Learning Materials
Features
Discover
Chapter 14: Q14.3-68 E (page 833)
Using the \({K_a}\) value of \(1.4 \times 1{0^{ - 5}}\), place \(Al\left( {{H_2}O} \right)_6^{3 + }\) in the correct location in Figure 14.8
In figure 14.8, it is located between \({H_2}O\) and \({H_3}{O^ + }.\)
Over 30 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
Explain why a sample of pure water at \({40^ \circ }{\rm{C}}\) is neutral even though \(\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right) = 1.7 \times {10^{ - 7}}M.\) \({K_{\rm{w}}}{\rm{\;is\;}}2.9 \times \)\({10^{ - 14}}{\rm{\;at\;}}{40^ \circ }{\rm{C}}.\)
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
\({\rm{\;(a)\;HN}}{{\rm{O}}_3} + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_3}{{\rm{O}}^ + } + {\rm{NO}}_3^ - \)
\({\rm{b) C}}{{\rm{N}}^ - } + {{\rm{H}}_2}{\rm{O}} \to {\rm{HCN}} + {\rm{O}}{{\rm{H}}^ - }\)
\({\rm{\;(c)\;}}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} + {\rm{C}}{{\rm{l}}^ - } \to {\rm{HCl}} + {\rm{HSO}}_4^ - \)
\({\rm{\;(d)\;HSO}}_4^ - + {\rm{O}}{{\rm{H}}^ - } \to {\rm{SO}}_4^{2 - } + {{\rm{H}}_2}{\rm{O}}\)
\({\rm{\;(e)\;}}{{\rm{O}}^{2 - }} + {{\rm{H}}_2}{\rm{O}} \to 2{\rm{O}}{{\rm{H}}^ - }\)
\({\rm{\;(f)\;}}{\left( {{\rm{Cu}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_3}({\rm{OH}})} \right)^ + } + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} \to {\left( {{\rm{Cu}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_4}} \right)^{2 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }}\)
\({\rm{\;(g)\;}}{{\rm{H}}_2}{\rm{S}} + {\rm{NH}}_2^ - \to {\rm{H}}{{\rm{S}}^ - } + {\rm{N}}{{\rm{H}}_3}\)
Salicylic acid, \(HO{C_6}{H_4}C{O_2}H\), and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.
(a) Both functional groups of salicylic acid ionize in water, with \({K_a} = 1.0 \times 1{0^{ - 3}}\)for the \( - C{O_2}H\) group and \(4.2 \times 1{0^{ - 13}}\) for the \( - OH \) group. What is the pH of a saturated solution of the acid (solubility \( = 1.8\;g/L)\).
(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, \(C{H_3}C{O_2}{C_6}{H_4}C{O_2}H\). The \(C{O_2}H\)functional group is still present, but its acidity is reduced, \({K_a} = 3.0 \times 1{0^{ - 4}}\). What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).
(c) Under some conditions, aspirin reacts with water and forms a solution of salicylic acid and acetic acid: \(C{H_3}C{O_2}{C_6}{H_4}C{O_2}H(aq) + {H_2}O(l) \to HO{C_6}{H_4}C{O_2}H(aq) + C{H_3}C{O_2}H(aq)\)
i. Which of the acids, salicylic acid or acetic acid, produces more hydronium ions in such a solution?
ii. What are the concentrations of molecules and ions in a solution produced by the hydrolysis of \(0.50\;g\)of aspirin dissolved in enough water to give \(75ml\) of solution?
A \({\bf{5}}{\bf{.36 g}}\) sample of \({\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl}}\) was added to \({\bf{25}}.{\bf{0}}{\rm{ }}{\bf{mL}}\) of \({\bf{1}}{\bf{.00 M NaOH}}\) and the resulting solution diluted to\({\bf{0}}.{\bf{100}}{\rm{ }}{\bf{L}}\).
(a) What is the pH of this buffer solution?
(b) Is the solution acidic or basic?
(c) What is the pH of a solution that results when \({\bf{3}}.{\bf{00}}{\rm{ }}{\bf{mL}}\) of \({\bf{0}}.{\bf{034}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{HCl}}\) is added to the solution?
Is the self ionization of water endothermic or exothermic? The ionization constant for water \(\left( {{K_W}} \right)\)is 2.9* \({10^{ - 14}}\)at \({40^ \circ }{\rm{C}}\)and 9.3 x \({10^{ - 14}}\)at \({60^ \circ }{\rm{C}}\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App