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Chapter 14: Problem 69

Monochloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}\), is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of \(K_{\mathrm{a}}\) for monochloroacetic acid is \(1.35 \times 10^{-3}\). Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of monochloroacetic acid.

Short Answer

Expert verified
The pH of a 0.10 M monochloroacetic acid solution is approximately 1.94.

Step by step solution

01

Write the dissociation equation for monochloroacetic acid.

Monochloroacetic acid \( (\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}) \), is a weak acid, and when it dissociates in water, it forms hydrogen ions and the monochloroacetate ion, as follows: \( \mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{C}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}^{-} (\mathrm{aq}) \).
02

Write the equation for the acid dissociation constant, \(K_a\).

The acid dissociation constant, \(K_a\), of the acid represents the balance between the reactants and products in the dissociation process and is given by the following equation: \(K_a = \frac{[\mathrm{H}^{+}][\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}^{-}]}{[\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}]}\).
03

Set up the initial and equilibrium concentrations of each species.

It is given that the initial concentration of monochloroacetic acid is 0.10 M. We know that initially, there is no dissociation, so the concentration of \(H^+\) and \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}^{-}\) ions will be zero. At equilibrium, let the concentration of the dissociated \(H^+\) and \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}^{-}\) ions be x. Hence, the equilibrium concentrations will be as follows: - Monochloroacetic acid: \(0.10 - x\) - Hydrogen ions: \(x\) - Monochloroacetate ions: \(x\)
04

Substitute the equilibrium concentrations into the \(K_a\) expression and solve for x.

Substituting the equilibrium concentrations into the \(K_a\) expression, we get: \(1.35 \times 10^{-3} = \frac{x^2}{0.10-x}\). Since the \(K_a\) value is small, we can assume that x will be very small, and thus, we can approximate that \(0.10 - x \approx 0.10\). Now the equation becomes \(1.35 \times 10^{-3} = \frac{x^2}{0.10}\), which can be solved for x: \(x^2 = 1.35 \times 10^{-3} \times 0.10\), and then \(x = \sqrt{1.35 \times 10^{-4}}\). After solving for x, we get \(x = 1.16 \times 10^{-2}\).
05

Calculate the pH of the solution.

Recall that the pH of a solution is given by: \(pH = -\log_{10} [\mathrm{H}^{+}]\). Since x represents the concentration of the \(H^+\) ions, we can plug-in x into the pH expression: \(pH = -\log_{10} (1.16 \times 10^{-2})\). After calculating, we find that the pH of the 0.10 M monochloroacetic acid solution is approximately 1.94.

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Most popular questions from this chapter

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{~kJ} / \mathrm{mol}\); \(\mathrm{H}-\mathrm{S}, 363 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{~kJ} / \mathrm{mol})\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}{ }^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H}\), \(322 \mathrm{~kJ} / \mathrm{mol}\) ) Give reasons for the orders you chose.

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

Students are often surprised to learn that organic acids, such as acetic acid, contain - OH groups. Actually, all oxyacids contain hydroxyl groups. Sulfuric acid, usually written as \(\mathrm{H}_{2} \mathrm{SO}_{4}\), has the structural formula \(\mathrm{SO}_{2}(\mathrm{OH})_{2}\), where \(\mathrm{S}\) is the central atom. Identify the acids whose structural formulas are shown below. Why do they behave as acids, while \(\mathrm{NaOH}\) and \(\mathrm{KOH}\) are bases? a. \(\mathrm{SO}(\mathrm{OH})_{2}\) b. \(\mathrm{ClO}_{2}(\mathrm{OH})\) c. \(\mathrm{HPO}(\mathrm{OH})_{2}\)

Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right]\), and the \(\mathrm{pH}\) of \(0.40 M\) solutions of each of the following amines (the \(K_{\mathrm{b}}\) values are found in Table \(14.3\) ). a. aniline b. methylamine

Rank the following \(0.10 M\) solutions in order of increasing \(\mathrm{pH}\). a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)
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