Question

What quantity (moles) of \(\mathrm{NaOH}\) must be added to \(1.0 \mathrm{~L}\) of \(2.0 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) to produce a solution buffered at each \(\mathrm{pH}\) ? a. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) b. \(\mathrm{pH}=4.00\) c. \(\mathrm{pH}=5.00\)

Short answer

In summary, the moles of sodium hydroxide (NaOH) needed to be added in each case are: a. \(2.0 \mathrm{moles}\) b. \(0.366 \mathrm{moles}\) c. \(3.64 \mathrm{moles}\)

Step-by-step solution

Recall the Henderson-Hasselbalch equation

Recall the Henderson-Hasselbalch equation: \[ \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \] Here, \(\mathrm{p}K_{\mathrm{a}}\) is the negative logarithm of the acid dissociation constant for acetic acid, \([\mathrm{A}^-]\) is the concentration of the conjugate base (\(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-\)), and \([\mathrm{HA}]\) is the concentration of acetic acid.

Find the \(\mathrm{p}K_{\mathrm{a}}\) for Acetic Acid

For acetic acid, \(\mathrm{K}_{\mathrm{a}} = 1.8 \times 10^{-5}\). Therefore, \(\mathrm{p}K_{\mathrm{a}} = -\log (\mathrm{K}_{\mathrm{a}}) = -\log (1.8 \times 10^{-5}) \approx 4.74\).

Calculate the Amount of Sodium Hydroxide to be Added for Each Case

Since we are solving for different pH values, let's go case by case: a. \(\mathrm{pH} = \mathrm{p}K_{\mathrm{a}}\) Plug in the given pH value into the Henderson-Hasselbalch equation: \[ 4.74 = 4.74 + \log \frac{[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]} \] b. \(\mathrm{pH} = 4.00\) \[ 4.00 = 4.74 + \log \frac{[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]} \] c. \(\mathrm{pH} = 5.00\) \[ 5.00 = 4.74 + \log \frac{[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]} \]

Calculate the Concentration of the Conjugate Base

For each case: a. \(\log \frac{[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]} = 0\), so \([\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-] = [\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]\) b. \(\log \frac{[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]} = -0.74\), so \([\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-] = 10^{-0.74} [\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]\) c. \(\log \frac{[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]} = 0.26\), so \([\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-] = 10^{0.26} [\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]\)

Calculate the Moles of Sodium Hydroxide to be Added

For each case: a. Since \([\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-] = [\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]\), we need equal amounts of acetic acid and its conjugate base. The initial concentration of acetic acid is \(2.0 \mathrm{M}\). Therefore, we need \(2.0 \mathrm{M} \times 1.0 \mathrm{L} = 2.0 \mathrm{moles}\) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-\), which are produced by adding \(2.0 \mathrm{moles}\) of \(\mathrm{NaOH}\). b. \([\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-] = 10^{-0.74} [\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}] = 0.183\times 2.0 \mathrm{M} = 0.366 \mathrm{M}\) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-\). So, we need to add \(0.366 \mathrm{M} \times 1.0 \mathrm{L} = 0.366 \mathrm{moles}\) of \(\mathrm{NaOH}\). c. \([\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-] = 10^{0.26} [\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}] = 1.82\times 2.0 \mathrm{M} = 3.64 \mathrm{M}\) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-\). So, we need to add \(3.64 \mathrm{M} \times 1.0 \mathrm{L} = 3.64 \mathrm{moles}\) of \(\mathrm{NaOH}\). So, the moles of \(\mathrm{NaOH}\) to be added for each case are: \(2.0 \mathrm{moles}\) (a), \(0.366 \mathrm{moles}\) (b), and \(3.64 \mathrm{moles}\) (c).

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a pivotal relationship in acid-base chemistry. It expresses the \textbf{pH} of a buffer solution in terms of the \textbf{pKa} (the acid dissociation constant) and the ratio of the concentrations of the conjugate base and the acid. Mathematically, it's represented as:
\[ \text{pH} = \text{pKa} + \log \frac{[\text{A}^-]}{[\text{HA}]} \]
This elegant formula allows us to manipulate buffer systems by calculating the required amounts of acid and base to achieve a desired pH level, making it a go-to tool in laboratory and industrial applications. It exemplifies the direct relationship between the pH of the solution and the proportion of the acid and its conjugate base, which is particularly crucial in buffering capacities – enabling solutions to resist changes in pH upon the addition of acids or bases.

Acid Dissociation Constant (Ka)

The acid dissociation constant, known as \( Ka \), is a qualitative measure of an acid's strength. In an aqueous solution, it signifies the tendency of an acid, \( HA \), to dissociate into a proton, \( H^+ \), and its conjugate base, \( A^- \). The larger the \( Ka \) value, the stronger the acid because it implies a greater extent of dissociation. Its negative logarithm is the \textbf{pKa}, which we often use in the Henderson-Hasselbalch equation. For instance, acetic acid (ethanoic acid) has a \( Ka \) value given, through which its \( pKa \) can be computed - a calculation shown in the first step of the solution. Understanding \textbf{Ka} is essential for grasping why certain acids and bases behave the way they do, which underpins the rational design of buffer systems.

Conjugate Base Concentration

The conjugate base concentration is a defining aspect of buffer solutions. When an acid, let's say acetic acid \( (HC_{2}H_{3}O_{2}) \), donates a proton, the remaining species \( (C_{2}H_{3}O_{2}^-) \) is its conjugate base. The ratio of the concentrations of this \textbf{conjugate base} to the original \textbf{acid} is a component of the Henderson-Hasselbalch equation and influences the pH of the buffer solution. Coordination of the conjugate base’s concentration is vital for setting up a buffer with a precise pH, which can be achieved by adding a strong base like \( NaOH \). As the base deprotonates the acid, it increases the conjugate base concentration relative to the acid, driving the pH to rise.

Acid-Base Titration

Acid-base titration is an analytical technique used to determine the concentration of an acid or a base in a solution. It involves the gradual addition of a solution of known concentration, the titrant, to a solution of the substance being analyzed, the analyte, until the reaction between the two is complete as indicated by a marked change in pH known as the equivalence point. This concept relates closely to our exercise problem where \( NaOH \), a strong base, is titrated into acetic acid. By calculating the amount of \( NaOH \) added to reach certain pH levels, we essentially perform a \textbf{titration calculation}. Understanding the principles of titration helps us interpret how the addition of \( NaOH \) shifts the acid/base equilibrium and changes the buffer pH, as demonstrated in the step-by-step solution with varying resulting pH conditions.