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. Its \(K_{\mathrm{a}}\) is \(9.8 \times 10^{-5}\). Calculate \(\left[\mathrm{H}^{+}\right]\) in solutions prepared by adding enough water to the following to make \(1.45 \mathrm{~L}\). (a) \(0.344 \mathrm{~mol}\) (b) \(28.9 \mathrm{~g}\)

Short answer

Question: Calculate the concentration of H⁺ ions in the following barbituric acid solutions: (a) 0.344 mol of barbituric acid in 1.45 L of solution, and (b) 28.9 g of barbituric acid in 1.45 L of solution. Given that the Ka of barbituric acid is 9.8 × 10⁻⁵. Answer: (a) The concentration of H⁺ ions in the first solution is approximately \(x = \sqrt{(9.8 \times 10^{-5})(\frac{0.344 \text{ mol}}{1.45\text{ L}})}\). (b) The concentration of H⁺ ions in the second solution is approximately \(x = \sqrt{(9.8 \times 10^{-5})(\frac{\text{moles}(\text{barbituric acid})}{1.45\text{ L}})}\), where moles of barbituric acid = \(\frac{28.9 \text{ g}}{128 \text{ g/mol}}\).

Step-by-step solution

Calculate the concentration of barbituric acid

To determine the concentration of barbituric acid in each solution, we need to divide the amount of substance (in moles) by the volume of the solution (in liters). (a) \(\text{conc}(\text{barbituric acid}) = \frac{0.344 \text{ mol}}{1.45\text{ L}}\) (b) We first find the moles of barbituric acid by its mass given. The molar mass of barbituric acid is $1(12.01)+4(1.01)+3(1.01)+ 2(14.01)+3(16)= 128\text{ g/mol}\(. Thus, \)\text{moles}(\text{barbituric acid})= \frac{28.9 \text{ g}}{128 \text{ g/mol}}\(. Now we can find the concentration: \)\text{conc}(\text{barbituric acid}) = \frac{\text{moles}(\text{barbituric acid})}{1.45\text{ L}}$

Use the Ka expression to find the concentration of H+ ions

For a weak acid, we can express the ionization as: \(K_a=\frac{[\mathrm{H}^+][\mathrm{A^-}]}{[\mathrm{HA}]}\) In this case, we are dealing with a monoprotic weak acid (barbituric acid), so we can assume the initial concentration of H+ and A- ions is negligible, and after dissociation, the concentrations are equal. Let's call x the concentration of these ions: \(K_a=\frac{x^2}{[\text{barbituric acid}]-x}\) Given that \(K_a\) is small, we can simplify this expression assuming that x is much smaller than the initial concentration of our weak acid: \(K_a \approx \frac{x^2}{[\text{barbituric acid}]}\) So, we need to solve for x in both cases: x = \(\sqrt{K_{a}\times \text{conc}(\text{barbituric acid})}\) (a) \(x = \sqrt{(9.8 \times 10^{-5})(\frac{0.344 \text{ mol}}{1.45\text{ L}})}\) (b) \(x = \sqrt{(9.8 \times 10^{-5})(\frac{\text{moles}(\text{barbituric acid})}{1.45\text{ L}})}\) The calculated x values represent the concentration of \(\mathrm{H}^{+}\) ions in each solution [H+].