Problem 107 - Chemistry and Chemical Reactivity

Question

Nicotinic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2},\) is found in minute amounts in all living cells, but appreciable amounts occur in liver, yeast, milk, adrenal glands, white meat, and corn. Whole-wheat flour contains about \(60 . \mu \mathrm{g}\) per gram of flour. One gram \((1.00 \mathrm{g})\) of the acid dissolves in water to give \(60 .\) mL of solution having a pH of \(2.70 .\) What is the approximate value of \(K_{\mathrm{a}}\) for the acid?

Step-by-Step Solution

Answer

The approximate value of \(K_a\) for nicotinic acid is \(3.01 \times 10^{-5}\).

Step 1: Write the Equation for Ionization

Nicotinic acid, \[\mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2\:(aq) + \mathrm{H}_2\mathrm{O}\rightarrow \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2^-\:(aq) + \mathrm{H}_3\mathrm{O}^+\:(aq)\]This is the ionization process of the acid in water. We aim to find the acid dissociation constant \(K_a\) for this reaction.

Step 2: Calculate the Concentration of H⁺ Ions

Given pH is 2.70, use the formula:\[pH = -\log[\mathrm{H}_3\mathrm{O}^+] \]Rearranging gives:\[[\mathrm{H}_3\mathrm{O}^+] = 10^{-2.70} = 2.00 \times 10^{-3}\, \mathrm{M}\]This is the concentration of \(\mathrm{H}_3\mathrm{O}^+\) ions in the solution.

Step 3: Determine Initial Concentration of Acid

One gram of nicotinic acid (molar mass \(123.11\, \mathrm{g/mol}\)) is dissolved in 60.0 mL, hence:\[\text{Moles of acid} = \frac{1.00\, g}{123.11\, g/mol} = 0.00812\, mol\]\[\text{Concentration} = \frac{0.00812\, mol}{0.060\, L} = 0.135\, M\]

Step 4: Generate the Expression for Ka

Assume \(x\) is the concentration of ionized acid:\[x = [\mathrm{H}_3\mathrm{O}^+] = 2.00 \times 10^{-3}\, M\] The initial concentration of acid is 0.135 M, and the concentration of ionized acid is also \(2.00 \times 10^{-3}\, M\), hence:\[K_a = \frac{(2.00 \times 10^{-3})^2}{0.135 - 2.00 \times 10^{-3}}\]

Step 5: Calculate Ka

Plug the values into the expression:\[K_a = \frac{(2.00 \times 10^{-3})^2}{0.133} = \frac{4.00 \times 10^{-6}}{0.133} = 3.01 \times 10^{-5} \]This is the acid dissociation constant \(K_a\) for nicotinic acid.

Key Concepts

Nicotinic Acid

Nicotinic acid, also known as niacin or vitamin B3, plays an essential role in human health, appearing naturally in small amounts in various food sources. It helps your body convert food into energy by aiding enzymes. This compound can be found in its free form in liver, yeast, and milk, to name a few. Given its biochemical importance, understanding its behavior in solution, like its dissociation, becomes essential, particularly in biological processes. This property ties directly into how we calculate its acid dissociation constant, a measure of its ability to donate hydrogen ions.

Ionization Process

Ionization is crucial to understanding acid behavior in water. For nicotinic acid, the ionization process can be expressed by the equation: \[\mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2\:(aq) + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2^-\:(aq) + \mathrm{H}_3\mathrm{O}^+\:(aq)\] This reaction shows nicotinic acid releasing a hydrogen ion to form a conjugate base while producing hydronium ions. Observing this equilibrium process gives insight into how weak acids like nicotinic acid interact with water. This reaction’s extent defines how weak or strong an acid is, influencing calculations for solutions' properties, including pH.

pH and Hydronium Concentration

The pH of a solution measures how acidic or basic it is. It is calculated using the concentration of hydronium ions ( \(\mathrm{H}_3\mathrm{O}^+\)). We use the equation: \[ \text{pH} = -\log[\mathrm{H}_3\mathrm{O}^+] \] In our case, with a pH of 2.70, reversing the equation (using antilog) finds the hydronium concentration: \([\mathrm{H}_3\mathrm{O}^+] = 10^{-2.70} = 2.00 \times 10^{-3}\, \mathrm{M}\) This concentration is closely tied to the ionization process and shows the equilibrium state of the acid in water. Observing such defined measures of hydronium in a solution helps us assess acidity and potential chemical behavior.

Molarity and Moles

Molarity is crucial when discussing solutions, representing the moles of a solute per liter of solution. Here, we focus on how to determine the initial concentration of nicotinic acid using its molarity. With 1 gram of nicotinic acid, using its molar mass (123.11 g/mol), we first calculate the moles: \[ 0.00812\, \text{mol} = \frac{1.00\, \text{g}}{123.11\, \text{g/mol}} \] Next, we convert this to concentration using the solution’s volume (60 mL or 0.060 L): \[ 0.135\, \text{M} = \frac{0.00812\, \text{mol}}{0.060\, \text{L}} \] Understanding moles and molarity allows you to appreciate how solutions' properties, such as concentration, influence chemical reactions, including equilibrium situations.

Equilibrium Expressions

Equilibrium expressions help predict the positions of equilibrium in reversible reactions. For ionization, the equilibrium constant, \(K_a\), denotes acid strength, calculated as: \[ K_a = \frac{[\mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2^-][\mathrm{H}_3\mathrm{O}^+]}{[\mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2]} \] Applying the values for nicotinic acid, we consider initial concentrations and ionization extents: - Ionized hydronium: \(2.00 \times 10^{-3}\, \text{M}\) - Initial acid concentration: \(0.135\, \text{M}\) Using \(x = [\mathrm{H}_3\mathrm{O}^+]\), the expression becomes: \[ K_a = \frac{(2.00 \times 10^{-3})^2}{0.135 - 2.00 \times 10^{-3}} = 3.01 \times 10^{-5} \] This calculation illuminates how partial ionization plays a role in determining this dissociation constant, a pillar for understanding the strength and behavior of acids in chemistry.