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Chapter 14: Problem 44

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

Expert verified
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 (HCHO鈧�) before titration is 0.186 M. At the equivalence point, the concentrations of potassium ions (K鈦�), hydroxide ions (OH鈦�), and formate ions (CHO鈧傗伝) are all 0.0718 M. The pH at the equivalence point is 12.86.

Step by step solution

01

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:$$ \text{moles of KOH} = 0.03974\,\text{L} \times 0.117\,\frac{\text{moles}}{\text{L}} = 0.00465\,\text{moles} $$
02

Determine the initial concentration of HCHO鈧�

At the equivalence point, the moles of HCHO鈧� and KOH are equal. So, we can write:$$ \text{moles of HCHO鈧倉 = \text{moles of KOH} = 0.00465\,\text{moles} $$To find the initial concentration of HCHO鈧�, divide the moles by the initial volume:$$ \left[\mathrm{HCHO}_{2}\right] = \frac{\text{moles of HCHO鈧倉}{\text{initial volume}} = \frac{0.00465\,\text{moles}}{0.025\,\text{L}} = 0.186\,\frac{\text{moles}}{\text{L}} $$
03

Calculate the concentrations of all species at the equivalence point

At the equivalence point, all the HCHO鈧� has reacted with the KOH to form CHO鈧傗伝 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 HCHO鈧� + volume of KOH):$$ \text{total volume} = 0.025\,\text{L} + 0.03974\,\text{L} = 0.06474\,\text{L} $$Now, we can find the concentrations:$$ \left[\mathrm{K}^{+}\right] = \left[\mathrm{OH}^{-}\right] = \left[\mathrm{CHO}_{2}^{-}\right] = \frac{0.00465\,\text{moles}}{0.06474\,\text{L}} = 0.0718\,\frac{\text{moles}}{\text{L}} $$
04

Find the pH at the equivalence point

At the equivalence point, we know [OH鈦籡 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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Formic Acid
Formic acid, chemically known as HCHO鈧�, 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 CHO鈧傗伝. This can be written as the reaction: HCHO鈧� 鈫� H鈦� + CHO鈧傗伝. This equilibrium is crucial because, as titration proceeds, more CHO鈧傗伝 is formed as HCHO鈧� 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 HCHO鈧�, indicating a stoichiometrically complete reaction.

For this reaction, the equivalence point means that all formic acid ( HCHO鈧� ) has reacted completely, and only formate ions ( CHO鈧傗伝 ) along with water and the remaining base ions are present in the solution. In mathematical terms, at the equivalence point:
  • [ HCHO鈧� ] = 0
  • [ CHO鈧傗伝 ] = [ K鈦� ] = [OH鈦籡
Reaching this point means that the solution's pH is then influenced by the formation of CHO鈧傗伝 (as it acts as a weak base), rather than by HCHO鈧�. 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鈦� ] ions, since the solution becomes basic.

To find the pH at the equivalence point of our related exercise, we first calculate the [ OH鈦� ] concentration using the relationship between moles of base and the total solution volume. Here, [ OH鈦� ] = 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.

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Most popular questions from this chapter

Four grams of a monoprotic weak acid are dissolved in water to make \(250.0 \mathrm{~mL}\) of solution with a pH of \(2.56\). The solution is divided into two equal parts, \(\mathrm{A}\) and \(\mathrm{B}\). Solution \(\mathrm{A}\) is titrated with strong base to its equivalence point. Solution B is added to solution A after solution \(\mathrm{A}\) is neutralized. The \(\mathrm{pH}\) of the resulting solution is \(4.26\). What is the molar mass of the acid?

Write a net ionic equation for the reaction between aqueous solutions of (a) sodium acetate \(\left(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) and nitric acid. (b) hydrobromic acid and strontium hydroxide. (c) hypochlorous acid and sodium cyanide. (d) sodium hydroxide and nitrous acid.

Which of the following would form a buffer if added to \(250.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{SnF}_{2} ?\) (a) \(0.100 \mathrm{~mol}\) of \(\mathrm{HCl}\) (b) \(0.060 \mathrm{~mol}\) of \(\mathrm{HCl}\) (c) \(0.040 \mathrm{~mol}\) of \(\mathrm{HCl}\) (d) \(0.040 \mathrm{~mol}\) of \(\mathrm{NaOH}\) (e) \(0.040 \mathrm{~mol}\) of \(\mathrm{HF}\)

A \(0.1375 \mathrm{M}\) solution of potassium hydroxide is used to titrate \(35.00 \mathrm{~mL}\) of \(0.257 M\) hydrobromic acid. (Assume that volumes are additive.) (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of potassium hydroxide is required to reach the equivalence point? (d) What is the \(\mathrm{pH}\) of the solution before any \(\mathrm{KOH}\) is added? (e) What is the \(\mathrm{pH}\) of the solution halfway to the equivalence point? (f) What is the \(\mathrm{pH}\) of the solution at the equivalence point?

A buffer solution is prepared by adding \(5.50 \mathrm{~g}\) of ammonium chloride and \(0.0188\) mol of ammonia to enough water to make \(155 \mathrm{~mL}\) of solution. (a) What is the \(\mathrm{pH}\) of the buffer? (b) If enough water is added to double the volume, what is the \(\mathrm{pH}\) of the solution?
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