For H3PO4 and H3BO3, does the subscript “3” of hydrogen in these two formulas seem to result in additional ions in solution as it did in Group A? Explain.

Respuesta :

[tex]H_{3}BO_{3} and H_{3}PO_{4}[/tex] are weak electrolytes. They partially ionize in solutions as below,

[tex]H_{3}BO_{3} (aq)<==>3H^{+}(aq)+BO_{3}^{3-}(aq)[/tex]

[tex]H_{3}PO_{4} (aq)<==>3H^{+}(aq)+PO_{4}^{3-}(aq)[/tex]

The number of ions increase in solution apparently but this increase will be very less as these compounds are weak electrolytes, there will be a little increase in conductivity of these compounds in solutions. Unlike strong electrolytes these compounds show a very little increase in the concentration of ions although there is a subscript of 3 on hydrogen as these compounds partially ionize in solution.

The additional ions found in the solutions are,

[tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}[/tex],[tex]{{\text{H}}_{\text{2}}}{\text{PO}}_4^ -[/tex],[tex]{\text{HPO}}_4^{2-}[/tex],[tex]{\text{PO}}_4^{3 - }[/tex] in case of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}[/tex] and [tex]{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_3}[/tex],[tex]{{\text{H}}_{\text{2}}}{\text{BO}}_3^ -[/tex],[tex]{\text{HBO}}_3^{2 - }[/tex],[tex]{\text{BO}}_3^{3 - }[/tex] in case of [tex]{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_3}[/tex].

Further explanation:

Weak electrolyte:

The electrolyte which does not dissociate completely in the aqueous solution is known as weak electrolyte. Weak electrolyte only ionizes from 1 % to 10 %. For example, HF (hydrofluoric acid) and  [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}[/tex] (acetic acid), [tex]{{\text{H}}_2}{\text{O}}[/tex] (water),   [tex]{\text{N}}{{\text{H}}_3}[/tex](ammonia).

Strong electrolyte:

The electrolyte which dissociates completely in the aqueous solution is known as strong electrolyte. Strong electrolyte 100 % ionizes in the water. For example, HCl (hydrochloric acid), NaCl (sodium chloride), HBr (hydrobromic acid), HI (hydroiodic acid).

The given compounds [tex]{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_3}[/tex] and [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}[/tex] are weak electrolytes thus they do not ionize in water completely. Therefore [tex]{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_3}[/tex] and [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}[/tex] dissociates partially in water.

The dissociation reaction of  [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}[/tex] is as follows:

[tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}+{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}+{{\text{H}}_{\text{2}}}{\text{PO}}_4^-\\ {{\text{H}}_{\text{2}}}{\text{PO}}_4^-+{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}+{\text{HPO}}_4^{2 - }\\{\text{HPO}}_4^-+ {{\text{H}}_{\text{2}}}{\text{O}}\rightleftharpoons {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}+{\text{PO}}_4^{3 - }[/tex]

The dissociation reaction of [tex]{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_3}[/tex] is as follow

 [tex]{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_{\text{3}}}+{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}+{{\text{H}}_{\text{2}}}{\text{BO}}_3^-\\ {{\text{H}}_{\text{2}}}{\text{BO}}_3^-+{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}+{\text{HBO}}_3^{2 - }\\{\text{HBO}}_3^-+ {{\text{H}}_{\text{2}}}{\text{O}}\rightleftharpoons {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}+{\text{BO}}_3^{3 - }[/tex]

Therefore, the species after partial dissociation can found in the aqueous solution are,

[tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}[/tex],[tex]{{\text{H}}_{\text{2}}}{\text{PO}}_4^ -[/tex],[tex]{\text{HPO}}_4^{2-}[/tex],[tex]{\text{PO}}_4^{3 - }[/tex] in case of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}[/tex] and [tex]{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_3}[/tex],[tex]{{\text{H}}_{\text{2}}}{\text{BO}}_3^ -[/tex],[tex]{\text{HBO}}_3^{2 - }[/tex],[tex]{\text{BO}}_3^{3 - }[/tex] in case of [tex]{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_3}[/tex].

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Strong and weak electrolytes

Keywords: H3PO4, H2PO4-, HPO42-, PO43-, weak electrolyte, strong electrolyte, H3BO3, H2BO3-, HBO32-, BO33-, boric acid, dissociation, completely dissociate, partial dissociate, aqueous solution, electrochemistry,