and manipulation of reaction pathways in various chemical processes.