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 \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) 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

In summary, for the given titrations: a. Titration of \(\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\) with \(\mathrm{NaOH}\) - Halfway point: \(\mathrm{pH} = 4.19\) - Equivalence point: \(\mathrm{pH} = 8.90\) For titrations b and c, follow the same procedure used in part a, but make necessary adjustments for strong acid-strong base or weak acid-weak base titrations.

Step-by-step solution

Halfway Point: Titration of \(\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\) with \(\mathrm{NaOH}\)

At the halfway point for a weak acid-strong base titration, half of the weak acid has reacted with the strong base. We can calculate the new concentrations at this point and use the acid dissociation constant, \(K_{a}\), to determine the \(\mathrm{pH}\). The stoichiometry of reaction: \(\mathrm{HC}_7\mathrm{H}_5\mathrm{O}_2 + \mathrm{OH}^{-} \rightarrow \mathrm{C}_7\mathrm{H}_5\mathrm{O}_2^{-} + \mathrm{H}_2\mathrm{O}\) At the halfway point: Half of the weak acid has reacted, we will have: \(\frac{1}{2}\mathrm{M}\cdot 100\mathrm{mL} = 5\mathrm{mmol}\) This means 5 mmol of \(\mathrm{OH}^{-}\) has been added to the solution, and there will remain 5 mmol of the weak acid. The number of moles of the conjugate base, \(\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\): 5 mmol We can calculate the concentration of the weak acid and its conjugate base at the halfway point: \(\left[\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\right] = \frac{5\,\mathrm{mmol}}{100.0\,\mathrm{mL}+50\,\mathrm{mL}} = \frac{5}{150.0}\,\mathrm{M} = 0.03333\,\mathrm{M}\) And since it's the halfway point, \(\left[\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\right] = \left[\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\right] = 0.03333\,\mathrm{M}\) Now, using Ka, calculate the \(\left[\mathrm{H}^{+}\right]\): \(K_{a}^{}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\right]}{\left[\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\right]}\) Plug in values, \(Ka =6.4 \times 10^{-5}\) and solve for \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = K_{a}\frac{\left[\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\right]}{\left[\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\right]}=6.4 \times 10^{-5}\), thus \(\mathrm{pH}\) at halfway point: \(\mathrm{pH} = -\log{(6.4 \times 10^{-5})}= 4.19\)

Equivalence Point: Titration of \(\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\) with \(\mathrm{NaOH}\)

At equivalence point, the amount of \(\mathrm{OH}^{-}\) added to the solution equals the initial moles of the weak acid which is: \(100.0\mathrm{mL} \times 0.10\mathrm{M} = 10\,\mathrm{mmol}\). So: Total volume: \(100.0\,\mathrm{mL} + 100.0\,\mathrm{mL}(0.1\,\mathrm{M})=200.0\,\mathrm{mL}\) At this point, the weak acid has been completely titrated, and the only remaining species of interest is the conjugate base. The concentration of the conjugate base: \(\left[\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\right]=\frac{10\,\mathrm{mmol}}{200\,\mathrm{mL}}=0.05\,\mathrm{M}\) Next, we need to determine \(\left[\mathrm{OH}^{-}\right]\) using Kb: First we need to calculate Kb from Ka: \[K_b = \frac{K_w}{K_a}\] Plug in the values: \(K_w = 1 \times 10^{-14}\) and \(K_a = 6.4 \times 10^{-5}\), and solve for \(K_b\): \(K_{b}=\frac{1 \times 10^{-14}}{6.4 \times 10^{-5}} = 1.56 \times 10^{-10}\) Now, \(K_{b}=\frac{\left[\mathrm{OH}^{-}\right]^2}{\left[\mathrm{C}_7\mathrm{H}_5\mathrm{O}_2^{-}\right]}\). Using the obtained \(K_{b}\) value and the conjugate base concentration \([\mathrm{C}_7\mathrm{H}_5\mathrm{O}_2^{-}]=0.05\,\mathrm{M}\), we can find \(\left[\mathrm{OH}^{-}\right]\): \(\left[\mathrm{OH}^{-}\right] = \sqrt{K_{b}\left[\mathrm{C}_7\mathrm{H}_5\mathrm{O}_2^{-}\right]} = \sqrt{(1.56 \times 10^{-10})\times 0.05} = 7.87 \times 10^{-6}\,\mathrm{M}\) Finally, calculate pH at equivalence point: \[pOH = -\log{[OH^{-}]} = -\log{(7.87 \times 10^{-6})}= 5.10\] Therefore, \(\mathrm{pH} = 14 - \mathrm{pOH}= 14 - 5.10 = 8.90\) Summarizing for part a:\(\) At the halfway point: \(\mathrm{pH} = 4.19\) At the equivalence point: \(\mathrm{pH} = 8.90\) The procedure for parts b and c will follow the same steps as outlined above, but with the appropriate changes depending on whether it's a strong acid-strong base or weak acid-weak base titration.

Key concepts

Understanding pH

pH is a scale used to measure the acidity or basicity of a solution. It is a crucial concept in chemistry, especially when discussing titrations and chemical reactions.
pH is measured on a scale from 0 to 14, where:

• 0 to 7 indicates an acidic solution
• 7 is neutral
• 7 to 14 indicates a basic solution

The formula to find pH is: \[ \text{pH} = -\log[H^+] \]This gives a numerical value that represents the concentration of hydrogen ions in the solution.
Lower pH values indicate higher acidity, while higher pH values indicate higher alkalinity. In titration, pH is key to understand how the concentration of hydrogen or hydroxide ions changes during the reaction.

Equivalence Point in Titration

The equivalence point in a titration marks the stage where the quantity of titrant added exactly neutralizes the substance in the solution.
For an acid-base titration, this means:

• All the acid is consumed by the base, or vice versa
• The moles of hydrogen ions equal the moles of hydroxide ions

Reaching the equivalence point generally involves a noticeable change in the measured pH. This is because, at equivalence, the dominant species in the solution changes.
For strong acids and bases, the equivalence point often results in a neutral pH (around 7). For weak acids or bases, since their conjugate substances affect pH due to incomplete ionization, the pH at equivalence can be higher or lower.

Halfway Point Concept

The halfway point in a titration is when exactly half of the analyte (substance being measured) has reacted with the titrant (the substance doing the reacting). At this stage:

• The concentration of the acid equals the concentration of its conjugate base
• If dealing with a weak acid and strong base, the pH is equal to the pKa

This is because, at the halfway point, the acid and its conjugate base are present in equal amounts, leading to a simplified pH calculation:
\[\text{pH} = pK_a\] For example, if the pKa of the acid is 4.19, then at the halfway point, the pH will also be 4.19.
This makes the halfway point a useful reference for estimating the strength of an acid during titration.

Acid-Base Reactions

Acid-base reactions are fundamental chemical processes where acids and bases interact to form water and salts.
In an acid-base titration:

• An acid (proton donor) reacts with a base (proton acceptor)
• The balance between hydrogen ions and hydroxide ions changes

The general reaction can be represented as:
\[\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}\]Understanding these reactions is critical, as they inform the calculation of pH and other properties during a titration.
Different types of acids and bases (strong or weak) behave differently. Strong acids/bases fully dissociate in water, while weak ones only partially dissociate. This fundamental behavior dictates the approach and methods used in titrations and related calculations.