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Chapter 17: Problem 90

Chloroacetic acid, \(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H}\), is a moderately weak acid \(\left(K_{\mathrm{a}}=1.40 \times 10^{-3}\right) .\) If you dissolve \(94.5 \mathrm{mg}\) of the acid in water to give 125 mL of solution, what is the pH of the solution?

Short Answer

Expert verified
The pH of the chloroacetic acid solution is approximately 2.48.

Step by step solution

01

Calculate the moles of chloroacetic acid

First, determine the molar mass of chloroacetic acid, \(\mathrm{ClCH_2CO_2H}\). By adding the atomic masses: \(\text{Cl: } 35.5\), \(\text{C: } 12.0\), \(\text{H: } 1.0\), \(\text{C: } 12.0\), \(\text{O: } 16.0\times2\), and \(\text{H: } 1.0\), we find that its molar mass is approximately \(94.5 \, \text{g/mol}\). Then, convert the given mass \(94.5 \, \text{mg}\) to grams: \(94.5 \, \text{mg} = 0.0945 \, \text{g}\). Next, calculate the moles: \(\text{moles} = \frac{0.0945 \, \text{g}}{94.5 \, \text{g/mol}} = 0.001 \, \text{mol}\).
02

Determine the concentration of the solution

To find the concentration, divide the number of moles of the acid by the volume of the solution in liters. The volume is \(125 \, \text{mL} = 0.125 \, \text{L}\). So, \(\text{Concentration} = \frac{0.001 \, \text{mol}}{0.125 \, \text{L}} = 0.008 \, \text{mol/L}\).
03

Set up the expression for the dissociation of chloroacetic acid

Chloroacetic acid can be described by the equilibrium dissociation equation:\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]The initial concentration of \(\text{HA}\) is \(0.008 \, \text{M}\). Upon dissociation, the change in concentration is described by \(-x\) for \(\text{HA}\) and \(+x\) for \(\text{H}^+\) and \(\text{A}^-\). So the equilibrium concentrations are \([\text{HA}] = 0.008-x\), \([\text{H}^+] = x\), and \([\text{A}^-] = x\).
04

Write the expression for the acid dissociation constant

Using the given \(K_a = 1.40 \times 10^{-3}\), we write:\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{x^2}{0.008-x} \]Assuming \(x\) is much smaller than 0.008, approximate \((0.008-x) \approx 0.008\). Then,\[ 1.40 \times 10^{-3} = \frac{x^2}{0.008} \]
05

Solve for the concentration of \(\text{H}^+\)

Solving for \(x\), we have:\[ x^2 = 1.40 \times 10^{-3} \times 0.008 \]\[ x^2 = 1.12 \times 10^{-5} \]Taking the square root of both sides gives:\[ x = \sqrt{1.12 \times 10^{-5}} \approx 3.34 \times 10^{-3} \text{ M} \]Since \(x\) represents the \([\text{H}^+]\) concentration, \([\text{H}^+] = 3.34 \times 10^{-3} \text{ M}\).
06

Calculate the pH of the solution

The pH is calculated using the formula:\[ \text{pH} = -\log([\text{H}^+]) = -\log(3.34 \times 10^{-3}) \approx 2.48 \]

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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 where one hydrogen atom in the methyl group is replaced by a chlorine atom. This gives chloroacetic acid the formula \( ext{ClCH}_2 ext{CO}_2 ext{H}\). The presence of the chlorine atom, a highly electronegative element, makes it a stronger acid than acetic acid. This substitution increases the acid's ability to donate protons by stabilizing the negative charge on the conjugate base, \( ext{ClCH}_2 ext{CO}_2^-\), through an effect known as the inductive effect.

Chloroacetic acid is categorized as a moderately weak acid. This categorization is based on its acid dissociation constant, \(K_a\). Generally, weak acids have small \(K_a\) values, indicating limited ionization in aqueous solutions. The \(K_a\) value for chloroacetic acid is given as \(1.40 \times 10^{-3}\), which reflects its ability to release hydrogen ions into the solution.
pH Calculation
To calculate the pH of a solution, we need to first determine the concentration of hydrogen ions, \([ ext{H}^+]\), in the solution. pH is the negative logarithm of the hydrogen ion concentration: \( ext{pH} = -\log([ ext{H}^+])\).

