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Chapter 17: Problem 17
A 0.2688 -g sample of a monoprotic acid neutralizes \(16.4 \mathrm{~mL}\) of \(0.08133 \mathrm{M}\) KOH solution. Calculate the molar mass of the acid.
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The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with the carbonate and then "back-titrating" the remaining acid with \(\mathrm{NaOH}\). (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a \(0.1022-\mathrm{g}\) sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).
Which of these solutions has the highest \(\left[\mathrm{H}^{+}\right]\) : (a) \(0.10 \mathrm{M} \mathrm{HF},\) (b) \(0.10 \mathrm{M} \mathrm{HF}\) in \(0.10 \mathrm{M} \mathrm{NaF},\) or (c) \(0.10 \mathrm{M}\) HF in \(0.10 \mathrm{M} \mathrm{SbF}_{5}\) ?
Specify which of these systems can be classified as a buffer system: (a) \(\mathrm{KCl} / \mathrm{HCl}\), (b) \(\mathrm{NH}_{3} / \mathrm{NH}_{4} \mathrm{NO}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{HPO}_{4} / \mathrm{NaH}_{2} \mathrm{PO}_{4}\)
For which of these reactions is the equilibrium constant called a solubility product? (a) \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Zn}(\mathrm{OH})_{4}^{2-}(a q)\) (b) \(3 \mathrm{Ca}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) (c) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightleftharpoons\) \(\mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)\)
If \(\mathrm{NaOH}\) is added to \(0.010 \mathrm{MAl}^{3+}\), which will be the predominant species at equilibrium: \(\mathrm{Al}(\mathrm{OH})_{3}\) or \(\mathrm{Al}(\mathrm{OH})_{4}^{-} ?\) The \(\mathrm{pH}\) of the solution is \(14.00 .\left[K_{\mathrm{f}}\right.\) for \(\mathrm{Al}(\mathrm{OH})_{4}^{-}=2.0 \times 10^{33}\)
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