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A \(0.10 \mathrm{M}\) solution of chloroacetic acid, \(\mathrm{Cl} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H},\) has a pH of \(1.95 .\) Calculate \(K_{\mathrm{a}}\) for the acid.

Short Answer

Expert verified
The ionization constant, \( K_a \), for chloroacetic acid is approximately \( 1.41 \times 10^{-3} \).

Step by step solution

01

Write the Ionization Equation

Chloroacetic acid (\( \text{ClCH}_2\text{CO}_2\text{H} \)) ionizes in water as follows:\[\text{ClCH}_2\text{CO}_2\text{H} \leftrightarrow \text{ClCH}_2\text{CO}_2^- + \text{H}^+\]We need to know the concentrations of these species to calculate the ionization constant, \( K_a \).
02

Determine the Hydrogen Ion Concentration

The pH of the solution is given as \( 1.95 \). Use the pH formula to find the concentration of hydrogen ions (\( \text{H}^+ \)):\[pH = -\log_{10}[\text{H}^+]\]Solving for \( [\text{H}^+] \):\[[\text{H}^+] = 10^{-1.95} \approx 1.12 \times 10^{-2} \, \text{M}\]
03

Use the ICE Table

Since chloroacetic acid is a weak acid, construct an ICE (Initial, Change, Equilibrium) table:- **Initial:** - \( [\text{ClCH}_2\text{CO}_2\text{H}] = 0.10 \text{ M} \) - \( [\text{ClCH}_2\text{CO}_2^-] = 0 \) - \( [\text{H}^+] = 0 \)- **Change:** - \( [\text{ClCH}_2\text{CO}_2\text{H}] = -x \) - \( [\text{ClCH}_2\text{CO}_2^-] = +x \) - \( [\text{H}^+] = +x \)- **Equilibrium:** - \( [\text{ClCH}_2\text{CO}_2\text{H}] = 0.10 - x \) - \( [\text{ClCH}_2\text{CO}_2^-] = x \) - \( [\text{H}^+] = x \approx 1.12 \times 10^{-2} \text{ M} \) (since \( x \)is quite small compared to 0.10)
04

Calculate the Acid Ionization Constant \( K_a \)

Using the Equilibrium concentrations from the ICE table, we can find \( K_a \):\[K_a = \frac{[\text{ClCH}_2\text{CO}_2^-][\text{H}^+]}{[\text{ClCH}_2\text{CO}_2\text{H}]}\]Substitute the values:\[K_a = \frac{(1.12 \times 10^{-2})(1.12 \times 10^{-2})}{0.10 - 1.12 \times 10^{-2}} \approx \frac{1.2544 \times 10^{-4}}{0.0888} \approx 1.41 \times 10^{-3}\]
05

Verify the Approximations

Since the calculated value of \( x \)was much smaller than the initial concentration of the acid (\( 0.10 \text{ M} \)), our approximation \( 0.10 - x \approx 0.10 \)is reasonable, verifying that the calculation of the ionization constant \( K_a \)is accurate.

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Key Concepts

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

Chloroacetic Acid
Chloroacetic acid, often represented as \( \text{ClCH}_2\text{CO}_2\text{H} \), is an interesting molecule because it's a weak acid, meaning it does not completely dissociate in water. This characteristic is crucial when considering its behavior in solutions, especially in pH calculations and assessing its ionization constant.
  • It contains a chlorine atom attached to the acetic acid structure, making it more acidic than acetic acid itself.
  • Its intended use ranges from being a precursor for various chemicals in industry to its applications in organic synthesis.

Because it only partially ionizes in water, the ionization constant \( K_a \) can help measure its strength as an acid and provide insights into how effectively it dissociates into hydrogen ions \( \text{H}^+ \) and chloroacetate ions \( \text{ClCH}_2\text{CO}_2^- \). Understanding chloroacetic acid's properties is essential in predicting reactions and equilibria in solutions.
pH Calculation
The concept of pH is a fundamental one in chemistry, representing the acidity or alkalinity of a solution. It's calculated as \( pH = -\log_{10}[\text{H}^+] \). For the given problem, knowing that the pH is \( 1.95 \) allows us to compute the concentration of hydrogen ions, \( [\text{H}^+] \).

By rearranging the pH formula, we get:\[ [\text{H}^+] = 10^{-1.95} \approx 1.12 \times 10^{-2} \, \text{M} \]
This calculation is crucial as it provides the concentration required for further steps in determining the ionization constant.

