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 \(M \mathrm{NaOH}\) 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 \(M \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

Short answer

a. Halfway point: \(pH \approx 3.79\) Equivalence point: \(pH \approx 7.87\) b. Halfway point: \(pH \approx 10.61\) Equivalence point: \(pH \approx 7.66\) c. Halfway point: \(pH \approx 7.0\) Equivalence point: \(pH \approx 7.0\)

Step-by-step solution

Titration a: Halfway Point#Region_content#To find the pH at the halfway point for titration a, we need to use the Henderson-Hasselbalch equation given below, which states: \(pH = pK_a + \log \frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\) Since we know the concentration of the weak acid (\(0.10 M\)), HC7H5O2, and that it is equal to the concentration of its conjugate base at the halfway point, we can plug these values into the Henderson-Hasselbalch equation, as well as the given \( K_a \) value. \( \)

Titration a: Halfway Point Calculations#Region_content#First, let's calculate \(pK_a\): \( pK_a = -\log{K_a} = -\log(6.4 \times 10^{-5}) \) Next, we will plug the values into Henderson-Hasselbalch equation: \( pH = (-\log(6.4 \times 10^{-5})) + \log \frac{[0.10]}{[0.10]} \) \( \)

Titration a: Equivalence Point#Region_content#At the equivalence point, the moles of acid (\(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\)) are equal to the moles of base (\(\mathrm{NaOH}\)). We first calculate the moles of acid and base before solving for the equilibrium concentrations and pH at the equivalence point. \( \)

Titration a: Equivalence Point Calculations#Region_content#Calculate the moles of \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(\mathrm{NaOH}\): Moles of acid: \((0.1 \textrm{ M})(0.1 \textrm{ L}) = 0.01 \textrm{ mol}\) Amount of \(\mathrm{NaOH}\) required to reach the equivalence point (equal to moles of acid): \(0.01 \textrm{ mol}\) Volume of \(\mathrm{NaOH}\) required to reach the equivalence point: \(\frac{0.01 \textrm{ mol}}{0.1 \textrm{ M}} = 0.1 \textrm{ L}\) Total volume at equivalence point: \(0.1 \textrm{ L} + 0.1 \textrm{ L} = 0.2 \textrm{ L}\) Now we find the pH at the equivalence point using the ion-product of water: \(K_w = [\mathrm{H}^+][\mathrm{OH}^-] \) And the \(K_a\) of the conjugate base (\(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2}^-\)): \(K_a = [\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}][\mathrm{OH}^{-}]/[\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2}^{-}]\) \( \)

Titration b: Halfway Point#Region_content#To find the pH at the halfway point for titration b, we need to use the equation for pOH: \(pOH = pK_b + \log \frac{[\mathrm{BH}^+]}{[\mathrm{B}]}\) Then convert the pOH value to pH value using: \(pH = 14 - pOH\) \( \)

Titration b: Halfway Point Calculations#Region_content#First, let's calculate \(pK_b\): \( pK_b = -\log{K_b} = -\log(5.6 \times 10^{-4}) \) Next, we will plug the values into pOH equation: \( pOH = (-\log(5.6 \times 10^{-4})) + \log \frac{[0.10]}{[0.10]} \) Finally, convert pOH to pH: \( pH = 14 - pOH \) \( \)

Titration b: Equivalence Point#Region_content#At the equivalence point, the moles of base (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\)) are equal to the moles of acid (\(\mathrm{HNO}_{3}\)). We first calculate the moles of base and acid before solving for the equilibrium concentrations and pH at the equivalence point. \( \)

Titration b: Equivalence Point Calculations#Region_content#First, calculate the moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) and \(\mathrm{HNO}_{3}\): Moles of base: \((0.1 \textrm{ M})(0.1 \textrm{ L}) = 0.01 \textrm{ mol}\) Amount of \(\mathrm{HNO}_{3}\) required to reach the equivalence point (equal to moles of base): \(0.01 \textrm{ mol}\) Volume of \(\mathrm{HNO}_{3}\) required to reach the equivalence point: \(\frac{0.01 \textrm{ mol}}{0.2 \textrm{ M}} = 0.05 \textrm{ L}\) Total volume at equivalence point: \(0.1 \textrm{ L} + 0.05 \textrm{ L} = 0.15 \textrm{ L}\) Now we find the pH at the equivalence point using the following equation: \(pH = 14 - pOH\) Where pOH can be calculated using the ion-product of water: \(K_w = [\mathrm{H}^+][\mathrm{OH}^-] \) \( \)

Titration c: Halfway Point#Region_content#In the titration of a strong acid (\(\textrm{HCl}\)) with a strong base (\(\textrm{NaOH}\)), pH at the halfway point can be directly determined by the average of their pKa values. \( \)

