Question

Barbituric acid, \(\mathrm{HC}_{4} \mathrm{H}_{3} \mathrm{~N}_{2} \mathrm{O}_{3},\) is used to prepare barbiturates, a class of drugs used as sedatives. A \(325-\mathrm{mL}\) aqueous solution of barbituric acid has a pH of 2.34 and contains \(9.00 \mathrm{~g}\) of the acid. What is \(K_{3}\) for barbituric acid?

Short answer

Answer: To find the Ka value of barbituric acid, follow these steps: 1. Calculate the concentration of H+ ions in the solution using the given pH: [H+] = 10^(-2.34). 2. Find the initial concentration of barbituric acid: moles = (9.00 g)/(128.09 g/mol) = 0.0703 mol; concentration = (0.0703 mol)/(0.325 L) = 0.216 M. 3. Use the Henderson-Hasselbalch equation to find Ka: 2.34 = pKa + log([(0.216 M) - [H+]]/[0.216 M]). Solve for pKa and then calculate Ka using Ka = 10^(-pKa).

Step-by-step solution

Calculate the concentration of H+ ions in the solution from the given pH

The formula to find the concentration of H+ ions is \(\mathrm{[H^{+}]} = 10^{-\mathrm{pH}}\). Given the pH of the solution is 2.34, we can calculate the \(\mathrm{[H^{+}]}\) using the formula: $$ \mathrm{[H^{+}]} = 10^{-2.34} $$

Calculate the initial concentration of barbituric acid

The initial concentration of barbituric acid can be determined using the mass of the acid and the volume of the solution. First, we need to convert the grams of barbituric acid into moles. To do this, we will use the molar mass of barbituric acid, which is calculated as follows: $$ \mathrm{Molar\, Mass} = (1\times1) +(12.01\times4) +(1\times3)+(14.01\times2) +(16.00\times3) = 128.09\, \mathrm{g/mol} $$The moles of barbituric acid can then be found by dividing the mass by the molar mass: $$ \mathrm{moles} = \frac{9.00\, \mathrm{g}}{128.09\, \mathrm{g/mol}} = 0.0703\, \mathrm{mol} $$The initial concentration (M) can be derived by dividing this number of moles by the volume of the solution (in L): $$ \mathrm{concentration} = \frac{0.0703\, \mathrm{mol}}{0.325\, \mathrm{L}} = 0.216\, \mathrm{M} $$

Use the Henderson-Hasselbalch equation to find Ka

The Henderson-Hasselbalch equation is given by: $$ \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{\mathrm{[A^{-}]}}{\mathrm{[HA]}} $$Where \(\mathrm{[A^{-}]}\) is the concentration of the conjugate base, and \(\mathrm{[HA]}\) is the concentration of the acid. In this case, the concentration of the conjugate base can be found from the concentration of H+ ions, as they are equal in a 1:1 ratio: $$ \mathrm{[A^{-}]} = \mathrm{[H^{+}]} $$Substitute the given values into the Henderson-Hasselbalch equation, and solve for the \(K_a\) value: $$ 2.34 = \mathrm{p}K_{\mathrm{a}} + \log \frac{0.216\, \mathrm{M} - \mathrm{[H^{+}]}}{0.216\, \mathrm{M}} $$Now you can solve for \(\mathrm{p}K_{\mathrm{a}}\), which can then be used to find the \(K_{a}\) value, using the relationship $$K_a = 10^{-\mathrm{p}K_a}$$

Key concepts

pH Calculation

Understanding how to calculate pH is fundamental in chemistry, especially when dealing with acid-base reactions. pH is the measure of how acidic or basic an aqueous solution is. The pH scale ranges from 0 to 14, with 7 being neutral, values less than 7 being acidic, and values greater than 7 being basic. Calculating the pH involves working with the concentration of hydrogen ions, \( [H^+] \), in the solution.

The relationship between pH and \( [H^+] \) concentration is given by the negative logarithm: \[ pH = -\log([H^+]) \] To find the pH, we simply take the negative log of the hydrogen ion concentration. In the example at hand, a pH of 2.34 indicates a relatively acidic solution, corresponding to particular \( [H^+] \) concentration. By applying this formula, we can start to understand the acidic nature of the solution and prepare for further calculations relating to the acid dissociation constant.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation plays a pivotal role in understanding the relationship between the pH of a solution and the pKa (acid dissociation constant) of a weak acid. This equation is particularly useful for buffer solutions, which resist changes in pH upon the addition of small amounts of acids or bases. It's formulated as: \[ pH = pKa + \log\frac{[A^-]}{[HA]} \] Here, \( pKa \) refers to the negative log of the acid dissociation constant \( Ka \), \( [A^-] \) is the concentration of the conjugate base, and \( [HA] \) represents the concentration of the weak acid.

The key to using the Henderson-Hasselbalch equation lies in accurately determining the concentration of the acid and its conjugate base at equilibrium. This allows the calculation of the pH given the \( pKa \) or vice versa. In the exercise, the equation is rearranged to solve for \( pKa \) using the given pH and concentrations, enabling the determination of the acid dissociation constant for barbituric acid.

Molar Mass

The molar mass of a compound is a physical property that is defined as the mass of one mole of its particles (atoms, molecules, or formula units). It is often expressed in grams per mole (g/mol) and can be calculated by summing the atomic masses of all atoms in the compound's chemical formula.

For instance, barbituric acid, \( \mathrm{HC}_{4} \mathrm{H}_{3} \mathrm{N}_{2} \mathrm{O}_{3} \), has a molar mass calculated by adding up the atomic masses of hydrogen (H), carbon (C), nitrogen (N), and oxygen (O) based on their respective quantities in the formula. This value is crucial when converting between grams of substance and moles, a common task in stoichiometry and chemical reaction quantification.

By knowing the molar mass, we are able to determine the number of moles present in a given mass of the compound, as done in the exercise, which is a fundamental step in calculating concentrations and ultimately, the dissociation constant of the acid.