Problem 27 - Analytical Chemistry

Question

A weak base, \(\mathrm{B}\), with a \(K_{\mathrm{b}}\) of \(1.0 \times 10^{-3}\) has a partition coefficient, \(K_{\mathrm{D}}\), of \(5.0 \times 10^{2}\) between water and an organic solvent. What restriction on the sample's \(\mathrm{pH}\) is necessary to ensure that \(99.9 \%\) of the weak base in a 50.0 -mL sample is extracted when using two 25.0 -mL portions of the organic solvent?

Step-by-Step Solution

Answer

The pH should be adjusted such that less ionized form remains, facilitating 99.9% extraction.

Step 1: Understanding the Problem

We have a weak base, \( \mathrm{B} \), with a given \( K_{\mathrm{b}} \) and partition coefficient \( K_{\mathrm{D}} \). We need to find the \( \mathrm{pH} \) condition to extract \( 99.9\% \) of the base using two portions of solvent.

Step 2: Calculate Base Ionization

The base ionization equilibrium is \( \mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH^+} + \mathrm{OH^-} \). The expression for \( K_{\mathrm{b}} \) is \( K_{\mathrm{b}} = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]} \). Given that \( K_{\mathrm{b}} = 1.0 \times 10^{-3} \), this implies the base is partially ionized.

Step 3: Calculate Distribution with Partition Coefficient

The partition coefficient \( K_{\mathrm{D}} = \frac{[\mathrm{B}]_{\text{organic}}}{[\mathrm{B}]_{\text{water}}} \) is given as \( 5.0 \times 10^{2} \). This tells us the ratio of the base distributed between the organic and aqueous phases.

Step 4: Setting up Extraction Equation

To ensure \( 99.9\% \) extraction, first find each extraction's efficienecy. Use \( q = \left( \frac{V_{\text{water}}}{K_{\mathrm{D}}V_{\text{organic}} + V_{\text{water}}} \right)^2 \) to find \( q \), the fraction remaining in the aqueous phase.

Step 5: Solve for Necessary Condition

The final fraction remaining after two extractions should satisfy \( (1-q)^2 = 0.001 \). Solve for \( q \) and then back-calculate using \( pH = pK_a + \log \left( \frac{[\mathrm{B}]}{[\mathrm{BH^+}]} \right) \) respecting the \( K_{\mathrm{D}} \).

Step 6: Calculate pH

After substituting values, determine the necessary \( \mathrm{pH} \) using known relationships between \( K_{\mathrm{b}} \), \( K_{\mathrm{a}} \), and ionization equilibrium.

Key Concepts

Weak Base

A **weak base** is a chemical that does not completely dissociate into its ions in a solution. This means that only a small fraction of the base actually reacts with water to form hydroxide ions (OH^-) and its conjugate acid (BH⁺).

Typically, weak bases are in equilibrium in aqueous solutions, represented by the equation:

• \( \mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH^+} + \mathrm{OH^-} \)

The equilibrium constant for this reaction is known as the base ionization constant, \( K_b \). A given weak base with \( K_b = 1.0 \times 10^{-3} \) indicates partial ionization, reflecting its weak status.

Understanding weak bases is crucial in calculating how they behave in various chemical reactions, especially when they are extracted into different solvents.

pH Condition

The **pH condition** is an important factor that dictates the ionization state of a compound in a solution. pH measures the acidity or basicity of a solution, and it plays a vital role in the extraction process of weak bases.

In the context of extraction, a specific pH can shift the equilibrium toward more of the ionized form (BH⁺) or the non-ionized form (B). This is necessary for maximizing the extraction of weak bases into an organic solvent. To ensure efficient extraction, the pH must be controlled so that most of the weak base remains in its unionized form, which is more soluble in the organic phase.

Using the relationship between pH and pKa, you can determine the degree of ionization of the base:

• \( \text{pH} = \text{pK}_a + \log \left( \frac{[ ext{B}]}{[ ext{BH}^+]} \right) \)

To extract 99.9% of a weak base from an aqueous phase, knowing the precise pH condition is crucial.

Base Ionization

**Base ionization** refers to the process in which a weak base accepts a proton (H⁺) from water, leading to the formation of the conjugate acid (BH⁺) and hydroxide ions (OH^-).

The extent of ionization can influence both the pH of the solution and the efficiency of the solvent extraction. For a weak base, the ionization is described by its base dissociation constant, \( K_b \), which provides insight into how much of the base is ionized at equilibrium.

Considerations around ionization are key when setting conditions for extraction processes, as the ionization state determines solubility and partitioning of the weak base between the aqueous and organic phases._RETURNS preserved"Base Ionization|implies that understanding base ionization helps to anticipate the behavior of weak bases during chemical reactions and extraction processes.

Extraction Efficiency

**Extraction efficiency** measures how well a substance is transferred from one phase to another during a liquid-liquid extraction process. In this context, it refers to how much of the weak base is removed from the aqueous phase into the organic solvent.

The partition coefficient, \( K_D \), indicates how a compound distributes itself between an organic solvent and water. A high \( K_D \), such as \( 5.0 \times 10^{2} \), shows that the base prefers the organic solvent, suggesting efficient extraction.

To calculate extraction efficiency, especially when multiple extractions are used, it's crucial to determine how much base remains in the aqueous phase:

• Use the formula \[ q = \left( \frac{V_{\text{water}}}{K_{\text{D}}V_{\text{organic}} + V_{\text{water}}} \right)^2 \] to find \( q \), the fraction remaining after two extractions.

Efficient extraction ensures minimal waste and maximizes the amount of substance extracted, which is vital in many analytical and preparative chemistry applications.