The \(K_{\mathrm{a}}\) for formic acid is \(1.78 \times 10^{-4} M.\) a. What is the pH of a \(0.1 \mathrm{M}\) solution of formic acid? b. \(150 \mathrm{mL}\) of \(0.1 \mathrm{MNaOH}\) is added to \(200 \mathrm{mL}\) of \(0.1 \mathrm{M}\) formic acid, and water is added to give a final volume of 1 L. What is the pH of the final solution?

Short Answer

Expert verified
a. The pH of the 0.1 M formic acid solution is calculated using an ICE table and a simplified Ka expression. The pH is then found using \(-log[H^+]\). b. After the addition of \(0.1 M NaOH\), we perform stoichiometry calculations to find the remaining moles of formic acid after the neutralization reaction. The problem simplifies back to part a, allowing us to use the ICE table and \(K_a\) expression introduced previously to calculate the pH of the final solution.

Step by step solution

01

- Calculation of the pH of formic acid

Firstly, recognize that formic acid (\(HCOOH\)) is a weak acid, it will partially disassociate into \(HCOO^-\) and \(H^+\). The reaction representing the dissociation of formic acid in water is: \(HCOOH = H^+ + HCOO^-\). Next, set up an ICE (Initial, Change, Equilibrium) table for the reaction. This is initial concentration: \(HCOOH\) - \(0.1M\), \(H^+\) - \(0M\), \(HCOO^-\) - \(0M\).Change in concentration: \(HCOOH\) - \(-xM\), \(H^+\) - \(+xM\), \(HCOO^-\) - \(+xM\).This is the final concentration: \(HCOOH\) - \(0.1-xM\), \(H^+\) - \(xM\), \(HCOO^-\) - \(xM\).Now, based on the value of \(K_{a}\), the expression for the dissociation constant can be written: \[K_{a} = \frac{[H^+][HCOO^-]}{[HCOOH]} = 1.78 \times 10^{-4}\]Now, utilize the assumption that since the acid is weak, \(x\) will be very small and hence \(0.1 - x \approx 0.1\) to simplify the equation:\[1.78 \times 10^{-4} = \frac{x \cdot x}{0.1}\]Solving the equation above for \(x = [H^+]\) by quadratic formula, hence the pH of the solution can be calculated as \(-log[H^+]\).
02

- pH of the solution after adding NaOH

The reaction of formic acid with NaOH is: \(HCOOH + OH^- = HCOO^- + H_2O\). It's a base-acid neutralization reaction, which goes to completion. Calculate the initial moles of acid and base: formic acid moles = 0.1M * 0.2L = 0.02molsodium hydroxide moles = 0.1M * 0.15L = 0.015molThe hydroxide ion will react with the formic acid, leaving an excess of formic acid. The moles of both formic acid and sodium hydroxide after the reaction would be:formic acid moles = 0.02mol - 0.015mol = 0.005molsodium hydroxide moles = 0The molarity of formic acid after dilution is \(M = \frac{n}{V} = \frac{0.005mol}{1L} = 0.005M\). The formic acid then ionizes to form the formate ion and \(H^+\). Now the problem is the same as part a, a new Ka expression can be written for the remaining formic acid and it can be solved in the same way to find the pH.

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Key Concepts

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

Understanding Ka (Acid Dissociation Constant)
When discussing weak acids, one important concept to grasp is the acid dissociation constant, symbolized as Ka. This reveals the extent to which an acid can donate protons to water, i.e., dissociate to form hydronium ions ()). The general equation for an acid, HA, dissociating in water is: ) in which ) and ) are the concentrations of the dissociated ions, respectively, and ) is the concentration of the undissociated acid. A small Ka value indicates a weak acid because it signifies less dissociation into ions. For example, formic acid ()) with a Ka of ) is considered a weak acid. Understanding the value of Ka helps predict the behavior of an acid in solution and, importantly, the pH of the solution – which is a measure of acidity.

