Problem 85 - Chemistry: The Central Science

Question

Furoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) 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 (a) a solution formed by adding \(25.0 \mathrm{~g}\) of furoic acid and \(30.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.250 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(20.0 \mathrm{~mL}\). of \(0.22 \mathrm{M} \mathrm{NaC} \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\) 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 \(0.500 \mathrm{~L}\) of \(0.0850 \mathrm{M} \mathrm{HC} \mathrm{H}_{3} \mathrm{O}_{3}\).

Step-by-Step Solution

Answer

The pH values of the three solutions are: (a) 3.22, (b) 2.92, and (c) 7.92.

Step 1: Calculate moles of furoic acid and sodium furoate

First, we must convert the masses of furoic acid and sodium furoate into moles. To do this, we will use the molar masses of each compound: Molar mass of furoic acid (HC₃H₃O₃): 112.06 g/mol Molar mass of sodium furoate (NaC₃H₃O₃): 128.06 g/mol Moles of furoic acid = (25 g) / (112.06 g/mol) = 0.223 mol Moles of sodium furoate = (30 g) / (128.06 g/mol) = 0.234 mol

Step 2: Calculate the concentrations of furoic acid and sodium furoate

Next, we calculate the molar concentrations of both furoic acid and sodium furoate by dividing the moles by the total volume of the solution: Volume of solution = 0.250 L Concentration of furoic acid = (0.223 mol) / (0.250 L) = 0.892 M Concentration of sodium furoate = (0.234 moles) / (0.250 L) = 0.936 M

Step 3: Calculate pH using the Henderson-Hasselbalch equation

We can now apply the Henderson-Hasselbalch equation to find the pH of the solution: pH = pKa + log ([A⁻]/[HA]) Given: Ka = 6.76 x 10⁻⁴ pKa = -log(Ka) = 3.17 pH = 3.17 + log (0.936 / 0.892) = 3.22 (b) Calculate pH of solution formed by mixing HC₃H₃O₃ and NaC₃H₃O₃ solutions

Step 1: Calculate the moles of furoic acid and sodium furoate

First, we must calculate the moles of furoic acid and sodium furoate in the initial solutions: Moles of furoic acid = (0.250 M) * (0.030 L) = 0.0075 mol Moles of sodium furoate = (0.22 M) * (0.020 L) = 0.0044 mol

Step 2: Calculate final concentrations after mixing and dilution

Next, we will calculate the final concentration of furoic acid (HA) and sodium furoate (A⁻) after mixing and diluting the total volume to 0.125 L: Final concentration of furoic acid = (0.0075 mol) / (0.125 L) = 0.0600 M Final concentration of sodium furoate = (0.0044 moles) / (0.125 L) = 0.0352 M

Step 3: Calculate pH using the Henderson-Hasselbalch equation

Now, we apply the Henderson-Hasselbalch equation to determine the pH of the solution: pH = 3.17 + log (0.0352 / 0.0600) = 2.92 (c) Calculate pH of solution prepared by adding NaOH to HC₃H₃O₃

Step 1: Perform stoichiometry to find the moles of furoic acid and sodium furoate after reaction

First, we need to calculate how much furoic acid reacts with the given amount of NaOH: Moles of NaOH = (1.65 M) * (0.050 L) = 0.0825 mol Moles of furoic acid = (0.0850 M) * (0.500 L) = 0.0425 mol Since all NaOH will react with the furoic acid: Moles of furoic acid after reaction = 0.0425 - 0.0825 = -0.040 mol This cannot be negative, so all the furoic acid is consumed and 0 mol remains. Moles of sodium furoate after reaction = 0.0425 + 0.0825 = 0.125 mol

Step 2: Calculate final concentrations after reaction

Next, find the final concentrations: Final concentration of furoic acid = 0 mol / (0.550 L) = 0 M Final concentration of sodium furoate = 0.125 mol / (0.550 L) = 0.227 M

