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The dissociation constant of acetic acid is \(0.000018\) and that for cyanoacetic acid is \(0.0036\) at \(298 \mathrm{~K}\). What would be the ratio of volumes of the two acid solutions, each containing equal moles of the acids, so that the solutions becomes isohydric? (a) \(1: 1\) (b) \(1: \sqrt{200}\) (c) \(1: 200\) (d) \(200: 1\)

Short Answer

Expert verified
The ratio of volumes of the two acid solutions to achieve isohydric conditions is \(1: \sqrt{200}\), which corresponds to option (b).

Step by step solution

01

Understanding isohydric solutions

Isohydric solutions are solutions that have the same hydrogen ion (H+) concentration. For two solutions to be isohydric, they must have equal moles of H+ ions in equal volumes of solution. Given the dissociation constants (Ka) for acetic acid and cyanoacetic acid, we have to find the ratio of volumes of the two solutions that would result in equal molarities of hydrogen ions released.
02

Determine the hydrogen ion concentration

The hydrogen ion concentration for a weak acid can be determined from its dissociation constant (Ka) and the initial concentration of the acid. Since equal moles of both acids are present, let's assume an equal initial concentration \( C \) for both acids. Using the formula \( [H^+] = \sqrt{Ka \cdot C} \), calculate the concentration of hydrogen ions released by each acid.
03

Equating hydrogen ion concentrations

For the solutions to be isohydric, the concentrations of hydrogen ions must be equal. This means \( [H^+]_{acetic} = [H^+]_{cyanoacetic} \). With the concentrations from the previous step, equate the two to find the ratio of their volumes.
04

Calculate the volume ratio

Since \( [H^+]_{acetic} \) and \( [H^+]_{cyanoacetic} \) are equal for isohydric solutions, from the equation \( \sqrt{Ka_{acetic} \cdot C} = \sqrt{Ka_{cyanoacetic} \cdot C} \), solve for the ratio of volumes (\( V_{acetic}:V_{cyanoacetic} \)). The volume ratio can be found by taking the ratio of the square roots of their dissociation constants since \( C \) cancels out.
05

Find the ratio of the dissociation constants

Calculate the ratio of the square roots of the dissociation constants of acetic acid and cyanoacetic acid. \( \frac{\sqrt{Ka_{acetic}}}{\sqrt{Ka_{cyanoacetic}}} = \sqrt{\frac{Ka_{acetic}}{Ka_{cyanoacetic}}} \).
06

Simplify the ratio to get the final answer

Substitute the given values of \( Ka_{acetic} = 0.000018 \) and \( Ka_{cyanoacetic} = 0.0036 \) into the ratio: \( \sqrt{\frac{0.000018}{0.0036}} = \sqrt{\frac{1}{200}} \). Simplify this to find the ratio of volumes of the two solutions to achieve isohydric conditions.

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Key Concepts

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

Dissociation Constant
The dissociation constant (Ka) of an acid reflects how fully the acid dissociates into its ions in water. For weak acids like acetic and cyanoacetic acid, this constant is particularly important because it does not dissociate completely. Instead, the weak acid establishes an equilibrium between the undissociated acid and its ions.

To understand it better, the dissociation reaction for a generic weak acid, HA, can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \].
The equilibrium constant for this reaction is given by:
\[ Ka = \frac{[H^+][A^-]}{[HA]} \].
The higher the value of Ka, the more the acid dissociates, meaning there are more hydrogen ions in the solution, contributing to its acidity. During the exercise, the respective dissociation constants directed us to the comparative strength of the acids' tendencies to donate protons.
Hydrogen Ion Concentration
The hydrogen ion concentration ([H+]) in a solution determines its acidity and is measured in moles per liter (M). In water, these ions are actually present as hydronium ions (H3O+), but it's common to refer to them simply as H+. The pH of a solution, which indicates its acidity or basicity, is the negative logarithm (base 10) of the hydrogen ion concentration.

Knowing the dissociation constant (Ka) of a weak acid and its initial concentration (C), we can estimate the hydrogen ion concentration using the formula:
\[ [H^+] = \sqrt{Ka \times C} \].
The square root arises because of the quadratic nature of the equilibrium expression for a mono-protic weak acid. For the textbook exercise, calculating [H+] for each acid was essential for comparing their acities and understanding how to create isohydric conditions.
Weak Acid Behavior
Weak acids, like acetic and cyanoacetic acid, only partially dissociate in water. This means they set up an equilibrium between the intact acid molecules and the ions they form upon dissociation.

