Question

Consider the titration of \(25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCO}_{2} \mathrm{H}\) with \(0.250\) M NaOH. How many milliliters of base are required to reach the equivalence point? Calculate the \(\mathrm{pH}\) at each of the following points: (a) After the addition of \(7.0 \mathrm{~mL}\) of base (b) Halfway to the equivalence point (c) At the equivalence point (d) After the addition of \(25.0 \mathrm{~mL}\) of base

Short answer

To reach the equivalence point, 20.0 mL of NaOH is required. pH values are: (a) 3.52, (b) 3.74, (c) 8.87, (d) 12.40.

Step-by-step solution

Calculate the moles of HCO2H

The initial moles of formic acid (\(\mathrm{HCO}_2\mathrm{H}\)) can be calculated by multiplying the volume of the acid solution by its molarity. We have \(25.0 \, \mathrm{mL} = 0.0250 \, \mathrm{L}\) and \(\mathrm{M} = 0.200 \, \mathrm{M}\).So, the moles of \(\mathrm{HCO}_2\mathrm{H} = 0.0250 \, \mathrm{L} \times 0.200 \, \mathrm{M} = 0.00500 \, \mathrm{mol}\).

Calculate the volume of NaOH at equivalence point

At the equivalence point, the moles of \(\mathrm{NaOH}\) added will equal the moles of \(\mathrm{HCO}_2\mathrm{H}\) initially present. Using \(0.250 \, \mathrm{M}\) \(\mathrm{NaOH}\), we find the required volume \(V\) by:\[V = \frac{\text{moles of } \mathrm{HCO}_2\mathrm{H}}{\text{molarity of } \mathrm{NaOH}} = \frac{0.00500 \, \mathrm{mol}}{0.250 \, \mathrm{M}} = 0.0200 \, \mathrm{L} = 20.0 \, \mathrm{mL}\].

Calculate the pH after addition of 7.0 mL of base

First, calculate the moles of \(\mathrm{NaOH}\) in \(7.0 \, \mathrm{mL}: \(7.0 \, \mathrm{mL} = 0.0070 \, \mathrm{L}\) \times 0.250 \, \mathrm{M} = 0.00175 \, \mathrm{mol}\).Moles of \(\mathrm{HCO}_2\mathrm{H}\) remaining: \(0.00500 - 0.00175 = 0.00325 \, \mathrm{mol}\).The concentration of \(\mathrm{HCO}_2\mathrm{H}\) is \(\frac{0.00325 \, \mathrm{mol}}{0.0250+0.0070 \, \mathrm{L}} = 0.0971 \, \mathrm{M}\).The concentration of \(\mathrm{HCO}_2^-\) (formate ion) is \(0.00175 \, \mathrm{mol} \div 0.0320 \, \mathrm{L} = 0.0547 \, \mathrm{M}\).Use the following equation of the Henderson-Hasselbalch equation:\[\mathrm{pH} = \mathrm{p}K_a + \log \left(\frac{\mathrm{[A^-]}}{\mathrm{[HA]}}\right)\]where \(\mathrm{p}K_a \) for \(\mathrm{HCO}_2\mathrm{H} \) is approximately \(3.74\).\[\mathrm{pH} = 3.74 + \log \left(\frac{0.0547}{0.0971}\right) = 3.52\].

pH halfway to the equivalence point

At the midpoint of titration, the concentration of \(\mathrm{HCO}_2^-\) equals the concentration of \(\mathrm{HCO}_2\mathrm{H}\), so the ratio in the Henderson-Hasselbalch equation is 1.\[\mathrm{pH} = \mathrm{p}K_a = 3.74\].

Calculate the pH at the equivalence point

Once \(20.0 \, \mathrm{mL}\) of \(\mathrm{NaOH}\) has been added, all \(\mathrm{HCO}_2\mathrm{H}\) has been converted to \(\mathrm{HCO}_2^-\). The formate ion \(\mathrm{HCO}_2^-\) is a weak base, and its concentration is\(\frac{0.00500 \, \mathrm{mol}}{0.0450 \, \mathrm{L}} = 0.111 \, \mathrm{M}\).The \(\mathrm{K_b}\) can be calculated using \(\mathrm{K_a}\) of \(\mathrm{HCO}_2\mathrm{H}: \mathrm{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}}\approx 5.56 \times 10^{-11}\). Use this to find \(\mathrm{OH}^-\) concentration via\[\mathrm{K_b} = \mathrm{[OH]^-}^2 = 5.56 \times 10^{-11}\rightarrow [\mathrm{OH}^-] = \sqrt{5.56 \times 10^{-11}} = 7.45 \times 10^{-6}\,\]\(\mathrm{pOH} = -\log(7.45 \times 10^{-6}) = 5.13\). Convert to pH:\[\mathrm{pH} = 14 - 5.13 = 8.87\].

