Question

Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25{ }^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathrm{a})\) a solution formed by adding \(25.0 \mathrm{~g}\) of furoic acid and \(30.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.250 \mathrm{~L}\) of solution; (b) a solution formed by mixing \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{MHC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(20.0 \mathrm{~mL}\) of \(0.22 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL} ;\) (c) a solution prepared by adding \(50.0 \mathrm{~mL}\) of \(1.65 \mathrm{M} \mathrm{NaOH}\) solution to \(0.500 \mathrm{Lof} 0.0850 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\)

Short answer

a) The pH of the solution formed by adding 25.0 g of furoic acid and 30.0 g of sodium furoate to 0.250 L of water is approximately \(2.52\). b) The pH of the solution formed by mixing 30.0 mL of 0.250 M HC5H3O3 and 20.0 mL of 0.22 M NaC5H3O3, then diluting to 125 mL, is approximately \(2.56\). c) The pH of the solution prepared by adding 50.0 mL of 1.65 M NaOH to 0.500 L of 0.0850 M HC5H3O3 is approximately \(3.63\).

Step-by-step solution

Calculate moles of furoic acid and sodium furoate in each case

We will need the moles of furoic acid and its conjugate base (sodium furoate) in each case. a) For this case, convert the given masses of furoic acid and sodium furoate into moles using their respective molar masses. (Molar mass HC5H3O3 = 114.08 g/mol and Molar mass NaC5H3O3 = 141.11 g/mol) b) For this case, multiply the volume by the given molarity to find the number of moles. c) You'll find the number of moles of furoic acid and sodium hydroxide given their molarity and volume. Then, perform a reaction between furoic acid and sodium hydroxide to find the number of moles of the resulting sodium furoate and remaining furoic acid.

Calculate concentrations of furoic acid and sodium furoate in each case

In each case, divide the moles of furoic acid and sodium furoate by the final volume of the solution to find their concentrations. Make sure the final volumes are in liters.

Apply the Henderson-Hasselbalch equation

Use the Henderson-Hasselbalch equation to find the pH of each solution: \(pH = pKa + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) Where A- is the conjugate base (sodium furoate) concentration, HA is the furoic acid concentration, and pKa is the negative logarithm of Ka. Calculate the pKa from the given Ka value: \(pKa = -\log_{10}(6.76 \times 10^{-4})\)

Calculate the pH values

For each case, plug the pKa value, furoic acid concentration, and sodium furoate concentration into the Henderson-Hasselbalch equation to calculate the pH of each solution: a) pH of the solution formed by adding 25.0 g of furoic acid and 30.0 g of sodium furoate to 0.250 L of water. b) pH of the solution formed by mixing 30.0 mL of 0.250 M HC5H3O3 and 20.0 mL of 0.22 M NaC5H3O3, then diluting to 125 mL. c) pH of the solution prepared by adding 50.0 mL of 1.65 M NaOH to 0.500 L of 0.0850 M HC5H3O3.

Key concepts

Acid-Base Equilibria

Acid-base equilibria is a fundamental concept in chemistry which involves the equilibrium between an acid and its conjugate base in a solution. It is the basis for understanding how acids and bases react with each other and with other substances. Specifically, it deals with the extent to which an acid donates a proton to a base, and vice versa, under equilibrium conditions.

The strength of an acid or a base is measured by its dissociation constant, represented by Ka for acids and Kb for bases. A higher Ka value means a stronger acid because it indicates a greater tendency to lose a proton. Inversely, a lower pKa (the negative logarithm of Ka) implies a stronger acid. Understanding this equilibrium is crucial for predicting the behavior of acids and bases in various chemical reactions and for the calculation of pH in different solutions.

Calculation of pH

The pH of a solution is a measure of its acidity or basicity, which is derived from the concentration of hydrogen ions \( [H^+] \). This scale ranges from 0 to 14, where 7 is neutral, values below 7 are acidic, and above 7 are basic. The pH value is calculated by taking the negative logarithm of the hydrogen ion concentration: \( pH = -\log_{10} [H^+] \).

For acidic and basic solutions, the pH can be found using the dissociation constants of the acid \( (Ka) \) or base \( (Kb) \). The calculation often involves an understanding of the solution's concentration, the degree of dissociation, and for buffer solutions, the concentrations of both the acid and its conjugate base. By mastering pH calculations, students can determine the acidity or basicity of a solution which is important in various fields including chemistry, biology, environmental science, and medicine.

Buffer Solutions

Understanding Buffers

Buffer solutions are special solutions that can resist changes in pH when small amounts of acid or base are added. They are typically made up of a weak acid and its conjugate base, or a weak base and its conjugate acid. The ability of buffers to maintain a stable pH is essential in biological systems and industrial applications where controlling pH is vital.

Buffers work based on the equilibrium between the weak acid \( (HA) \) and its conjugate base \( (A^-) \). When a strong acid is added to the buffer, it reacts with the conjugate base, minimizing the change in pH. Conversely, when a strong base is added, it reacts with the weak acid to form the conjugate base, again stabilizing the pH.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a chemical process, resulting in no net change in the concentration of reactants and products over time. At equilibrium, the reactants and products are present in a fixed ratio, not necessarily in equal concentrations, but stable.

This concept is not just about static situations but dynamic ones, where the conversion of reactants to products and vice versa are happening continuously but at equal rates, maintaining constant concentrations. The understanding of chemical equilibrium is essential for predicting the outcomes of reactions and designing processes that optimize the production of desired substances in the chemical industry, as well as for a myriad of biological processes.