Question #3d25a

Problem

What is the equivalent weight of #"H"_3"PO"_2# when it disproportionates to #"PH"_3# and #"H"_3"PO"_3#?

Solution

The equivalent weight is the molecular weight (#"MM"#) divided by the number of electrons transferred per mole of reactant (#n#).

#color(blue)(|bar(ul(color(white)(a/a) "Equivalent weight" = "MM"/ncolor(white)(a/a)|)))" "#

We need the moles of electrons, so let's balance the equation by the ion-electron method.

Unbalanced equation: #"H"_3"PO"_2 → "PH"_3 + "H"_3"PO"_3#

Reduction half-reaction: #1 × ["H"_3"PO"_2 + "4H"^+ + "4e"^"-" → "PH"_3 +"2H"_2"O"]#
Oxidation half-reaction: #2 × ["H"_3"PO"_2 + "H"_2"O" → "H"_3"PO"_3 + "2H"^+ + "2e"^"-"]#
Balanced equation: #3"H"_3"PO"_2 → "PH"_3 + 2"H"_3"PO"_3#

We see that 4 mol of electrons are transferred in the reaction.

Since 3 mol of #"H"_3"PO"_2# are involved, the number of electrons transferred per mole is

#n = ("4 mol e"^"-")/("3 mol H"_3"PO"_2) = 4/3color(white)(l) "mol electrons per mole of H"_3"PO"_2"#

∴ #"Equivalent weight" = "MM"/n = "66.00 g"/(4/3) = "66.00 g"× 3/4 = "49.50 g"#

The equivalent weight of #"H"_3"PO"_2# in this reaction is 49.50 g.