Problem 25 - Analytical Chemistry

Question

A weak acid, HA, with a \(K_{\mathrm{a}}\) of \(1.0 \times 10^{-5}\) has a partition coefficient, \(K_{\mathrm{D}}\), of \(1.2 \times 10^{3}\) between water and an organic solvent. What restriction on the sample's \(\mathrm{pH}\) is necessary to ensure that \(99.9 \%\) of the weak acid in a \(50.0-\mathrm{mL}\) sample is extracted using a single \(50.0-\mathrm{mL}\) portion of the organic solvent?

Step-by-Step Solution

Answer

The sample's pH must be less than approximately 2.10.

Step 1: Identify the fraction extracted (f)

To ensure that 99.9% of the weak acid is extracted, we need the fraction extracted, \(f\), to be 0.999. The fraction extracted can be calculated by the formula \(f = \frac{DK_D}{1 + DK_D}\), where \(D\) is the distribution ratio.

Step 2: Derive expression for distribution ratio (D)

The distribution ratio \(D\) for a weak acid is defined as \(D = \frac{[HA]_o}{[HA] + [A^-]}\) in terms of the concentrations in the solvent. Given the conditions for \(f\), and knowing \(K_D = 1.2 \times 10^3\), solve for \(D\) such that \(f = 0.999\).

Step 3: Calculate required distribution ratio (D)

Substituting \(f = 0.999\) and \(K_D = 1.2 \times 10^3\) into the fraction extracted equation: \( 0.999 = \frac{DK_D}{1 + DK_D} \), solve for \(D\):\[ 0.999(1 + D \times 1.2 \times 10^3) = D \times 1.2 \times 10^3 \]\[ 0.999 + 0.999 \times D \times 1.2 \times 10^3 = D \times 1.2 \times 10^3 \]\[ D \times 1.2 \times 10^3 - 0.999 \times D \times 1.2 \times 10^3 = 0.999 \]\[ D(1.2 \times 10^3 - 1.2 \times 10^3 \times 0.999) = 0.999 \]\[ D = \frac{0.999}{1.2} = 0.8325 \]

Step 4: Relate distribution ratio to pH and solve

The distribution ratio \(D\) implies \(D = [HA]_o/[HA] +[A^-] = [HA]_o / ([HA]_1 + 10^{-pH}) \). Rearranging for pH:\[ pH = pK_a + \log \left(\frac{D}{K_D - D}\right) \]Substituting \(K_a = 1.0 \times 10^{-5}\) which means \(pK_a = 5\), \(D = 0.8325\), and using \(K_D = 1.2 \times 10^3\), we have:\[ pH = 5 + \log \left(\frac{0.8325}{1.2 \times 10^3 - 0.8325}\right) \]Simplify to find:\[ pH \approx 2.10 \]

Step 5: Conclusion on the necessary pH

Therefore, to ensure that 99.9% of the weak acid is extracted, the sample's pH must be less than approximately 2.10.

Key Concepts

Partition Coefficient

The partition coefficient, often denoted as \( K_D \), is a fundamental concept in chemistry that helps us understand how a solute distributes itself between two immiscible solvents. In our exercise, the partition coefficient is given as \( 1.2 \times 10^3 \), reflecting the distribution of a weak acid between water and an organic solvent.

• A high partition coefficient indicates a higher concentration of the compound in the organic solvent compared to water.
• This coefficient helps in determining how much of the acid remains or is extracted into the organic layer.
• It is a crucial factor when planning extraction processes, as it affects the efficiency of the separation.

Understanding the partition coefficient's role allows us to predict how substances will behave during solvent extraction, which is essential especially in pharmaceutical and chemical industries.

Distribution Ratio

The distribution ratio, denoted as \( D \), is closely related to the partition coefficient. It is a measure that considers both the ionized and non-ionized forms of a substance, like our weak acid HA, in a solvent.

• The formula for the distribution ratio is given by \( D = \frac{[HA]_o}{[HA] + [A^-]} \), where \([HA]_o\) is the concentration in the organic layer and \([HA] + [A^-]\) is the total concentration in the aqueous phase.
• In extractions, \( D \) is crucial because it can vary with \( pH \), as the proportion of ionized species changes.
• To maximize extraction efficiency, maintaining an optimal \( pH \) to achieve the desired \( D \) value is important.

For our problem, it was calculated that \( D \) should be approximately 0.8325 to ensure effective extraction, based on initial conditions of \( f = 0.999 \) and \( K_D = 1.2 \times 10^3 \).

pH Calculation

Calculating the appropriate \( pH \) is often necessary when dealing with weak acid extractions. The \( pH \) affects both the distribution ratio and the degree of ionization of the acid.

• The relationship between \( pH \) and the distribution ratio \( D \) is given by: \[ pH = pK_a + \log \left( \frac{D}{K_D - D} \right) \]
• Using the given \( K_a \) as \( 1.0 \times 10^{-5} \) (with \( pK_a = 5 \)), \( D = 0.8325 \), and \( K_D = 1.2 \times 10^3 \), one can plug these into the equation to find the \( pH \).
• In this scenario, a \( pH \) of approximately 2.10 was determined to be necessary for the extraction to be 99.9% efficient.

Adjusting the \( pH \) allows control over the systems equilibria, thus fine-tuning the efficiency of the extraction.

Weak Acid Extraction

Weak acid extraction is a common technique used to separate acids from mixtures, particularly in organic chemistry and biochemistry.

• The process relies on the distribution of the acid between an aqueous solution and an organic solvent, driven by differing solubilities and \( pH \).
• Here, the aim was to extract 99.9% of a weak acid. To achieve this, the extraction efficiency is influenced by both the partition coefficient and the distribution ratio.
• Managing the \( pH \) precisely is paramount as it decides the extent to which the acid remains ionized and hence less extractable into the organic phase.

Understanding the parameters controlling weak acid extraction helps refine laboratory practices, ensuring that processes are both effective and economical, important for applications like drug synthesis and purification.