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Cyanic acid, HOCN, is a weak acid with a \(K_{a}\) value of \(3.5 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). In a \(0.293 M\) solution of the acid, the degree of ionization is \(3.5 \times 10^{-2}\). Calculate the degree of ionization in a \(0.293 \mathrm{M}\) solution to which sufficient \(\mathrm{HCl}\) is added to make it \(4.19 \times 10^{-2} M \mathrm{HCl}\) in the given volume.

Short Answer

Expert verified
The degree of ionization decreases due to the common ion effect from added HCl.

Step by step solution

01

Understanding the Problem

The problem provides us with the degree of ionization of cyanic acid, HOCN, in a 0.293 M solution. It asks us to determine how this degree of ionization changes when HCl is added to the solution, increasing the concentration of HCl to 0.0419 M.
02

Calculate Initial H鈦 Concentration

First, we need to find the initial concentration of H鈦 ions in the solution before adding HCl. Given the ionization degree is 0.035, and the initial concentration is 0.293 M, the initial concentration of H鈦 from HOCN ionization is: \[ [H^+]_{initial} = 0.293 imes 0.035 = 0.010255 ext{ M} \]
03

New H鈦 Concentration After HCl Addition

Adding HCl will increase the concentration of H鈦 ions. The concentration of H鈦 ions after adding HCl is the sum of the initial H鈦 concentration from the ionization and the concentration of H鈦 from HCl:\[ [H^+]_{new} = 0.010255 ext{ M} + 0.0419 ext{ M} = 0.052155 ext{ M} \]
04

Apply the Ionization Equation With Added HCl

Use the equilibrium expression for the ionization of cyanic acid involving the acid dissociation constant \( K_a \):\[ K_a = \frac{[H^+][OCN^-]}{[HOCN]} \]Substitute the expressions after adding HCl:\[ 3.5 \times 10^{-4} = \frac{[HOCN]_{ionized} imes 0.052155}{0.293 - [HOCN]_{ionized}} \]
05

Solve for Degree of Ionization

Assuming the ionization due to the added HCl is negligible, calculate the ionization due to HOCN using the adjusted H鈦 concentration. Rearrange:\[ [HOCN]_{ionized} = \frac{3.5 \times 10^{-4} imes (0.293 - [HOCN]_{ionized})}{0.052155} \]Let \( x = [HOCN]_{ionized} \), and solve this equation to find the new degree of ionization.

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Key Concepts

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

Weak Acids
Weak acids are fascinating compounds with unique properties that set them apart from strong acids. Unlike strong acids, which completely dissociate into ions in water, weak acids only partially ionize. This means that when a weak acid is dissolved in water, only a small fraction of its molecules turn into hydrogen ions (\( H^+ \) or hydronium ions) and anions. The rest remain as complete molecules.
Understanding the behavior of weak acids involves recognizing that they exist in equilibrium. This balance is influenced by factors like concentration and temperature. Adding more acid to the solution doesn't linearly increase the concentration of ions, because the equilibrium constant of the reaction governs the extent of ionization. It's important to know that the degree of ionization does not exceed 100%, which is a key point differentiating weak acids from their strong counterparts.
Examples of weak acids include acetic acid (\( CH_3COOH \)) and cyanic acid (\( HOCN \)). These acids are usually described by their ability to donate protons (hydrogen ions) to a base, shown by their acid dissociation constant, or \( K_a \). This constant is a measure of the strength of the weak acid, indicating how well it can donate protons.
Degree of Ionization
The degree of ionization is an important concept when studying weak acids. It tells us the fraction of the total acid molecules that have ionized in solution. This is vital for calculating the pH of the solution or determining how certain changes, like adding another acid, will affect the system.
In a solution, the degree of ionization can be calculated as the ratio of ionized acid concentration to the initial acid concentration, usually expressed as a percentage. For instance, if \( x \) moles of an acid initially present in \( 1 M \) solution ionizes into its constituent ions, the degree of ionization is \( \frac{x}{1} \times 100\% \) .
Consider cyanic acid from our exercise. Initially, it has a degree of ionization calculated based on its \( H^+ \) concentration before any HCl is added. The presence of additional HCl introduces more hydrogen ions into the solution, altering the dynamics of ionization. As a rule, when more protons are present in the solution, the degree of ionization typically decreases because the equilibrium shifts to favor the non-ionized form of the acid, thanks to Le Chatelier's principle.
Equilibrium Constants
Equilibrium constants such as \( K_a \) are essential for understanding the chemistry of weak acids. The acid dissociation constant, \( K_a \), directly measures a weak acid's strength by indicating the extent to which it ionizes in solution. The greater the \( K_a \) value, the stronger the acid, meaning it's more likely to donate \( H^+ \) ions.
The value of \( K_a \) is calculated based on the concentrations of the products and reactants at equilibrium. In reactions involving weak acids, it takes into account both the dissociated ions and the undissociated acid molecule. For cyanic acid, the expression for its \( K_a \) is: \[ K_a = \frac{[H^+][OCN^-]}{[HOCN]} \]
This formula helps explain the relationship between the \( H^+ \) concentration in the solution and the degree of ionization. While standard calculations often assume \( H^+ \) ions are solely from the weak acid, added acids like HCl complicate the equilibrium by adding extra \( H^+ \). Thus, careful calculation is necessary to correctly interpret the scenario.
Learning how equilibrium constants function is crucial for solving problems in acid-base chemistry and predicting how a system will react under various conditions.

