Chapter 14: Problem 27
What is a buffer?
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Chapter 14: Problem 27
What is a buffer?
These are the key concepts you need to understand to accurately answer the question.
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Determine the \(\mathrm{pH}\) of each solution and classify it as acidic, basic, or neutral. (a) \(\mathrm{pOH}=12.5\) (b) \(\mathrm{pOH}=5.5\) (c) \(\mathrm{pOH}=0.55\) (d) \(\mathrm{pOH}=7.98\)
Calculate the \(\mathrm{pH}\) of each solution: (a) \(1.34 \times 10^{-3} \mathrm{M} \mathrm{HClO}_{4}\) (b) \(0.0211 \mathrm{M} \mathrm{NaOH}\) (c) \(0.0109 \mathrm{MHBr}\) (d) \(7.02 \times 10^{-5} \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)
Calculate \(\left[\mathrm{OH}^{-}\right]\)given \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)in each aqueous solution and classify the solution as acidic or basic. (a) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.3 \times 10^{-3} \mathrm{M}\) (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.1 \times 10^{-12} \mathrm{M}\) (c) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5.2 \times 10^{-4} \mathrm{M}\) (d) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=6.1 \times 10^{-9} \mathrm{M}\)
For each reaction, identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid, and the conjugate base. (a) \(\mathrm{HBr}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{Br}^{-}(a q)\) (b) \(\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\mathrm{HNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\)
For each \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\), determine the \(\mathrm{pH}\) and state whether the solution is acidic or basic. (a) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.0025 \mathrm{M}\) (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.8 \times 10^{-12} \mathrm{M}\) (c) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.6 \times 10^{-9} \mathrm{M}\) (d) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.0195 \mathrm{M}\)
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