Remember:
  • Lower pH values indicate higher acidity.
  • The pH scale is logarithmic, so each unit change represents a tenfold change in \( [\text{H}^+] \).

Understanding pH calculation is key for accurate analysis of weak acids like chloroacetic acid.
Ionization Equation
To understand an acid's behavior in solution, writing and interpreting its ionization equation is essential. For chloroacetic acid, the equation is: \[\text{ClCH}_2\text{CO}_2\text{H} \leftrightarrow \text{ClCH}_2\text{CO}_2^- + \text{H}^+\]
This equation showcases the equilibrium between the undissociated acid and its ionized form. Knowing how to set up this equilibrium equation is half the battle in solving for the acid ionization constant \( K_a \).

Some key points:
  • The double arrows indicate the acid does not completely dissociate and exists in equilibrium.
  • Understanding the stoichiometry here is crucial for establishing relationships between concentrations.

Writing correct ionization equations helps predict how much of an acid ionizes in a particular solution, impacting calculations like \( K_a \). It represents the fundamental chemical relationship needed for analyzing weak acids.
ICE Table
An ICE table, short for Initial, Change, Equilibrium, is a systematic way to keep track of the concentrations of reactants and products in an equilibrium reaction. They are particularly useful for weak acids such as chloroacetic acid, where equilibrium plays a significant role.

Let's break down the use of an ICE table:
  • **Initial:** Start with known initial concentrations. For chloroacetic acid, it begins at \( 0.10 \text{ M} \), and the ions start at \( 0 \text{ M} \).
  • **Change:** Account for the shift in equilibrium. As the reaction proceeds, the concentration of the acid decreases by \( x \), while the ions increase by \( x \).
  • **Equilibrium:** Determine the concentrations when equilibrium is reached. We find \( x \) to be approximately \( 1.12 \times 10^{-2} \text{ M} \), calculated from the \( [\text{H}^+] \).

By putting these insights together, we see how small the change \( x \) is compared to the initial concentration, supporting our calculation and assumption that \( 0.10 - x \approx 0.10 \). This is essential in applying the solution to calculate \( K_a \), confirming that ICE tables are powerful tools for dealing with reaction equilibria, especially with acid and base solutions.

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Most popular questions from this chapter

What is the pH of a 0.0015 M solution of \(\mathrm{Ba}(\mathrm{OH})_{2} ?\)

You have three solutions labeled \(A, B\), and \(C\). You know only that each contains a different cation \(-\mathrm{Na}^{+}, \mathrm{NH}_{4}^{+},\) or \(\mathrm{H}_{3} \mathrm{O}^{+} .\) Each has an anion that does not contribute to the solution \(\mathrm{pH}\) (eg. \(\mathrm{Cl}^{-}\) ). You also have two other solutions, Y and \(\mathrm{Z}\) each containing a different anion, \(\mathrm{Cl}^{-}\) or \(\mathrm{OH}^{-}\) with a cation that does not influence solution \(\mathrm{pH}\) \(\left(\mathrm{e} \cdot \mathrm{g} \cdot, \mathrm{K}^{+}\right) .\) If equal amounts of \(\mathrm{B}\) and \(\mathrm{Y}\) are mixed, the result is an acidic solution. Mixing A and Z gives a neutral solution, whereas \(\mathrm{B}\) and Z give a basic solution. Identify the five unknown solutions. (Adapted from D. H. Barouch: Voyages in Conceptual Chemistry, Boston, Jones and Bartlett, \(1997 .)\)

Prove that \(K_{\mathrm{a} 3} \times K_{\mathrm{b} 1}=K_{\mathrm{w}}\) for phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4},\) by adding the chemical equilibrium expressions that correspond to the third ionization step of the acid in water with the first of the three successive steps of the reaction of phosphate ion, \(\mathrm{PO}_{4}^{3-},\) with water.

Which is the stronger of the following two acids? (a) acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=1.8 \times 10^{-5}\) (b) chloroacetic acid, \(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{p} K_{a}=2.85\)

Ascorbic acid (vitamin \(\mathrm{C}, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\) ) is a diprotic acid \(\left(K_{\mathrm{al}}=6.8 \times 10^{-5} \text {and } K_{\mathrm{a} 2}=2.7 \times 10^{-12}\right)\) What is the pH of a solution that contains 5.0 \(\mathrm{mg}\) of acid per milliliter of solution?

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