Problem 107 - Chemistry: An Atoms-Focused Approach

Question

Identify the conjugate base for the following weak acids and calculate \(K_{\mathrm{b}}\) for each. Which conjugate base is the strongest? a. \(\mathrm{ClCH}_{2} \mathrm{COOH}\) b. \(\mathrm{NH}_{4}^{+}\) c. HCN d. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Step-by-Step Solution

Answer

In summary, the strongest conjugate base among the given options is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\), with a \(K_{\mathrm{b}}\) value of \(5.88 \times 10^{-1}\).

Step 1: Identifying the conjugate base of each acid

To identify the conjugate base for each weak acid, remove one hydrogen ion (H\(^+\)) from the molecular formula of the acid. The resulting species is the conjugate base. The conjugate bases are: a. \(\mathrm{ClCH}_{2} \mathrm{COO}^{-}\) b. \(\mathrm{NH}_{3}\) c. \(\mathrm{CN}^{-}\) d. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\)

Step 2: Calculating \(K_{\mathrm{b}}\) of each conjugate base

Using the relationship between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\), compute the \(K_{\mathrm{b}}\) for each conjugate base. The \(K_{\mathrm{a}}\) values for the corresponding weak acids can be found in a reference table, and the \(K_{\mathrm{w}}\) value is \(1.0 \times 10^{-14}\) at 25°C. a. \(K_{\mathrm{b}(ClCH_2COO^{-})} = \frac{K_{\mathrm{w}}}{K_{\mathrm{a}(ClCH_2COOH)}} = \frac{1.0 \times 10^{-14}}{1.4 \times 10^{-3}} = 7.14 \times 10^{-12}\) b. \(K_{\mathrm{b}(NH_3)} = \frac{K_{\mathrm{w}}}{K_{\mathrm{a}(NH_4^+)}} = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-10}} = 1.79 \times 10^{-5}\) c. \(K_{\mathrm{b}(CN^{-})} = \frac{K_{\mathrm{w}}}{K_{\mathrm{a}(HCN)}} = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}} = 1.61 \times 10^{-5}\) d. \(K_{\mathrm{b}(CH_3CH_2O^{-})} = \frac{K_{\mathrm{w}}}{K_{\mathrm{a}(CH_3CH_2OH)}} = \frac{1.0 \times 10^{-14}}{1.7 \times 10^{-16}} = 5.88 \times 10^{-1}\)

Step 3: Identifying the strongest conjugate base

To determine the strongest conjugate base, compare the \(K_{\mathrm{b}}\) values calculated in the previous step. The highest \(K_{\mathrm{b}}\) value corresponds to the strongest conjugate base: a. \(K_{\mathrm{b}(ClCH_2COO^{-})} = 7.14 \times 10^{-12}\) b. \(K_{\mathrm{b}(NH_3)} = 1.79 \times 10^{-5}\) c. \(K_{\mathrm{b}(CN^{-})} = 1.61 \times 10^{-5}\) d. \(K_{\mathrm{b}(CH_3CH_2O^{-})} = 5.88 \times 10^{-1}\) Since the \(K_{\mathrm{b}}\) of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) is the highest, it can be concluded that \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) is the strongest conjugate base among the given options.

Key Concepts

Acid-Base Chemistry

In the realm of chemistry, understanding acids and bases is crucial. Acids are substances that can donate a proton ( H^+ ), while bases can accept a proton. When an acid donates its proton, it forms a conjugate base. Simply put, a conjugate base is what's left of the acid after it has released a proton.

For example, look at this transformation: HCl (donates H^+ ) → Cl^{-} . Here, Cl^{-} is the conjugate base of HCl . This relationship is critical in understanding how substances interact in chemical reactions. Knowing the conjugate base helps predict the behavior of substances in solutions and their capabilities in different chemical environments.

Identifying conjugate bases is straightforward:

• Remove one H^+ from the acid’s formula.
• The remaining structure is the conjugate base.

This simple step allows us to study the properties and strength of the base, giving more insight into acid-base chemistry dynamics.

Strongest Conjugate Base

Conjugate base strength is an essential part of acid-base chemistry. It determines how effectively a base can accept a proton. In the given examples, we compute the K_b values to establish strength.

The strength of a conjugate base is directly linked to its K_b value. A higher K_b value indicates a stronger base because it means the base dissociates more in water and is more prone to accept protons.

Let's review the compounds:

• \[\mathrm{ClCH_2COO^{-}}: K_b = 7.14 \times 10^{-12}\]
• \[\mathrm{NH_3}: K_b = 1.79 \times 10^{-5}\]
• \[\mathrm{CN^{-}}: K_b = 1.61 \times 10^{-5}\]
• \[\mathrm{CH_3CH_2O^{-}}: K_b = 5.88 \times 10^{-1}\]

Clearly, \mathrm{CH_3CH_2O^{-}} stands out, having the highest K_b value, making it the strongest conjugate base.

This understanding aids in anticipating reactions, especially in synthesis and analytical processes.

Ka and Kb Relationship

The relationship between K_a, the acid dissociation constant, and K_b, the base dissociation constant, is fundamental to understanding acid-base equilibrium.

The formula that describes this relationship is:\[K_a \times K_b = K_w\]where K_w is the ion-product constant of water, 1.0 \times 10^{-14} at 25°C. This equation tells us that if we know the K_a value of an acid, we can find the K_b of its conjugate base, and vice versa.

Here’s how it works:

• Find the K_a value from a reference table.
• Use the formula to calculate K_b: \[K_b = \frac{K_w}{K_a}\]

This straightforward calculation lets us understand both the acidity and basicity of a compound.

This concept helps predict the behavior of acids and bases in various conditions, making it crucial for designing and conducting experiments where pH control is necessary.