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Chapter 14: Q14.1-11 E (page 827)

What are amphiprotic species? Illustrate with suitable equations.

Short Answer

Expert verified

Amphiprotic species are the compounds that can act both as a Bronsted-Lowry's acid and base. Example of water acting as an amphiprotic species:

\(\begin{aligned}{{\rm{HCl}}(aq) + {{\rm{H}}_2}{\rm{O}}(l) \to {\rm{C}}{{\rm{l}}^ - }(aq) + {{\rm{H}}_3}{{\rm{O}}^ + }(aq)}\\{{{\rm{S}}^{2 - }} + {{\rm{H}}_2}{\rm{O}}(l) \to {\rm{H}}{{\rm{S}}^ - }(aq) + {\rm{O}}{{\rm{H}}^ - }(aq)}\end{aligned}\)

Step by step solution

01

Amphiprotic species

Amphiprotic species are the compounds that can act both as a Bronsted-Lowry's acid and base. (Reminder: Bronsted-Lowry's acid is proton \(\left( {{H^ + }} \right)\)donor and its base is proton acceptor).

02

Explanation using equation

One example is water - here are following equations:

\({\rm{HCl}}(aq) + {{\rm{H}}_2}{\rm{O}}(l) \to {\rm{C}}{{\rm{l}}^ - }(aq) + {{\rm{H}}_3}{{\rm{O}}^ + }(aq)\)

Here water acts as a proton acceptor, hence it is Bronsted-Lowry's base.

\({S^{2 - }} + {H_2}O(l) \to {\rm{H}}{{\rm{S}}^ - }(aq) + O{H^ - }(aq)\)In this case water donates a proton to the hence it is Bronsted-Lowry's acid. With these two equations we see that water is amphiprotic - can act as an acid or a base.

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Most popular questions from this chapter

Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic 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 \({60^ \circ }{\rm{C}}\)

Question: Write equations that show \(N{H_3}\) as both a conjugate acid and a conjugate base.

Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.

Identify and label the Br酶nsted-Lowry acid, its conjugate base, the Br酶nsted-Lowry base, and its conjugate acid in each of the following equations:

\({\rm{\;(a)\;NO}}_2^ - + {{\rm{H}}_2}{\rm{O}} \to {\rm{HN}}{{\rm{O}}_2} + {\rm{O}}{{\rm{H}}^ - }\).

\({\rm{\;(b)\;HBr}} + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_3}{{\rm{O}}^ + } + {\rm{B}}{{\rm{r}}^ - }\)

\({\rm{\;(c)\;H}}{{\rm{S}}^ - } + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_2}{\rm{S}} + {\rm{O}}{{\rm{H}}^ - }\)

\({\rm{\;(d)\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - + {\rm{O}}{{\rm{H}}^ - } \to {\rm{HP}}{{\rm{O}}_4}^{2 - } + {{\rm{H}}_2}{\rm{O}}\)

\({\rm{\;(e)\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - + {\rm{HCl}} \to {{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} + {\rm{C}}{{\rm{l}}^ - }\)

\({\rm{\;(f)\;}}{\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} \to {\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }}\)

\({\rm{\;(g)\;C}}{{\rm{H}}_3}{\rm{OH}} + {{\rm{H}}^ - } \to {\rm{C}}{{\rm{H}}_3}{{\rm{O}}^ - } + {{\rm{H}}_2}\)
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