Question

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

Short answer

For the titrations in this problem: a. At the halfway point, the pH is equal to 4.19, and at the equivalence point, the pH is found by first calculating the pOH using the ionization equation and then finding the pH using \(pH=14-pOH\). b. At the halfway point, find the pOH using the \(pK_{b}\), then calculate the pH using \(pH=14-pOH\). At the equivalence point, find the pOH using the concentration of the acid and \(K_a\), then calculate the pH using \(pH=14-pOH\). c. At the halfway point, find the pH using the initial concentrations of the strong acid and base. At the equivalence point, the pH is 7 since all the strong acid is neutralized by the strong base.

Step-by-step solution

At the halfway point of the titration, we have added exactly half of the required base to neutralize the weak acid. The ratio of the acid to its conjugate base will be equal to 1 and we can use the Henderson-Hasselbalch equation to calculate the pH: \(\mathrm{pH}=pK_{a}+\log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\). In this case, since the ratio of \([\mathrm{A}^-]\) to \([\mathrm{HA}]\) is 1, the pH is equal to the \(pK_a\) of our acid. #Step 2: Calculate pKa#

To find the pH at the halfway point, we need to find the \(pK_a\) value of the acid first. We can find the \(pK_a\) by taking the negative logarithm of the given \(K_a\): \(pK_{a}=-\log(K_a)=-\log(6.4 \times 10^{-5})\). #Step 3: Calculate pH at halfway point#

Plugging the given values into the equation for pH at the halfway point, we get: \(pH=4.19\) #Step 4: Equivalence point#

At the equivalence point, all of the weak acid is neutralized, and we have only the conjugate base in solution. To find the pH, we will use the hydrolysis reaction of the conjugate base, \(\mathrm{A}^{-}\), with water and solve for the concentration of hydroxide ion. #Step 5: Calculate pOH#

Write an ionization equation involving the conjugate base: \(\mathrm{A}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HA} + \mathrm{OH}^-\). By using the fact that \(K_w = K_a K_b\), solving for the concentration of hydroxide ion, and using the equivalence point volume, we can find the pOH. #Step 6: Calculate pH at equivalence point#

Using the calculated pOH, we can find the pH at the equivalence point by subtracting the calculated pOH from 14: \(pH = 14 - pOH\). Following the same procedure for the other titrations, we have: b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5}\mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\)

Halfway point (pOH)#

At the halfway point: \(\mathrm{pOH}=pK_{b}=-\log(K_{b})=-\log(5.6 \times 10^{-4})\)

Halfway point (pH = 14 - pOH)#

Calculate the pH using the relationship pH and pOH: \(pH = 14 - pOH\)

Equivalence point (pOH)#

At the equivalence point, find the pOH using \(K_a\) and the concentration of the acid.

Equivalence point (pH = 14 - pOH)#

Calculate the pH using the relationship pH and pOH: \(pH = 14 - pOH\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25\mathrm{M} \mathrm{NaOH}\)

Halfway point (pH)#

At the halfway point, find the pH by using the initial concentrations of the strong acid and base involved in the titration.

Equivalence point (pH)#

At the equivalence point, all the strong acid is neutralized by the strong base, forming a neutral solution with a pH of 7.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a crucial mathematical tool used to estimate the pH of a buffer solution, composed of a weak acid and its conjugate base. This equation is expressed as: \[ \mathrm{pH} = pK_a + \log\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right) \]where:

• \(\mathrm{pH}\) is the measure of acidity or alkalinity of a solution,
• \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\),
• \([\mathrm{A}^-]\) is the concentration of the conjugate base,
• \([\mathrm{HA}]\) is the concentration of the acid.

When the concentrations of the acid and its conjugate base are equal, the pH of the buffer solution is equal to the \(pK_a\) of the acid. This occurs at the halfway point of the titration, making it a common application of the equation. Hence, it is a helpful equation for predicting how the pH of a solution can be affected during processes like titration.
By using the Henderson-Hasselbalch equation, students can appreciate the delicate balance a buffer system maintains between an acid and its conjugate base.

Equivalence point

During a titration, the equivalence point refers to the exact moment at which the amount of titrant added is chemically equivalent to the amount of substance initially in the solution. This can be either a weak acid being neutralized by a base or vice versa.

• For a strong acid-strong base titration, the pH at the equivalence point is neutral, commonly around 7.
• In the case of a weak acid titrated by a strong base, the solution is typically basic at the equivalence point due to the production of the conjugate base.
• Conversely, titrating a weak base with a strong acid results in an acidic pH at the equivalence point due to the conjugate acid formation.

Recognizing the equivalence point is key to understanding the completion of the reaction. It signifies that stoichiometrically equivalent amounts of acid and base have reacted, and it often leads to significant changes in the pH of the solution.

pH calculation

Calculating the pH of a solution during a titration requires an understanding of both the amount of titrant added and the acid-base balance at that specific moment.
At different points in the titration, the solution's pH can vary dramatically:

• Before addition of titrant, pH depends on the weak acid or base's initial concentration.
• At the halfway point, the Henderson-Hasselbalch equation is used, where pH \( = pK_a\) for weak acids.
• At the equivalence point, all original acid or base is neutralized, leaving the pH dependent on the resulting conjugate.

Each of these stages involves different calculations, ranging from simple logarithmic functions to balancing chemical equations. These calculations explain how the solution transitions from its initial state to the equivalency and beyond.

Weak acid

A weak acid is a type of acid that does not completely dissociate in solution. In other words, it only partially ionizes, producing a limited amount of hydrogen ions (\(\mathrm{H}^+\)). Common examples include acetic acid (\(\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\)) or benzoic acid. Because of their partial dissociation, weak acids lead to an equilibrium that can be represented by:\[\mathrm{HA}_{(aq)} \leftrightharpoons \mathrm{H}^+_{(aq)} + \mathrm{A}^-_{(aq)}\]where \([\mathrm{H}^+]\) is much lower compared to strong acids.
The measure of this partial dissociation is the acid dissociation constant, \(K_a\). A small \(K_a\) value indicates a weaker acid, i.e., a lower tendency to give up \(\mathrm{H}^+\) ions. Understanding weak acids helps interpret the behaviour of a solution during titration, particularly why the pH changes slowly at the start and sharpens near the equivalence point.

Conjugate base

In the context of an acid-base reaction, a conjugate base is the species formed when an acid donates a proton (\(\mathrm{H}^+\)). Essentially, it's what remains after the acid has given up a hydrogen ion.
For example, if we consider acetic acid \(\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\), when it donates a \(\mathrm{H}^+\) ion, it forms acetate \(\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-\), which is its conjugate base. In a titration, understanding the role of the conjugate base is crucial:

• At the halfway point, an equal amount of conjugate base and acid exist.
• At equivalence, the solution's characteristics are dominated by the conjugate base.
• It determines the basicity of the solution post equivalence point.

Conjugate bases play a pivotal role in buffering systems and are a key concept in understanding acidity shifts during titration, making them fundamental in pH calculations.