Calculate the pH after the addition of 25.0 mL of base

After adding 25.0 \, \mathrm{mL} of \(\mathrm{NaOH}\): \(\mathrm{NaOH}\) over equivalence point is \(0.00500 - 0.00625 = -0.00125 \, \mathrm{mol} \text{ excess }\mathrm{OH}^-\).The total volume is \(50.0 \, \mathrm{mL} = 0.0500 \, \mathrm{L}\), so the concentration of excess \(\mathrm{OH}^-\) is \(0.00125 \, \mathrm{mol}/0.0500 \, \mathrm{L} = 0.0250 \, \mathrm{M}\).Find \(\mathrm{pOH}\): \[\mathrm{pOH} = -\log(0.0250) = 1.60\].Convert to pH:\[\mathrm{pH} = 14 - 1.60 = 12.40\].

Key concepts

Equivalence Point

During a titration, the equivalence point is a pivotal moment. It is when the moles of titrant added equal the moles of the substance being titrated. In a titration of a weak acid like formic acid (\( \text{HCO}_2\text{H} \)) with a strong base like sodium hydroxide (\( \text{NaOH} \)), the reaction reaches the equivalence point when all of the acid has reacted with the base. This moment is key because it signifies that the acid has been completely neutralized.

In our specific example, for the equivalence point:

• We have an initial amount of 0.0050 mol of \( \text{HCO}_2\text{H} \).
• When exactly 0.020 L (20.0 mL) of \( \text{NaOH} \) is added (from a 0.250 M solution), this amount matches the moles of \( \text{HCO}_2\text{H} \) initially present.

At this point, the \( \text{HCO}_2\text{H} \) has been entirely converted to its conjugate base, \( \text{HCO}_2^- \), marking the completion of the acid-base reaction.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a valuable tool for estimating the pH of a solution during titration. It provides a link between the \( \text{pH} \), \( \text{pK}_a \) of the acid, and the concentration ratio of the deprotonated (A\^-) and protonated (HA) forms. The equation is expressed as:

\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]This formula is especially useful when titrating weak acids or bases because it accounts for the reversible nature of the protonation and deprotonation processes.

In our exercise:

• When 7.0 mL of base has been added, the concentrations of formic acid and formate ion are known, enabling the use of the Henderson-Hasselbalch equation to find the pH, calculated to be \( 3.52 \).
• At the halfway point to the equivalence point, the concentrations of \( \text{HA} \) and \( \text{A}^- \) are equal, simplifying our equations as \( \text{pH} = \text{pK}_a \).

pH Calculation

Calculating \( \text{pH} \) at different points in a titration allows us to understand the progress of the reaction. The pH value alters significantly based on how much titrant has been added because it reflects the relative concentration of hydronium (\( \text{H}^+ \)) or hydroxide (\( \text{OH}^- \)) ions in solution.

• Before the equivalence point, the \( \text{pH} \) is determined considering the existing acid presence and any conjugate base formed. This is calculated particularly during the addition of smaller volumes of base.
• At the equivalence point, the only species present from the original weak acid are its conjugate base, requiring different calculations such as solving for \( \text{pOH} \) and then converting it to \( \text{pH} \).
• Beyond the equivalence point, the solution mainly contains excess base, and \( \text{pH} \) is found by considering this additional \( \text{OH}^- \) contribution.

For example, when 25.0 mL of base was added, the resulting pH was high (\( 12.40 \)), indicating a basic solution beyond unity of the equivalence point.

Weak Acid Titration

Titrating a weak acid, such as formic acid, with a strong base involves multiple stages where the chemical dynamics evolve remarkably. Weak acids only partially dissociate in water, which means that at various stages of the titration, both the undissociated acid and its conjugate base coexist.

In a titration setup:

• The initial solution mostly contains the weak acid, and its pH can be calculated using the weak acid dissociation constant (\( \text{K}_a \)).
• As titrant is added, the weak acid reacts to form its conjugate base, and this creates a buffer system that resists significant pH changes until closer to the equivalence point.
• Past the equivalence point, the solution transitions into a base-dominated solution where any excess strong base determines the pH.

This process of titration not only measures the amount of acid but also gives insight into the acid's strength and behavior in an aqueous solution.