In this problem, we need to calculate the pH of a chloroacetic acid solution. First, we find that dissolving 94.5 mg of chloroacetic acid in 125 mL of water provides a molarity of 0.008 M. The solution's \(K_a\) helps relate the molar concentration of the acid, its ions, and \(x\), the change in concentration due to dissociation.

By solving the equilibrium expression \(K_a = \frac{x^2}{0.008-x}\) and considering \(x\) negligible compared to 0.008, we find \(x\), or \([ ext{H}^+]\), to be approximately \(3.34 \times 10^{-3}\) M. Finally, we calculate pH as the negative logarithm of \([ ext{H}^+]\), giving us a pH of about 2.48, indicating an acidic solution.
Equilibrium Constant
The equilibrium constant \(K_a\) is a crucial parameter in acid-base chemistry, quantifying an acid's propensity to dissociate in aqueous solution. For chloroacetic acid, \(K_a = 1.40 \times 10^{-3}\), which reflects its moderate acidity.

The expression for \(K_a\) is derived from the equilibrium concentrations of the species involved in the dissociation reaction: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \). The formula is \(K_a = \frac{[ ext{H}^+][ ext{A}^-]}{[ ext{HA}]}\).

In the case of chloroacetic acid, solving for \([ ext{H}^+]\) involves an assumption that \(x\), the change in concentration due to dissociation, is small relative to the initial concentration. This simplification allows us to easily calculate \([ ext{H}^+]\), and hence the pH of the solution. Understanding \(K_a\) is essential for predicting how an acid behaves in various aqueous environments and is foundational for pH calculation.

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Most popular questions from this chapter

Describe an experiment that will allow you to place the following three bases in order of increasing base strength: \(\mathrm{NaCN}, \mathrm{CH}_{3} \mathrm{NH}_{2}, \mathrm{Na}_{2} \mathrm{CO}_{3}\)

A hydrogen atom in the organic base pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\), can be substituted by various atoms or groups to give \(\mathrm{XC}_{5} \mathrm{H}_{4} \mathrm{N},\) where \(\mathrm{X}\) is an atom such as \(\mathrm{Cl}\) or a group such as \(\mathrm{CH}_{3}\). The following table gives \(K_{\mathrm{a}}\) values for the conjugate acids of a variety of substituted pyridines. $$ \begin{array}{ll} \hline \text { Atom or Group } \mathrm{X} & K_{\mathrm{a}} \text { of Conjugate Acid } \\ \hline \mathrm{N} 0_{2} & 5.9 \times 10^{-2} \\ \mathrm{Cl} & 1.5 \times 10^{-4} \\ \mathrm{H} & 6.8 \times 10^{-6} \\ \mathrm{CH}_{3} & 1.0 \times 10^{-6} \\ \hline \end{array} $$ (a) Suppose each conjugate acid is dissolved in sufficient water to give a 0.050 M solution. Which solution would have the highest pH? The lowest pH? (b) Which of the substituted pyridines is the strongest Bronsted base? Which is the weakest Br酶nsted base?

A You mix \(30.0 \mathrm{mL}\) of \(0.15 \mathrm{M} \mathrm{NaOH}\) with \(30.0 \mathrm{mL}\) of \(0.15 \mathrm{M}\) acetic acid. What molecules and ions exist in this solution? List them in order of decreasing concentration.

Write the formula and give the name of the conjugate base of each of the following acids. (a) HCN (b) \(\mathrm{HSO}_{4}^{-}\) (c) HF

Decide whether each of the following substances should be classified as a Lewis acid or a Lewis base. (a) \(\mathrm{BCl}_{3}\) (Hint: Draw the electron dot structure.) (b) \(\mathrm{H}_{2} \mathrm{NNH}_{2},\) hydrazine (Hint: Draw the electron dot structure. (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (Hint: Draw the electron dot structure.)
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