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Most popular questions from this chapter

a. When \(0.10 \mathrm{~mol}\) of the ionic solid \(\mathrm{NaX},\) where \(\mathrm{X}\) is an unknown anion, is dissolved in enough water to make \(1.0 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is 9.12. When \(0.10 \mathrm{~mol}\) of the ionic solid \(\mathrm{ACl}\), where \(\mathrm{A}\) is an unknown cation, is dissolved in enough water to make \(1.0 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is 7.00. What would be the \(\mathrm{pH}\) of \(1.0 \mathrm{~L}\) of solution that contained 0.10 mol of \(A X ?\) Be sure to document how you arrived at your answer. b. In the AX solution prepared above, is there any \(\mathrm{OH}^{-}\) present? If so, compare the \(\left[\mathrm{OH}^{-}\right]\) in the solution to the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) c. From the information presented in part a, calculate \(K_{b}\) for the \(\mathrm{X}^{-}(a q)\) anion and \(K_{a}\) for the conjugate acid of \(\mathrm{X}^{-}(a q)\) d. To \(1.0 \mathrm{~L}\) of solution that contains \(0.10 \mathrm{~mol}\) of \(\mathrm{AX}\) you add 0.025 mol of \(\mathrm{HCl}\). How will the \(\mathrm{pH}\) of this solution compare to that of the solution that contained only \(\mathrm{NaX}\) ? Use chemical reactions as part of your explanation; you do not need to solve for a numerical answer. e. Another \(1.0 \mathrm{~L}\) sample of solution is prepared by mixing \(0.10 \mathrm{~mol}\) of \(\mathrm{AX}\) and \(0.10 \mathrm{~mol}\) of \(\mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is found to be 3.12 . Explain why the \(\mathrm{pH}\) of this solution is 3.12 . f. Finally, consider a different 1.0 - \(\mathrm{L}\) sample of solution that contains \(0.10 \mathrm{~mol}\) of \(\mathrm{AX}\) and \(0.1 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The \(\mathrm{pH}\) of this solution is found to be 13.00 . Explain why the \(\mathrm{pH}\) of this solution is 13.00 . g. Some students mistakenly think that a solution that contains \(0.10 \mathrm{~mol}\) of \(\mathrm{AX}\) and \(0.10 \mathrm{~mol}\) of \(\mathrm{HCl}\) should have a pH of 1.00 . Can you come up with a reason why students have this misconception? Write an approach that you would use to help these students understand what they are doing wrong.

a Draw a pH titration curve that represents the titration of \(25.0 \mathrm{~mL}\) of \(0.15 \mathrm{M}\) propionic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH},\) by the addition of \(0.15 M \mathrm{KOH}\) from a buret. Label the axes and put a scale on each axis. Show where the equivalence point and the buffer region are on the titration curve. You should do calculations for the \(0 \%, 50 \%, 60 \%,\) and \(100 \%\) titration points. \(D\) Is the solution neutral, acidic, or basic at the equivalence point? Why?

Weak base \(\mathrm{B}\) has a \(\mathrm{p} K_{b}\) of 6.78 and weak acid \(\mathrm{HA}\) has a p \(K_{a}\) of 5.12 . a. Which is the stronger base, \(\mathrm{B}\) or \(\mathrm{A}^{-} ?\) b. Which is the stronger acid, HA or \(\mathrm{BH}^{+} ?\) c. Consider the following reaction: \(\mathrm{B}(a q)+\mathrm{HA}(a q) \rightleftharpoons \mathrm{BH}^{+}(a q)+\mathrm{A}^{-}(a q)\) Based on the information about the acid/base strengths for the species in this reaction, is this reaction favored to proceed more to the right or more to the left? Why? d. An aqueous solution is made in which the concentration of weak base \(\mathrm{B}\) is one half the concentration of its acidic salt, \(\mathrm{BHCl}\), where \(\mathrm{BH}^{+}\) is the conjugate weak acid of B. Calculate the \(\mathrm{pH}\) of the solution. e. An aqueous solution is made in which the concentration of weak acid HA is twice the concentration of the sodium salt of the weak acid, NaA. Calculate the \(\mathrm{pH}\) of the solution. f. Assume the conjugate pairs \(\mathrm{B} / \mathrm{BH}^{+}\) and \(\mathrm{HA} / \mathrm{A}^{-}\) are capable of being used as color-based end point indicators in acid-base titrations, where \(\mathrm{B}\) is the base form indicator and \(\mathrm{BH}^{+}\) is the acid form indicator, and HA is the acid form indicator and \(\mathrm{A}^{-}\) is the base form indicator. Select the indicator pair that would be best to use in each of the following titrations: (1) Titration of a strong acid with a strong base. (i) \(\mathrm{B} / \mathrm{B} \mathrm{H}^{+}\) (ii) \(\mathrm{HA} / \mathrm{A}^{-}\) (2) Titration of a weak base with a strong acid. (i) \(\mathrm{B} / \mathrm{B} \mathrm{H}^{+}\) (ii) \(\mathrm{HA} / \mathrm{A}^{-}\)

Obtain a the \(K_{b}\) value for \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}_{2}^{-} ;\) b the \(K_{a}\) value for \(\mathrm{NH}_{3} \mathrm{OH}^{+}\) (hydroxylammonium ion).

Ionization of the first proton from \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is complete \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right.\) is a strong acid); the acid-ionization constant for the second proton is \(1.1 \times 10^{-2}\). a What would be the approximate hydronium-ion concentration in \(0.100 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) if ionization of the second proton were ignored? The ionization of the second proton must be considered for a more exact answer, however. Calculate the hydronium-ion concentration in \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4},\) accounting for the ionization of both protons.

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