Question

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A \(25.0-\mathrm{mL}\) sample of \(0.100 M\) lactic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}, \mathrm{p} K_{\mathrm{a}}=\right.\) \(3.86\) ) is titrated with \(0.100 M\) NaOH solution. Calculate the \(\mathrm{pH}\) after the addition of \(0.0 \mathrm{~mL}, 4.0 \mathrm{~mL}, 8.0 \mathrm{~mL}, 12.5 \mathrm{~mL}, 20.0 \mathrm{~mL}\), \(24.0 \mathrm{~mL}, 24.5 \mathrm{~mL}, 24.9 \mathrm{~mL}, 25.0 \mathrm{~mL}, 25.1 \mathrm{~mL}, 26.0 \mathrm{~mL}, 28.0 \mathrm{~mL}\) and \(30.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}\). Plot the results of your calculations as pH versus milliliters of \(\mathrm{NaOH}\) added.

Short answer

The pH at various stages of titrating a 25.0 mL, 0.100 M lactic acid sample (pKa = 3.86) with 0.100 M NaOH can be calculated using the Henderson-Hasselbalch equation. The initial pH before adding NaOH is approximately 3.86. During titration, calculate the moles of lactic acid and NaOH and derive the amounts of HC3H5O3 and C3H5O3- at each stage. Calculate the pH at the given volumes (0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL) and plot the results as a titration curve of pH vs. milliliters of NaOH added. The plot will show the initial pH, buffer region, equivalence point, and pH when excess NaOH is added.

Step-by-step solution

1. Calculate initial pH before titration

Before any NaOH has been added, the pH of the solution is solely determined by the dissociation of the lactic acid. Using the pKa value and the initial concentration of lactic acid, we can estimate the pH. The ionization equilibrium for lactic acid can be written as: \(HC_3H_5O_3 \rightleftharpoons H^+ + C_3H_5O_3^-\) The equilibrium expression of the acid dissociation constant (Ka) is: \(Ka = \frac{[H^+][C_3H_5O_3^-]}{[HC_3H_5O_3]}\) Since pKa = -log(Ka), we can solve for Ka: \(Ka = 10^{-pKa} = 10^{-3.86}\) Let's represent the change in lactic acid concentration as x. At equilibrium, the concentrations of the dissociated H+ and the conjugate base C3H5O3- will be x and the concentration of the undissociated lactic acid will be (0.100 - x). Thus, the Ka expression becomes: \(Ka = \frac{x^2}{0.100 - x}\) Using the Ka value found previously, we can solve for x, which represents the [H+] concentration: \(x = [H^+] = 10^{-3.86} \approx 1.38 \times 10^{-4}\) Now, we can calculate the initial pH using the formula \(pH = -\log[H^{+}]\) \(pH = -\log(1.38 \times 10^{-4}) \approx 3.86\)

2. Calculate pH during titration

During titration, as NaOH is added to the solution, it reacts with the lactic acid, HC3H5O3, neutralizing it, and producing water and the conjugate base C3H5O3- as NaC3H5O3. At each time point, we will calculate the concentrations of the weak acid and its conjugate base and use the Henderson-Hasselbalch equation to calculate the pH. The Henderson-Hasselbalch equation is given by: \(pH = pKa + \log\frac{[C_3H_5O_3^-]}{[HC_3H_5O_3]}\) At each time point during titration, the volume and concentration of NaOH added so far can be calculated and used to find out how much of the weak acid remains non-titrated. First, calculate the moles of lactic acid and NaOH for each volume of NaOH added using the molarity and volume: moles of lactic acid = 0.100 * 25.0mL = 2.5 mmol moles of NaOH = 0.100 * volume of NaOH in mL For each step of the titration, we can derive the amounts of HC3H5O3 and C3H5O3- by considering the limiting reactant. The limiting reactant can either be the lactic acid or the NaOH. #2a. Before the equivalence point (NaOH added < 25.0 mL): In this case, the lactic acid will be the limiting reactant (lactic acid will remain un-neutralized). moles of non-titrated lactic acid = 2.5 mmol - moles of NaOH moles of C3H5O3- = moles of NaOH #2b. At the equivalence point (NaOH added = 25.0 mL): In this case, there is no more lactic acid available as it is all neutralized. moles of non-titrated lactic acid = 0 mmol moles of C3H5O3- = 2.5 mmol (all 2.5 mmol of lactic acid have been converted into the conjugate base) #2c. After the equivalence point (NaOH added > 25.0 mL): In this case, there is excess NaOH. moles of non-titrated lactic acid (HC3H5O3) = 0 mmol (no lactic acid is left) moles of C3H5O3- more than 2.5 mmol Remaining moles of NaOH that didn't react with lactic acid = moles of NaOH - 2.5 mmol Again, for each volume of NaOH added, use the ratios above in the Henderson-Hasselbalch equation to calculate the pH at that point.

