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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 \(0.0020 M\) solution of chloroacetic acid. \(K_{a}\) is \(1.3 \times 10^{-3}\).

Short Answer

Expert verified
The hydronium-ion concentration is 0.00115 M, and the pH is approximately 2.94.

Step by step solution

01

Write the ionization equation

Write the ionization equation for chloroacetic acid, \( \mathrm{HC}_2\mathrm{H}_2\mathrm{ClO}_2 \) in water: \[ \mathrm{HC}_2\mathrm{H}_2\mathrm{ClO}_2(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{C}_2\mathrm{H}_2\mathrm{ClO}_2^-(aq) \]
02

Set up the equilibrium concentration table

Assume the initial concentration of \( \mathrm{HC}_2\mathrm{H}_2\mathrm{ClO}_2 \) is \([\mathrm{HC}_2\mathrm{H}_2\mathrm{ClO}_2] = 0.0020 \) M. At equilibrium, the concentrations will be:- \([\mathrm{H}^+] = x \)- \([\mathrm{C}_2\mathrm{H}_2\mathrm{ClO}_2^-] = x \)- \([\mathrm{HC}_2\mathrm{H}_2\mathrm{ClO}_2] = 0.0020 - x \) (where \(x\) is the change in concentration of \([\mathrm{H}^+]\) and \([\mathrm{C}_2\mathrm{H}_2\mathrm{ClO}_2^-] \)).
03

Write the expression for the acid dissociation constant

The expression for the acid dissociation constant \( K_a \) is:\[ K_a = \frac{[\mathrm{H}^+][\mathrm{C}_2\mathrm{H}_2\mathrm{ClO}_2^-]}{[\mathrm{HC}_2\mathrm{H}_2\mathrm{ClO}_2]} \]Substituting the equilibrium concentrations:\[ K_a = \frac{x^2}{0.0020 - x} = 1.3 \times 10^{-3}\]
04

Approximate and solve for \(x\)

Since \( K_a \) is relatively large, we cannot assume \( x \ll 0.0020 \). Solve the quadratic equation \( x^2 + 1.3 \times 10^{-3}x - 2.6 \times 10^{-6} = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting \(a = 1, b = 1.3\times10^{-3}, c = -2.6\times10^{-6}\), solve to find \( x \approx 0.00115 \).
05

Calculate the hydronium-ion concentration

The equilibrium concentration of \( \mathrm{H}^+ \) (hydronium-ion concentration) is \( x = 0.00115 \) M.
06

Calculate the pH

Calculate the pH using the formula:\[ \mathrm{pH} = -\log([\mathrm{H}^+]) \]Substitute \([\mathrm{H}^+] = 0.00115\) M:\[ \mathrm{pH} = -\log(0.00115) \approx 2.94 \]

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Key Concepts

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

Acid Dissociation Constant (Ka)
The acid dissociation constant, denoted as \( K_a \), is a crucial parameter in understanding the strength of an acid. It measures how easily an acid donates a proton to a base, dissolving in water to form ions. The higher the \( K_a \), the stronger the acid.

In the case of chloroacetic acid \( \text{HC}_2\text{H}_2\text{ClO}_2 \), the \( K_a \) value is \( 1.3 \times 10^{-3} \). This high \( K_a \) compared to other acids like acetic acid, indicates that more chloroacetic acid molecules will dissociate into ions in solution, increasing its acidity.

For computing \( K_a \), the dissociation of the acid is expressed as:
  • \( \text{HC}_2\text{H}_2\text{ClO}_2 \rightleftharpoons \text{H}^+ + \text{C}_2\text{H}_2\text{ClO}_2^- \)
  • The formula for \( K_a \) is: \( K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_2\text{ClO}_2^-]}{[\text{HC}_2\text{H}_2\text{ClO}_2]} \)
In this example, \( K_a \) helps us understand that chloroacetic acid dissociates significantly, affecting the solution's hydrogen ion concentration.
Equilibrium Concentration
When an acid like chloroacetic acid ionizes, not all its molecules dissociate completely. Instead, they reach an equilibrium where some molecules remain un-ionized. This state is captured by the term "equilibrium concentration."

