Chapter 13: Problem 14
Why is the \(\mathrm{pH}\) of water at \(25^{\circ} \mathrm{C}\) equal to \(7.00 ?\)
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Chapter 13: Problem 14
Why is the \(\mathrm{pH}\) of water at \(25^{\circ} \mathrm{C}\) equal to \(7.00 ?\)
These are the key concepts you need to understand to accurately answer the question.
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Consider a \(0.60-M\) solution of \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3},\) lactic acid \(\left(K_{\mathrm{a}}=\right.\) \(\left.1.4 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) ii. \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) iii. \(\mathrm{H}^{+}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) \(\mathbf{v} . \mathrm{OH}^{-}\) b. Complete the following ICE table in terms of \(x,\) the amount (mol/L) of lactic acid that dissociates to reach equilibrium. c. What is the equilibrium concentration for \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-} ?\) d. Calculate the \(p H\) of the solution.
What are the major species present in a 0.150-M \(\mathrm{NH}_{3}\) solution? Calculate the \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of this solution.
Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\mathrm{a}}\) value for hypochlorous acid is \(3.5 \times 10^{-8},\) which is the stronger base, \(\mathrm{OCl}^{-}\) or \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} ?\)
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 \(K_{\mathrm{b}}\)
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.
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