3. Plot the results

Now that we have calculated all the pH values for each volume of NaOH added, we can create a plot of pH versus milliliters of NaOH added. This plot will show how the pH changes during the titration process. To create this plot, simply label the x-axis as the "Volume of NaOH added (mL)" and the y-axis as "pH", and plot the pH values calculated for each specific volume of NaOH added on the graph. Remember to connect the points to form a titration curve which should feature an inflexion point around 25 mL, the equivalence point. By examining the titration curve, one can determine the specific points in the titration process, such as the initial pH, the buffer region, the equivalence point, and the pH when excess NaOH is added.

Key concepts

Lactic Acid

Lactic acid, chemically known as lactate or 2-hydroxypropanoic acid, is an organic acid with the formula \( HC_3H_5O_3 \) that plays a vital role in various biochemical processes. It is produced in muscles during intense exercise when oxygen levels are low, and it contributes to the muscle 'burn' sensation. Lactic acid is a weak acid, meaning it does not fully ionize in water, which is an essential factor when considering its behavior during acid-base titration.

Lactic acid in solution exists in equilibrium with its dissociated form, producing hydrogen ions \( H^+ \) and lactate ions \( C_3H_5O_3^- \). This behavior is described by an acid dissociation constant, \( K_a \), which is a quantitative measure of the acid's strength. The corresponding \( pK_a \) value of lactic acid is 3.86, indicating its relative tendency to donate a proton in aqueous solutions.

pH Calculation

pH is a measure of the hydrogen ion concentration in a solution and thus its acidity or alkalinity. The pH scale runs from 0 to 14, with 7 being neutral. A pH less than 7 indicates acidity, while a pH greater than 7 indicates alkalinity. To calculate the pH of a solution, you can use the formula \( pH = -\text{log}[H^+] \), where \( [H^+] \) is the concentration of hydrogen ions.

In the context of titration, initial pH calculation is critical as it sets the baseline for understanding how the titration process will affect the solution's acidity. Calculating the pH of a weak acid like lactic acid involves setting up an equilibrium expression and solving for the hydrogen ion concentration. It's a crucial step in understanding the behavior of the acid under titration and interpreting the resulting titration curve.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is indispensable for pH calculation during the titration of a weak acid with a strong base. It provides a direct connection between the pH of a solution and its acid-base composition. The equation is given by \( pH = pK_a + \text{log} \frac{[A^-]}{[HA]} \), where \( [A^-] \) is the concentration of the conjugate base and \( [HA] \) is the concentration of the weak acid.

This relationship is crucial when you are in the buffer region of a titration curve, where the addition of the titrant does not significantly change the pH of the solution because the weak acid and its conjugate base are present in similar concentrations. The equation is named after its developers, Lawrence Joseph Henderson and Karl Albert Hasselbalch.

Titration Curve

A titration curve is a graph that displays the change in the solution's pH as a titrant is added. For lactic acid titration, the curve typically starts at a lower pH (acidic), gradually increases through the buffer region, and jumps sharply at the equivalence point before leveling out again.

During the early stages of titration, the curve reflects the initial pH, which is important for understanding the strength and concentration of the acid. As more base is added, the curve moves through the buffer region, where the pH does not change much because the acid and its conjugate base are neutralizing each other. The sharp increase in the titration curve near the equivalence point is noteworthy, as this is where equal amounts of acid and base have reacted.

Equivalence Point

The equivalence point in a titration is the juncture at which the amount of titrant added exactly neutralizes the substance present in the solution being titrated. In acid-base titration, it refers to the point where the moles of acid equal the moles of base added. At the equivalence point, the pH of the solution often experiences a rapid change, which can be easily identified on a titration curve by the inflection point.

For a weak acid-strong base titration, the equivalence point pH is typically greater than 7, resulting in a slightly basic solution due to the production of the weak acid's conjugate base. Accurately identifying the equivalence point is vital because it enables the determination of the unknown concentration of an acid or base in a solution. In the provided example, 25.0 mL of NaOH is required to reach the equivalence point with 25 mL of 0.100 M lactic acid, indicating a stoichiometric reaction where the moles of acid and base are equal.