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.

Relax

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].

Learn more:

1. Identify oxidation numbers: https://brainly.com/question/2086855

2. Calculation of volume of gas: https://brainly.com/question/3636135

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,