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Chapter 14: Problem 127

Calculate the \(\mathrm{pH}\) of a \(0.050-\mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.4 \times 10^{-5}\).

Short Answer

Expert verified
The pH of the \(0.050 - \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution is approximately 2.58.

Step by step solution

01

Write the dissociation reaction for the given species

The dissociation of the hydrolyzed aluminium complex is written as: \( \mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+} (aq) \rightleftharpoons \mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}^{2+} (aq) + \mathrm{H}_{3} \mathrm{O}^{+} (aq) \)
02

Write the expression for the acid dissociation constant, Kₐ

Using the equilibrium constant expression, we can write the \(K_{\mathrm{a}}\) as: \( K_{\mathrm{a}} = \frac{[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}^{2+}][\mathrm{H}_{3} \mathrm{O}^{+}]}{[\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}]} \)
03

Set up the initial and equilibrium concentrations

Let x be the equilibrium concentration of \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}^{2+}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\). Then, the initial concentration of \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(0.050 \mathrm{M}\). At equilibrium, the concentrations can be written as: \( [\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}^{2+}] = x \) \( [\mathrm{H}_{3} \mathrm{O}^{+}] = x \) \( [\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}] = 0.050 - x \)
04

Solve for x using the Kₐ value

We can now substitute the equilibrium concentrations into the \(K_{\mathrm{a}}\) expression and solve for x: \( 1.4 \times 10^{-5} = \frac{x \times x}{(0.050 - x)} \) Since the \(K_{\mathrm{a}}\) value is very small, we can assume that x << 0.050, so: \( 1.4 \times 10^{-5} \approx \frac{x^{2}}{0.050} \) Solving for x, we get: \( x \approx \sqrt{(1.4 \times 10^{-5}) \times 0.050} \approx 2.66 \times 10^{-3} \) This means that the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) at equilibrium is approximately \(2.66 \times 10^{-3} \mathrm{M}\).
05

Calculate the pH of the solution

Finally, we use the definition of \(\mathrm{pH}\) to find the \(\mathrm{pH}\) of the solution: \( \mathrm{pH} = -\log_{10}[\mathrm{H}_{3} \mathrm{O}^{+}] \) Substituting the value we found for the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration, \( \mathrm{pH} \approx -\log_{10}(2.66 \times 10^{-3}) \approx 2.58 \) So, the \(\mathrm{pH}\) of the \(0.050-\mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution is approximately 2.58.

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Most popular questions from this chapter

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives \(\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{~N}_{2} \mathrm{O}_{10} \mathrm{~S}\) as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise 155.) Why is codeine sulfate used instead of codeine?

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.

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but is now available only by prescription because of its addictive properties. If the \(\mathrm{pH}\) of a \(1.7 \times 10^{-3}-M\) solution of codeine is \(9.59\), calculate \(K_{\mathrm{b}}\).

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

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.
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