For the calculation of pH, the dissociation constant provides the necessary relationship to evaluate how much an acid dissociates. The formula ) can be used to determine ), the concentration of hydronium ions ()), which is essential in calculating the pH as pH = -log ). With the given Ka and the acid's initial concentration, the equation enables us to solve for ) after setting up the problem with an ICE table and applying appropriate approximations.
Deciphering the ICE Table
To calculate the pH of a weak acid solution, chemists often use an ICE table, which stands for Initial, Change, Equilibrium. It is a systematic way to organize what happens during a reaction. Initially, you list the concentrations of all reactants and products. For weak acids like formic acid in water, this would look like: ), ), ). From there, you specify the changes that occur as the acid dissociates (), ), )). Finally, the table reflects the new equilibrium concentrations after accounting for the change ().

Using the ICE table alongside the equilibrium constant equation simplifies solving for ), the change in concentration for the reactants and products. It assumes that the concentration of undissociated acid remains almost the same () as its initial value, which is reasonable when dealing with weak acids and small dissociation rates, thus simplifying the algebra required for finding ). This approximation is fundamental for calculating the pH in part a of our exercise.
Weak Acid Behavior
Understanding weak acid behavior is crucial in pH calculations. Unlike strong acids, which completely ionize in water, weak acids only partially dissociate. This means that when adding a weak acid like formic acid () to water, not all formic acid molecules will become ions; instead, a dynamic equilibrium is established between the undissociated acid and the ions produced. The extent of this dissociation is given by the Ka value of the acid.

Weak acids also have a less pronounced change in pH when neutralized with a base like NaOH than strong acids would. Part b of the textbook problem demonstrates this concept by having ) react with ), where the reaction proceeds to completion, and uses the mole ratio to find the remaining concentration of the acid. The subsequent change in pH is then calculated, considering the weak acid’s equilibrium behavior, which is not as straightforward as the immediate and total neutralization seen with strong acids.
Neutralization Reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt. In the context of the exercise, the reaction between ) (a weak acid) and ) (a strong base) demonstrates a neutralization reaction where the result is a formate ion ()) and water ()) This kind of reaction typically goes to completion, meaning all the base reacts with the acid available until one of the reactants is fully consumed.

The textbook example illustrates that the moles of ) are less than the moles of ), indicating that upon reaction completion, there will be some leftover formic acid. It is essential to understand this concept because it affects the pH calculation post-reaction, as any excess acid or base would impact the hydrogen ion concentration in the final solution. In our case, we calculate the remaining concentration of the acid after the neutralization to determine the final pH of the solution.

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Most popular questions from this chapter

Given a solution of \(0.1 \mathrm{M}\) HEPES in its fully protonated form, and ready access to \(0.1 \mathrm{M} \mathrm{HCl}, 0.1 \mathrm{M} \mathrm{NaOH}\) and distilled water, describe the preparation of 1 L of 0.025 M HEPES buffer solution, \(\mathrm{pH} 7.8\)

Given \(0.1 \mathrm{M}\) solutions of acetic acid and sodium acetate, describe the preparation of \(1 \mathrm{L}\) of \(0.1 \mathrm{M}\) acetate buffer at a pH of 5.4.

Given \(0.1 M\) solutions of acetic acid and sodium acetate, describe the preparation of \(1 \mathrm{L}\) of \(0.1 \mathrm{M}\) acetate buffer at a pH of 5.4.

Tris-hydroxymethyl aminomethane (TRIS) is widely used for the preparation of buffers in biochemical research. Shown here is the structure of TRIS in its protonated form: Its acid dissociation constant, \(K_{\mathrm{a}},\) is \(8.32 \times 10^{-9} M .\) You have available at your lab bench a \(0.1 \mathrm{M}\) solution of TRIS in its protonated form, 0.1 \(M\) solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), and ample distilled water. Describe the preparation of a 1 L solution of 0.02 M TRIS buffer, pH 7.8.

Shown here is the structure of triethanolamine in its fully protonated form: Its \(\mathrm{p} K_{\mathrm{a}}\) is \(7.8 .\) You have available at your lab bench \(0.1 \mathrm{M}\) solutions of \(\mathrm{HCl}, \mathrm{NaOH}\), and the uncharged (free base) form of triethanolamine, as well as ample distilled water. Describe the preparation of a 1 L solution of 0.05 M triethanolamine buffer, pH 7.6.

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