Step 3: Calculate pH using the Henderson-Hasselbalch equation

Finally, apply the Henderson-Hasselbalch equation to find the pH: Note that since there is no furoic acid left in the solution, the pH will be determined solely by the presence of the sodium furoate in the solution and the hydrolysis of the resulting ion. pH = 3.17 + log ([A⁻]/[HA]) Since [HA] is 0, we cannot use the equation directly. Instead, we must use the conjugate base ionization reaction: C₃H₃O₃⁻ + H₂O ⇌ HC₃H₃O₃ + OH⁻ This reaction leads to a relation between Kb and Ka for a conjugate acid-base pair: Kb = Kw / Ka = (1 x 10⁻¹⁴) / (6.76 x 10⁻⁴) = 1.48 x 10⁻¹¹ Now, we use the Kb value and the concentration of sodium furoate to find the pOH: Kb = ([OH⁻]²) / (0.227 M) [OH⁻]² = (1.48 x 10⁻¹¹) * (0.227 M) [OH⁻] = 8.29 x 10⁻⁷ pOH = -log(8.29 x 10⁻⁷) = 6.08 Now we can find the pH by taking the difference from 14: pH = 14 - pOH = 14 - 6.08 = 7.92

Key Concepts

Understanding the Henderson-Hasselbalch Equation in pH Calculation

The Henderson-Hasselbalch equation is a key player when it comes to calculating the pH of buffer solutions in acid-base chemistry. It correlates the pH, pKa (the negative logarithm of the acid dissociation constant, Ka), and the molar concentrations of an acid and its conjugate base. Presented as \( pH = pKa + \log(\frac{[A^-]}{[HA]}) \), the equation serves as a simplified approach for assessing the acidity or basicity of a solution.

This relationship is especially useful when dealing with a mixture of a weak acid and its salt form, as seen in textbook problems regarding buffer solutions. In our exercise, the Henderson-Hasselbalch equation was employed to predict the pH of various mixtures involving furoic acid and sodium furoate. By understanding pKa and how the log function translates the ratio of the acid and base forms to a scale of pH, students can better grasp the dynamics of buffers in maintaining pH stability. It's also important to note that this equation assumes that the activity coefficients are equal, which is generally a good approximation for dilute solutions.

Exploring the Fundamentals of Acid-Base Chemistry

Acid-base chemistry is central to understanding the behavior of molecules in solution and their reaction to pH levels. Acids are substances that can donate protons (H+), whereas bases accept protons. The strength of an acid or base is determined by its ability to dissociate in water—a process quantitatively described by its dissociation constant (Ka for acids and Kb for bases).

Key Concepts of Acidity and Basicity

Acids and bases come in strong and weak forms. Strong acids and bases dissociate completely in solution, while weak ones do not. This fundamental characteristic influences how substances like furoic acid (a weak acid) interact with bases (like NaOH, a strong base) in our textbook exercise. When a strong base is added to a weak acid, it fully dissociates, neutralizing it and forming water and a salt—in this case, sodium furoate. The pH of the resulting solution will then depend on the concentration and nature of the remaining ionic species.

Molar Concentration and Its Role in pH Calculations

Molar concentration, commonly referred to as molarity, is a measure of the amount of a substance per unit volume of solution, recorded in moles per liter (M). It's a vital concept in stoichiometry and chemistry as a whole, which allows scientists to quantify and compare the amounts of reactants and products in a given reaction.

In the realm of buffering solutions, understanding the molar concentration is essential for applying the Henderson-Hasselbalch equation effectively. As depicted in the exercise, after determining the amount of acid and its conjugate base in moles, their respective molar concentrations were calculated by dividing by the volume of the solution. This step is critical as molarity directly affects the ratio in the logarithmic term of the equation, thereby influencing the pH value. Thus, a firm grasp of how to determine molar concentrations is crucial for students tackling pH and buffer problems.

Stoichiometry: The Quantitative Story of Chemical Reactions

Stoichiometry is the bread and butter of quantitative chemistry. It refers to the calculation of reactants and products in chemical reactions and is integral for predictions about the yield of products or the required quantity of a reactant for a desired reaction outcome. In the context of our problem involving acid-base reactions, stoichiometry helps us determine how much acid will react with a given amount of base.

The Balance in Stoichiometry

In the example of NaOH reacting with furoic acid, stoichiometry helps us outline the extent of the reaction—the amount of furoic acid that will be converted to sodium furoate and any excess base or acid left. This quantitative snapshot provided by stoichiometry, including the moles before and after the reaction, is what feeds into the Henderson-Hasselbalch equation for the final pH calculation. Without a proper stoichiometric analysis, accurately determining the pH of a solution would be challenging, if not impossible.