This behavior is governed by the equilibrium constant (Ka) for the acid dissociation reaction. The value of Ka can be influenced by various factors including temperature. A weak acid will have a smaller Ka value compared to a strong acid, which dissociates completely and for which the concept of a dissociation constant is not ordinarily applied. Weak acids also exhibit a pH-dependent dissociation—meaning their ionization increases with increasing pH. This characteristic was at the heart of our exercise to establish the isohydric point.
Equilibrium Constant Calculations
The equilibrium constant calculations for weak acids are vital in determining the concentration of ions in solution. In the context of our textbook problem, we used the dissociation constants (Ka) to work out the volume ratio for isohydric solutions.

The calculations involve setting up an equilibrium expression, inserting the appropriate concentrations, and solving for the unknown. In cases where the concentrations of products and reactants are equal, the expression simplifies as it did in our exercise, allowing to find the volume ratio effortlessly by comparing the square roots of the two dissociation constants.
Acetic Acid
Acetic acid, a common weak acid known for its presence in vinegar, has a characteristic formula of CH3COOH. Acetic acid's behavior in solution is typical of weak acids; it does not fully dissociate, which means in aqueous solution, it establishes an equilibrium between the undissociated acetic acid molecules and the acetate ions, creating a relatively small Ka value compared to strong acids.

This property of acetic acid affects its hydrogen ion concentration, and therefore, its influence on the pH of a solution was directly examined in our exercise.
Cyanoacetic Acid
Cyanoacetic acid features a cyano group (CN) attached to acetic acid, altering its acid dissociation behavior. With the formula C3H3NO2, cyanoacetic acid exhibits a stronger acidic nature than acetic acid, evidenced by its larger Ka value. This results in a higher concentration of hydrogen ions compared to acetic acid for the same molarity.

Understanding the dissociation properties of cyanoacetic acid allowed us to evaluate the conditions needed for creating isohydric solutions and was crucial in determining the correct volume ratio with acetic acid in the solved exercise.

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Most popular questions from this chapter

An amount of \(0.15\) mole of pyridinium chloride has been added into \(500 \mathrm{ml}\) of 0.2 M pyridine solution. Calculate pH and hydroxyl ion concentration in the resulting solution assuming no change in volume. \(K_{\mathrm{b}}\) for pyridine \(=1.5 \times 10^{-9}\). \((\log 2=0.3, \log 0.3=0.48)\) (a) \(9.0\) (b) \(5.0\) (c) \(8.64\) (d) \(5.36\)

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The solubility of \(\mathrm{Li}_{3} \mathrm{Na}_{3}\left(\mathrm{AlF}_{6}\right)_{2}\) is \(0.0744 \mathrm{~g}\) per \(100 \mathrm{ml}\) at \(298 \mathrm{~K}\). Calculate the solubility product of the salt. (Atomic masses: \(\mathrm{Li}=7, \mathrm{Na}=23, \mathrm{Al}=27, \mathrm{~F}=19)\) (a) \(2.56 \times 10^{-22}\) (b) \(2 \times 10^{-3}\) (c) \(7.46 \times 10^{-19}\) (d) \(3.46 \times 10^{-12}\)

What is the \(\mathrm{pH}\) of \(4 \times 10^{-3} \mathrm{M}-\mathrm{Y}(\mathrm{OH})_{2}\) solution assuming the first dissociation to be \(100 \%\) and second dissociation to be \(50 \%\), where \(Y\) represents a metal cation? \((\log 2=0.3, \log 3=0.48)\) (a) \(11.78\) (b) \(11.22\) (c) \(2.22\) (d) \(2.78\)

Silver ions are slowly added in a solution with \(\left[\mathrm{Br}^{-}\right]=\left[\mathrm{Cl}^{-}\right]=\left[\mathrm{CO}_{3}^{2-}\right]=\left[\mathrm{AsO}_{4}^{3-}\right]\) \(=0.1 \mathrm{M}\). Which compound will precipitate first? (a) \(\operatorname{AgBr}\left(K_{\mathrm{sp}}=5 \times 10^{-13}\right)\) (b) \(\mathrm{AgCl}\left(K_{\mathrm{sp}}=1.8 \times 10^{-10}\right)\) (c) \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\left(K_{\mathrm{sp}}=8.1 \times 10^{-12}\right)\) (d) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=1 \times 10^{-22}\right)\)

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