Problem 96 - Chemistry: The Central Science in SI Units, Global Edition

Question

How many microliters of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution must be added to \(25.00 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce a buffer with \(\mathrm{pH}=3.75 ?\)

Step-by-Step Solution

Answer

To create a buffer with a pH of 3.75, 6.99 µL of 1.000 M NaOH solution must be added to the 25.00 mL of the 0.1000 M lactic acid solution.

Step 1: Calculate the concentration of lactic acid initially

The given concentration of the lactic acid solution is \(0.1000 \mathrm{M}\). So, Initial concentration of lactic acid, \([\mathrm{HA}]_{\mathrm{initial}} =0.1000 \mathrm{M}\)

Step 2: Calculate the moles of \(\mathrm{NaOH}\) needed using the Henderson-Hasselbalch equation

We are given that the \(\mathrm{pH}=3.75\). Now, we use the Henderson-Hasselbalch equation: We have, \(\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\mathrm{A}^-_{\mathrm{eq}}]}{[\mathrm{HA}_{\mathrm{eq}}]}\) Rearranging for \([\mathrm{A}^-_{\mathrm{eq}}]\), $ [\mathrm{A}^-_{\mathrm{eq}}]= [\mathrm{HA}_{\mathrm{eq}}] \cdot 10^{\left(\mathrm{pH}-\mathrm{p}K_\mathrm{a}\right)} $ Now, we can calculate the moles of the conjugate base \(\mathrm{A}^-\) formed: moles of \(\mathrm{A}^-\) formed \(= \mathrm{moles} \;[\mathrm{HA}]_{\mathrm{initial}} - \mathrm{moles}\; [\mathrm{HA}]_{\mathrm{eq}}\) The moles of \(\mathrm{OH}^-\) ions required \(= \mathrm{moles} \; \mathrm{A}^-\).

Step 3: Calculate the volume of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution required

Using the concentration of \(\mathrm{NaOH}\), we can calculate the volume of the \(\mathrm{NaOH}\) solution required to form the desired buffer solution: Volume of \(\mathrm{NaOH}\) solution \(= \dfrac{\mathrm{moles} \; \mathrm{OH}^-}{[\mathrm{OH}^-]}\) Now, we plug in the values to find the volume of the \(\mathrm{NaOH}\) solution. First, let's find the moles of \(\mathrm{HA}_{\mathrm{eq}}}\): \([\mathrm{HA}_{\mathrm{eq}}] = \left( 0.1000 - \rm{moles \; {A}^-}\right)\phantom{.}\mathrm{M}\) Now, we can use the Henderson-Hasselbalch equation to find \(\mathrm{moles} \; \mathrm{A}^-\): $ [\mathrm{A}^-_{\mathrm{eq}}]= [\mathrm{HA}_{\mathrm{eq}}] \cdot 10^{\left(\mathrm{pH}-\mathrm{p}K_\mathrm{a}\right)} $ \(\Longrightarrow \mathrm{moles} \; \mathrm{A}^- = \frac{0.1000 - \mathrm{moles} \; \mathrm{A}^-}{\mathrm{L}} \cdot 10^{\left(3.75-\log 1.38 \cdot 10^{-4}\right)}\) Solving the equation, we get: \(\mathrm{moles} \; \mathrm{A}^- = 6.989 \cdot 10^{-4}\;\mathrm{mol}\) Finally, we calculate the volume of \(\mathrm{NaOH}\) solution: Volume of \(\mathrm{NaOH}\) solution \(= \dfrac{6.989 \cdot 10^{-4}\;\mathrm{mol}}{1.000 \phantom{.}\mathrm{M}}\) \(\Longrightarrow V_\mathrm{NaOH} = 6.99\;\mathrm{µL}\) Therefore, \(6.99\;\mathrm{µL}\) of \(1.000\;\mathrm{M}\;\mathrm{NaOH}\) solution must be added to the \(25.00\;\mathrm{mL}\) of the \(0.1000\;\mathrm{M}\) lactic acid solution to produce a buffer with \(\mathrm{pH}=3.75\).

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a fundamental part of understanding how buffers work. It's a handy equation used in chemistry to calculate the pH of a buffer solution. The equation is expressed as follows:\[ \mathrm{pH} = \mathrm{pK}_\mathrm{a} + \log \left( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right) \]In this equation,

• \( \mathrm{pH} \) is the measure of acidity/basicity of the solution,
• \( \mathrm{pK}_\mathrm{a} \) is the negative logarithm of the acid dissociation constant, which gives you an idea of how strong the acid is,
• \( [\mathrm{A}^-] \) is the concentration of the conjugate base,
• \( [\mathrm{HA}] \) is the concentration of the undissociated acid.

This equation helps to determine the pH of a solution by considering how much of the acid and its conjugate base are present. It is especially useful for constructing buffer solutions, where maintaining a steady pH is essential. In the problem, we utilized this equation to find the amount of base needed to create a buffer at a specific pH level.
Understanding how to manipulate this equation is crucial in fields such as biochemistry and pharmacology, where precise pH levels need to be maintained.

pH calculation

Calculating the pH of a solution is a core skill in acid-base chemistry. The pH is a way to quantify the acidity or basicity of a solution, with usually a scale ranging from 0 to 14.
The pH is calculated based on the concentration of hydrogen ions (\([\mathrm{H}^+]\)) in the solution, using the formula:\[ \mathrm{pH} = -\log_{10}([\mathrm{H}^+]) \]However, the problem involves creating a buffer solution, we used the Henderson-Hasselbalch equation to indirectly calculate the pH by comparing the concentration ratio of lactic acid and its conjugate base after adding a specific volume of NaOH.
To achieve the desired pH, we first needed to find the pKa of lactic acid, which is a measure of its dissociation. Once known, you can plug it into the Henderson-Hasselbalch equation, along with your target pH value and solved for the concentrations needed.
This calculation enabled us to determine how much of the strong base, NaOH, was needed to alter the chemical equilibrium, ensuring the buffer had a pH of 3.75.
Understanding pH calculation is vital for many scientific and practical applications, like agriculture to maintain soil pH, or in medicine for drug formulation.

Acid-base chemistry

Acid-base chemistry involves understanding the properties and behaviors of acids and bases within solutions. Lactic acid, in the given problem, is a weak acid that partially dissociates into its ions when dissolved in water.
In acid-base reactions, acids donate protons (\( \mathrm{H}^+ \)), while bases accept them, resulting in the formation of water and salts.
Buffer solutions, such as the one we created in the problem, are essential in acid-base chemistry. They resist changes in pH when small amounts of acids or bases are added. Thus, they consist of a weak acid and its conjugate base, or a weak base and its conjugate acid.

• In the problem, lactic acid and its conjugate base form the buffer pair.
• The strong base NaOH was used to convert portions of the lactic acid into its conjugate base, allowing us to create a stable buffer system.

The NaOH acts as a proton acceptor, allowing the reaction to maintain its equilibrium, which dictated how much was needed to achieve the desired pH.
This concept is vital in many natural and industrial processes, including maintaining human blood pH and in chemical manufacturing plants.