Question

WEB A 25.00-mL sample of formic acid, \(\mathrm{HCHO}_{2}\), is titrated with \(39.74 \mathrm{~mL}\) of \(0.117 \mathrm{M} \mathrm{KOH}\). (a) What is \(\left[\mathrm{HCHO}_{2}\right]\) before titration? (b) Calculate \(\left[\mathrm{HCHO}_{2}\right],\left[\mathrm{CHO}_{2}^{-}\right],\left[\mathrm{OH}^{-}\right]\), and \(\left[\mathrm{K}^{+}\right]\) at the equivalence point. (Assume that volumes are additive.) (c) What is the \(\mathrm{pH}\) at the equivalence point?

Short answer

Based on the given problem and step-by-step solution, provide a short answer regarding the concentration of formic acid before titration, the concentrations of different species at the equivalence point, and the pH at the equivalence point. The initial concentration of formic acid (HCHO₂) before titration is 0.186 M. At the equivalence point, the concentrations of potassium ions (K⁺), hydroxide ions (OH⁻), and formate ions (CHO₂⁻) are all 0.0718 M. The pH at the equivalence point is 12.86.

Step-by-step solution

Calculate the moles of KOH used in the titration

To find the moles of KOH, we can use the given volume and molarity:$$ \text{moles of KOH} = \text{volume} \times \text{molarity} = 39.74\,\text{mL} \times 0.117\,\frac{\text{moles}}{\text{L}}$$We need to convert mL to L:$$ 39.74\,\text{mL} \times \frac{1\,\text{L}}{1000\,\text{mL}} = 0.03974\,\text{L}$$Now, we multiply the volume by the molarity:$$ \text{moles of KOH} = 0.03974\,\text{L} \times 0.117\,\frac{\text{moles}}{\text{L}} = 0.00465\,\text{moles} $$

Determine the initial concentration of HCHO₂

At the equivalence point, the moles of HCHO₂ and KOH are equal. So, we can write:$$ \text{moles of HCHO₂} = \text{moles of KOH} = 0.00465\,\text{moles} $$To find the initial concentration of HCHO₂, divide the moles by the initial volume:$$ \left[\mathrm{HCHO}_{2}\right] = \frac{\text{moles of HCHO₂}}{\text{initial volume}} = \frac{0.00465\,\text{moles}}{0.025\,\text{L}} = 0.186\,\frac{\text{moles}}{\text{L}} $$

Calculate the concentrations of all species at the equivalence point

At the equivalence point, all the HCHO₂ has reacted with the KOH to form CHO₂⁻ and water. So, we have:$$ \left[\mathrm{HCHO}_{2}\right] = 0 \\ \left[\mathrm{K}^{+}\right] = \left[\mathrm{OH}^{-}\right] = \left[\mathrm{CHO}_{2}^{-}\right] \\ $$To find the concentrations of these ions, we can use the moles of KOH and the total volume at the equivalence point (initial volume of HCHO₂ + volume of KOH):$$ \text{total volume} = 0.025\,\text{L} + 0.03974\,\text{L} = 0.06474\,\text{L} $$Now, we can find the concentrations:$$ \left[\mathrm{K}^{+}\right] = \left[\mathrm{OH}^{-}\right] = \left[\mathrm{CHO}_{2}^{-}\right] = \frac{0.00465\,\text{moles}}{0.06474\,\text{L}} = 0.0718\,\frac{\text{moles}}{\text{L}} $$

Find the pH at the equivalence point

At the equivalence point, we know [OH⁻] and can use it to find the pOH. Then, we can find the pH using the relationship between pH and pOH:$$ \text{pOH} = -\log\left[\mathrm{OH}^{-}\right] = -\log(0.0718) \approx 1.14 \\ \text{pH} = 14 - \text{pOH} = 14 - 1.14 = 12.86 $$So, the pH at the equivalence point is 12.86.

Key concepts

Formic Acid

Formic acid, chemically known as HCHO₂, is the simplest carboxylic acid and is typically found in ant venom. It's a weak acid, which means it doesn't completely dissociate in water. This partial dissociation is important to understand when dealing with titrations, as the concentration of hydrogen ions ext([H+] ) is not as straightforward to calculate as with strong acids.

When formic acid is in an aqueous solution, it partially ionizes, releasing a proton ext([H+] ) and forming the formate ion CHO₂⁻. This can be written as the reaction: HCHO₂ ↔ H⁺ + CHO₂⁻. This equilibrium is crucial because, as titration proceeds, more CHO₂⁻ is formed as HCHO₂ is consumed. Understanding this concept is key to successfully navigating problems involving weak acid titration like the one in our exercise.

In the context of titration, knowing that formic acid is a weak acid also implies that we reach an equivalence point where amounts of acid and base react completely.

Equivalence Point

The equivalence point in a titration is an exciting phase where the amount of titrant added is just enough to completely neutralize the analyte solution. It signifies that, in our exercise, mol KOH = mol HCHO₂, indicating a stoichiometrically complete reaction.

For this reaction, the equivalence point means that all formic acid ( HCHO₂ ) has reacted completely, and only formate ions ( CHO₂⁻ ) along with water and the remaining base ions are present in the solution. In mathematical terms, at the equivalence point:

• [ HCHO₂ ] = 0
• [ CHO₂⁻ ] = [ K⁺ ] = [OH⁻]

Reaching this point means that the solution's pH is then influenced by the formation of CHO₂⁻ (as it acts as a weak base), rather than by HCHO₂. It's essential to understand this shift of species because it directly impacts calculations like the pH at the equivalence point.

pH Calculation

pH calculation involves determining the acidity or basicity of a solution. For titrations involving weak acids like formic acid, upon reaching the equivalence point, we rely on the concentrations of ions present, specifically the [ OH⁻ ] ions, since the solution becomes basic.

To find the pH at the equivalence point of our related exercise, we first calculate the [ OH⁻ ] concentration using the relationship between moles of base and the total solution volume. Here, [ OH⁻ ] = 0.0718 M, leads us to find the pOH:

pH and pOH have a simple relationship: pH = 14 - 1.14 = 12.86.

This high pH confirms the solution's basic nature at the equivalence point, typical when titrating a weak acid with a strong base.