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Chapter 16: Problem 44

Chloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2},\) has a greater acid strength than acetic acid, because the electronegative chlorine atom pulls electrons away from the \(\mathrm{O}-\mathrm{H}\) bond and thus weakens it. Calculate the hydronium-ion concentration and the \(\mathrm{pH}\) of a \(0.0020 \mathrm{M}\) solution of chloroacetic acid. \(K\) is \(1.3 \times 10^{-3}\)

Short Answer

Expert verified
[H鈧僌鈦篯 鈮� 1.61 x 10鈦宦� M, pH 鈮� 2.79.

Step by step solution

01

Write the dissociation equation

The dissociation of chloroacetic acid (\(\mathrm{HC}_2 \mathrm{H}_2 \mathrm{ClO}_2\)) in water is represented as:\[ \mathrm{HC}_2 \mathrm{H}_2 \mathrm{ClO}_2 (aq) + \mathrm{H}_2\mathrm{O} (l) \rightleftharpoons \mathrm{H}_3\mathrm{O}^+ (aq) + \mathrm{C}_2 \mathrm{H}_2 \mathrm{ClO}_2^- (aq) \] This equation shows how the acid donates a proton to form the hydronium ion \(\mathrm{H}_3\mathrm{O}^+\).
02

Set up the expression for the acid dissociation constant, \(K_a\)

The acid dissociation constant \(K_a\) is calculated using:\[ K_a = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{C}_2 \mathrm{H}_2 \mathrm{ClO}_2^-]}{[\mathrm{HC}_2 \mathrm{H}_2 \mathrm{ClO}_2]} \]Plug given value of \(K_a = 1.3 \times 10^{-3}\).
03

Assume initial conditions for concentrations

At the start, the concentration of chloroacetic acid (\([\mathrm{HC}_2 \mathrm{H}_2 \mathrm{ClO}_2]\)) is 0.0020 M, and the concentrations of \([\mathrm{H}_3\mathrm{O}^+]\) and \([\mathrm{C}_2 \mathrm{H}_2 \mathrm{ClO}_2^-]\) are initially zero. Let \(x\) be the change in concentration at equilibrium.
04

Establish the equilibrium concentrations

At equilibrium, the concentrations will be:- \([\mathrm{HC}_2 \mathrm{H}_2 \mathrm{ClO}_2]\) = 0.0020 M - x- \([\mathrm{H}_3\mathrm{O}^+]\) = x- \([\mathrm{C}_2 \mathrm{H}_2 \mathrm{ClO}_2^-]\) = x.
05

Substitute into \(K_a\) expression and solve for \(x\)

Substitute the equilibrium concentrations into the \(K_a\) expression:\[ 1.3 \times 10^{-3} = \frac{x^2}{0.0020 - x} \]Assume \(x\) is small, so \(0.0020 - x \approx 0.0020\).Thus, \(1.3 \times 10^{-3} \approx \frac{x^2}{0.0020}\).Solve for \(x\):\[ x^2 = 1.3 \times 10^{-3} \times 0.0020 \]\[ x^2 = 2.6 \times 10^{-6} \]\[ x = \sqrt{2.6 \times 10^{-6}} = 1.61 \times 10^{-3} \]
06

Calculate hydronium-ion concentration

The hydronium-ion concentration \([\mathrm{H}_3\mathrm{O}^+]\) is equal to \(x\), which is \(1.61 \times 10^{-3}\) M.
07

Calculate pH

The \(\mathrm{pH}\) is calculated as:\[ \mathrm{pH} = -\log([\mathrm{H}_3\mathrm{O}^+]) = -\log(1.61 \times 10^{-3}) \]Use a calculator to find \(\mathrm{pH} \approx 2.79\).

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Key Concepts

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

Chloroacetic Acid
Chloroacetic acid is a derivative of acetic acid with one chlorine atom replacing a hydrogen atom. The presence of the chlorine atom increases the acid's strength. This happens because chlorine is an electronegative element, meaning it attracts electrons more strongly than other elements.
This electronegative property weakens the bond between hydrogen and oxygen in the acid, making it easier for the acid to release a proton.
  • **Greater Acid Strength:** The weakening of the O-H bond by the chlorine atom makes chloroacetic acid stronger than acetic acid.
  • **Molecular Structure:** The formula for chloroacetic acid is HC鈧侶鈧侰lO鈧�, showcasing its chloride substitution which significantly impacts acid behavior.
Understanding these structural and molecular nuances helps explain why chloroacetic acid has a higher acidity level compared to its non-chlorinated counterpart.
pH Calculation
The pH is a measure of how acidic or basic a solution is, with lower pH values indicating stronger acids. To calculate the pH of a solution, we use the concentration of hydronium ions, [H鈧僌鈦篯.
Once we know [H鈧僌鈦篯, the pH can be found using the formula: pH = -log([H鈧僌鈦篯).

