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What is the hydronium-ion concentration of a \(2.00 M\) solution of 2,6 -dinitrobenzoic acid, \(\left(\mathrm{NO}_{2}\right)_{2} \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{COOH},\) for which \(K_{a}=7.94 \times 10^{-2} ?\)

Short Answer

Expert verified
The hydronium-ion concentration is approximately \(0.3985\, M\).

Step by step solution

01

Understanding the Dissociation Equation

2,6-dinitrobenzoic acid, \(\text{(NO}_2\text{)}_2\text{C}_6\text{H}_3\text{COOH}\), is a weak acid. When it dissociates in water, it forms hydronium ions (\(\text{H}_3\text{O}^+\)) and the corresponding anion ([\(\text{(NO}_2\text{)}_2\text{C}_6\text{H}_3\text{COO}^-\)]). The equation of dissociation is given as: \[\text{(NO}_2\text{)}_2\text{C}_6\text{H}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{(NO}_2\text{)}_2\text{C}_6\text{H}_3\text{COO}^-(aq)\]
02

Apply the Expression for the Acid Dissociation Constant

The dissociation constant \(K_a\) for the acid can be expressed as: \[K_a = \frac{[\text{H}_3\text{O}^+][\text{(NO}_2\text{)}_2\text{C}_6\text{H}_3\text{COO}^-]}{[\text{(NO}_2\text{)}_2\text{C}_6\text{H}_3\text{COOH}]}\]Here, the initial concentration of the acid before dissociation is given as \(C = 2.00\, M\).
03

Define Changes in Concentration

At the start, before dissociation, the concentration of 2,6-dinitrobenzoic acid is \(2.00\, M\), while concentrations of hydronium ions and anions are \(0\, M\). If \(x\) is the concentration of \(\text{H}_3\text{O}^+\) ions formed at equilibrium, then:- \([\text{(NO}_2\text{)}_2\text{C}_6\text{H}_3\text{COO}^-] = x\)- \([\text{(NO}_2\text{)}_2\text{C}_6\text{H}_3\text{COOH}] = 2.00 - x\) Since \(K_a\) is quite high, let's assume \(x\) is not too small.
04

Solve for Hydronium Ion Concentration

Now substitute these relationships into the \(K_a\) expression: \[7.94 \times 10^{-2} = \frac{x^2}{2.00 - x}\]Assume \(x\) is small compared to \(2.00\), thus \(2.00 - x \approx 2.00\). Then,\[7.94 \times 10^{-2} \approx \frac{x^2}{2.00}\]Solving for \(x\):\[x^2 = 2.00 \times 7.94 \times 10^{-2}\]\[x^2 = 0.1588\]\[x = \sqrt{0.1588} \approx 0.3985\, M\]Thus, the hydronium ion concentration \([\text{H}_3\text{O}^+]\) is approximately \(0.3985\, M\).

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Key Concepts

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

Acid Dissociation Constant
The acid dissociation constant, often symbolized as \( K_a \), is fundamental in understanding the behavior of weak acids in solution. It is a quantitative measure of the strength of an acid in solution. More specifically, \( K_a \) is the equilibrium constant for the dissociation (ionization) of the acid into its ions. This constant equation, \[ K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \], tells us about the balance between the non-ionized form of the acid \( \text{HA} \) and its ionized form \( \text{H}_3\text{O}^+ \) and \( \text{A}^- \).
  • Higher \( K_a \) value indicates a stronger acid which dissociates more in solution.
  • Lower \( K_a \) implies that the acid is weak and dissociates less.
Understanding \( K_a \) is crucial for calculating the concentrations of components in a weak acid solution at equilibrium and therefore determining the hydronium ion concentration.
Weak Acid
A weak acid is one that only partially dissociates into ions when dissolved in water. This means that not all of the acid molecules donate protons to water molecules to form hydronium ions \( \text{H}_3\text{O}^+ \). In the dissociation of weak acids, the equilibrium lies to the left, favoring the non-dissociated form.
  • Examples of weak acids include acetic acid and, as mentioned in this exercise, 2,6-dinitrobenzoic acid.
  • Due to partial dissociation, weak acids have specific \( K_a \) values that help to predict the degree of ionization in solution.
The behavior of weak acids is significant in various fields, including chemistry and biology, where buffering and pH control are vital.
Chemical Equilibrium
Chemical equilibrium refers to a state in a reversible reaction where the rates of the forward and reverse reactions are equal. At this point, the concentrations of reactants and products remain unchanged over time. For the dissociation of a weak acid, this concept is essential as it determines the specific concentrations of all species involved at equilibrium.
  • In the acid dissociation, both the non-ionized acid and the ionized products are present in a balance at equilibrium.
  • This can be represented by a balanced chemical reaction: \( \text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^- \).
Understanding chemical equilibrium helps in solving equilibrium problems like finding the hydronium ion concentration using the known \( K_a \) value.
Concentration Changes
In a reaction involving a weak acid, such as the dissociation equation for 2,6-dinitrobenzoic acid, concentration changes occur from the initial state to equilibrium. Initially, the concentration of the acid is known, but the concentrations of ions form as the reaction progresses to equilibrium.
  • The initial concentration of the weak acid before dissociation is called \( C \).
  • As the dissociation reaches equilibrium, changes in concentration are small relative to the initial concentration.
  • These changes can be expressed through variables: let \( x \) represent the change in concentration of ions formed.
This concept is the basis for establishing the concentrations in the equilibrium expression and is crucial for calculating the actual concentration of the hydronium ions in the solution.

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Most popular questions from this chapter

A chemist wanted to determine the concentration of a solution of lactic acid, \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3} .\) She found that the \(\mathrm{pH}\) of the solution was 2.60 . What was the concentration of the solution? The \(K_{a}\) of lactic acid is \(1.4 \times 10^{-4}\).

Write the equation for the acid ionization of the \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ion.

A buffer solution is prepared by mixing equal volumes of \(0.10 \mathrm{M} \mathrm{NaNO}_{2}\) and \(0.10 \mathrm{M} \mathrm{HNO}_{2}\) solutions.

Butylamine, \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NH}_{2}\) is a weak base. A \(0.47 M\) aqueous solution of butylamine has a pH of \(12.13 .\) What is \(K_{b}\) for butylamine? Calculate the \(\mathrm{pH}\) of a \(0.35 \mathrm{M}\) aqueous solution of butylamine.

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.

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