Initially, the concentration of chloroacetic acid is \( 0.0020 \) M. As the acid dissociates, the concentrations at equilibrium are:
  • \([\text{H}^+] = x\)
  • \([\text{C}_2\text{H}_2\text{ClO}_2^-] = x\)
  • \([\text{HC}_2\text{H}_2\text{ClO}_2] = 0.0020 - x\)
This demonstrates the shift in concentrations as the solution finds balance. Solving for \( x \), which represents the concentration of ions at equilibrium, involves using the \( K_a \) expression. By substituting back, we see \( x = 0.00115 \), meaning this is the equilibrium concentration of hydrogen ions and the acid's conjugate base in the solution.
pH Calculation
One of the most common ways to express the acidity of a solution is by calculating its pH, which is directly related to the hydronium-ion concentration \([\text{H}^+]\). A lower pH value indicates higher acidity.

The formula used to calculate pH is:
  • \( \text{pH} = -\log([\text{H}^+]) \)
For the chloroacetic acid solution, once we determine that \([\text{H}^+] = 0.00115\) M, we substitute this into the pH formula:
  • \( \text{pH} = -\log(0.00115) \)
Computing this gives a pH of approximately 2.94. This confirms that the solution is quite acidic, as expected from an acid with a relatively high \( K_a \). Understanding pH calculations is essential for anyone studying acid-base chemistry, as it links concentration to the acid's strength.

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Most popular questions from this chapter

A 25.00-mL sample contains \(0.562 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}\). This sample is used to standardize an NaOH solution. At the equivalence point, \(42.36 \mathrm{~mL}\) of \(\mathrm{NaOH}\) has been added. a. What was the concentration of the \(\mathrm{NaOH}\) ? b. What is the \(\mathrm{pH}\) at the equivalence point? C. Which indicator, bromthymol blue, methyl violet, or alizarin yellow R, should be used in the titration? Why?

Note whether the aqueous solution of each of the following salts will be acidic, basic, or neutral. a. \(\mathrm{Na}_{2} \mathrm{~S}\) b. \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(\mathrm{KClO}_{4}\) d. \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\)

a. For each of the following salts, write the reaction that occurs when it dissociates in water: \(\mathrm{NaCl}(s), \mathrm{NaCN}(s)\), \(\mathrm{KClO}_{2}(s), \mathrm{NH}_{4} \mathrm{NO}_{3}(s), \operatorname{KBr}(a q)\), and \(\mathrm{NaF}(s)\) b. Consider each of the reactions that you wrote above, and identify the aqueous ions that could be proton donors (acids) or proton acceptors (bases). Briefly explain how you decided which ions to choose. c. For each of the acids and bases that you identified in part b, write the chemical reaction it can undergo in aqueous solution (its reaction with water). d. Are there any reactions that you have written above that you anticipate will occur to such an extent that the \(\mathrm{pH}\) of the solution will be affected? As part of your answer, be sure to explain how you decided. e. Assume that in each case above, \(0.01 \mathrm{~mol}\) of the salt was dissolved in enough water at \(25^{\circ} \mathrm{C}\) to make \(1.0 \mathrm{~L}\) of solution. In each case, what additional information would you need in order to calculate the \(\mathrm{pH}\) ? If there are cases where no additional information is required, be sure to state that as well. f. Say you take \(0.01\) mol of \(\mathrm{NH}_{4} \mathrm{CN}\) and dissolve it in enough water at \(25^{\circ} \mathrm{C}\) to make \(1.0 \mathrm{~L}\) of solution. Using chemical reactions and words, explain how you would go about determining what effect this salt will have on the \(\mathrm{pH}\) of the solution. Be sure to list any additional information you would need to arrive at an answer.

The equilibrium equations and \(K_{a}\) values for three reaction systems are given below. \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{NH}_{3}(a q) ; K_{a}=5.6 \times 10^{-10}\) \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \(\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) ; K_{a}=4.3 \times 10^{-7}\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}\) $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HPO}_{4}^{2-}(a q) ; K_{a}=6.2 \times 10^{-8} $$ a. Which conjugate pair would be best for preparing a buffer with a pH of \(6.96\) ? Why? b. How would you prepare \(100 \mathrm{~mL}\) of a buffer with a \(\mathrm{pH}\) of \(6.96\) assuming that you had available \(0.10 \mathrm{M}\) solutions of each pair?

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