A calculator is typically used to compute this logarithmic expression.
  • **Low pH Value (below 7):** Indicates an acidic solution.
  • **Calculation Steps:** Given the solution of chloroacetic acid with a concentration of 1.61 脳 10鈦宦� M hydronium ions, the pH can be calculated as approximately 2.79.
By understanding this concept, you can determine the acidity of any solution by first finding the hydronium ion concentration, and then applying it to the pH formula.
Hydronium Ion Concentration
Hydronium ions are formed when an acid dissociates in water. For chloroacetic acid, the dissociation involves the donation of a proton to water, creating hydronium ions H鈧僌鈦�.
The concentration of hydronium ions is crucial for calculating the pH and understanding the acid's strength.
  • **Formation:** In the dissociation equation of chloroacetic acid, HC鈧侶鈧侰lO鈧�, the acid releases a proton which forms H鈧僌鈦� in solution.
  • **Equilibrium:** At equilibrium, the concentration of hydronium ions is equal to the amount of acid that dissociates, denoted by the variable "x" in equilibrium calculations.
  • **Final Concentration:** In the case of chloroacetic acid, the [H鈧僌鈦篯 is found to be 1.61 脳 10鈦宦� M.
Knowing how to calculate [H鈧僌鈦篯 provides a deeper insight into the behavior of acids in various conditions.
Chemical Equilibrium
Chemical equilibrium is the state where the concentrations of all reactants and products remain constant over time. It occurs when the rate of the forward reaction equals the rate of the reverse reaction. In the case of chloroacetic acid, equilibrium describes the balance between the undissociated acid and the ions formed in solution.
  • **Equilibrium Expression:** The expression for the equilibrium condition of chloroacetic acid is given by the equation: K鈧� = [H鈧僌鈦篯[C鈧侶鈧侰lO鈧傗伝] / [HC鈧侶鈧侰lO鈧俔.
  • **Role of K鈧�:** The acid dissociation constant, K鈧�, measures the strength of the acid and how much it dissociates in water.
  • **Assumptions in Calculations:** Often, "x," representing the change in concentration, is assumed small compared to initial concentrations for simplification.
Understanding chemical equilibrium helps in predicting the extent of a reaction and calculating the concentrations of reactants and products in solutions.

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Most popular questions from this chapter

Barbituric acid, \(\mathrm{HC}_{4} \mathrm{H}_{3} \mathrm{~N}_{2} \mathrm{O}_{3},\) is used to prepare various barbiturate drugs (used as sedatives). Calculate the concentrations of hydronium ion and barbiturate ion in a \(0.10 M\) solution of the acid. The value of \(K_{a}\) is \(9.8 \times 10^{-5}\).

Briefly describe two methods for determining \(K_{a}\) for a weak acid.

A solution of weak base is titrated to the equivalence point with a strong acid. Which one of the following statements is most likely to be correct? a. The \(\mathrm{pH}\) of the solution at the equivalence point is 7.0 . b. The \(\mathrm{pH}\) of the solution is greater than 13.0 . c. The \(\mathrm{pH}\) of the solution is less than 2.0 . d. The \(\mathrm{pH}\) of the solution is between 2.0 and 7.0 . e. The \(\mathrm{pH}\) of the solution is between 7.0 and 13.0 .

A 0.239 -g sample of unknown organic base is dissolved in water and titrated with a \(0.135 \mathrm{M}\) hydrochloric acid solution. After the addition of \(18.35 \mathrm{~mL}\) of acid, a \(\mathrm{pH}\) of 10.73 is recorded. The equivalence point is reached when a total of \(39.24 \mathrm{~mL}\) of \(\mathrm{HCl}\) is added. The base and acid combine in a 1: 1 ratio.

Write the chemical equation for the base ionization of aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} .\) Write the \(K_{b}\) expression for aniline.
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