Titration c: Halfway Point Calculations#Region_content#Determine the average pKa value of HCl and NaOH: pH = \( \frac{pK_{a, \textrm{HCl}} + pK_{a, \textrm{NaOH}}}{2}\) Since \(\textrm{HCl}\) and \(\textrm{NaOH}\) are strong acids and bases, their pKa values are negligible, and the pH at the halfway point is about 7.0. \( \)

Titration c: Equivalence Point#Region_content#To find the pH at the equivalence point for a strong acid-strong base titration, we can directly determine the concentration of \([\mathrm{H}^+]\) ions in the solution. \( \)

Titration c: Equivalence Point Calculations#Region_content#First, calculate the moles of \(\textrm{HCl}\) and \(\textrm{NaOH}\): Moles of acid: \((0.5 \textrm{ M})(0.1 \textrm{ L}) = 0.05 \textrm{ mol}\) Amount of \(\textrm{NaOH}\) required to reach the equivalence point (equal to moles of acid): \(0.05 \textrm{ mol}\) Volume of \(\textrm{NaOH}\) required to reach the equivalence point: \(\frac{0.05 \textrm{ mol}}{0.25 \textrm{ M}} = 0.2 \textrm{ L}\) Total volume at equivalence point: \(0.1 \textrm{ L} + 0.2 \textrm{ L} = 0.3 \textrm{ L}\) Now we find the concentration of \([\mathrm{H}^+]\) and \([\mathrm{OH}^-]\) ions in the solution: \([\mathrm{H}^+] = \frac{\text{moles of } \mathrm{H}^+}{\textrm{total volume}} = \frac{0.05 \textrm{ mol}}{0.3 \textrm{ L}}\) \([\mathrm{OH}^-] = \frac{\text{moles of } \mathrm{OH}^-}{\textrm{total volume}} = \frac{0.05 \textrm{ mol}}{0.3 \textrm{ L}}\) Finally, calculate pH using the following equation: \(pH = -\log [\mathrm{H}^+]\)

Key concepts

Equivalence Point

In titration, the equivalence point is the moment when the amount of acid equals the amount of base added, resulting in a neutralization reaction. At this point, the moles of hydrogen ions (\( \mathrm{H}^+\)) exactly equal the moles of hydroxide ions (\(\mathrm{OH}^-\)). This doesn't always mean the pH is 7.
The pH at the equivalence point depends on the nature of the acid and base involved:

• For strong acid-strong base titrations, the equivalence pH is around 7 because both ions neutralize each other perfectly.
• For weak acid-strong base titrations, the pH tends to be greater than 7 as the conjugate base formed generates (\(\mathrm{OH}^-\)).
• Weak base-strong acid titrations will have an equivalence point pH less than 7 because the conjugate acid formed generates (\(\mathrm{H}^+\)).

To calculate the pH at the equivalence point, determine the solution's total volume and the new concentrations of ions, using the ion-product constant for water (\(K_w = 1.0 \times 10^{-14}\)). Then, evaluate the pH based on whether excess (\(\mathrm{H}^+\)) or (\(\mathrm{OH}^-\)) ions predominate.

Halfway Point

The halfway point in a titration is when half of the acid or base has been neutralized. This point is crucial because the concentration of the weak acid equals its conjugate base. At this juncture, the pH can be directly determined using the pKa of the acid in the Henderson-Hasselbalch equation:\[pH = pK_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]Since \([\text{A}^-] = [\text{HA}]\), the equation simplifies to \(pH = pK_a\).
This simplification means at the halfway point, the pH directly corresponds to the pKa value of the weak acid. This characteristic is helpful because it provides a quick calculation and a simple way to discern the acid’s dissociation constant (\(K_a\)).

• For weak acid titrations, this point is particularly relevant and instrumental for various calculations and characterizations.
• It's a reliable spot for identifying the acid or base strength, significance in buffer solutions, and in understanding how titration curves are shaped.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a vital tool in chemistry for simplifying the calculation of pH in buffer solutions. It relates pH to pKa and the relative concentrations of a weak acid (\(\text{HA}\)) and its conjugate base (\(\text{A}^-\)):\[pH = pK_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\].
This equation proves especially useful in titrations because:

• It provides a straightforward way to calculate pH during the titration process before and after reaching the equivalence point.
• While technically derived from the acid dissociation equation, it simplifies mathematical handling by using logarithmic relationships applicable in diluted solutions.

In a practical sense, this equation helps chemists understand the behavior of acids and bases in buffer solutions, allowing them to design systems that maintain a consistent pH level even when small amounts of acid or base are added.