What is the van't Hoff factor for a binary ionic compound #"MX"# for which ion pairing has led to a percent dissociation of #50%#?

I'm asking this as a fun question. I know it's supposed to be #1.5#.

1 Answer
Truong-Son N.
May 14, 2018

The dissociation of #"MX"# in water would be:

#"MX"(aq) -> "M"^(z_+)(aq) + "A"^(z_-)(aq)#

  • A #100%# dissociation would lead to a van't Hoff factor of #2#, since the total concentration would just be that of the two ions produced per one solute particle.

  • A #0%# dissociation would obviously lead to a van't Hoff factor of #1#.

So the quick answer is just #i = 1.5#. (This is only that easy because the coefficients are all #1#. If they were not, it would not be #1.5# per se.)

A full ICE table would also yield the same result.

#"MX"(aq) " " -> " ""M"^(z_+)(aq) + "A"^(z_-)(aq)#

#"I"" "["MX"]_i" "" "" "" "0" "" "" "" "0#
#"C"" "-["MX"]_i/2" "+["MX"]_i/2" "+["MX"]_i/2#
#"E"" "["MX"]_i/2" "" "" "["MX"]_i/2" "" "["MX"]_i/2#

From this, the total ion concentration is:

#["ions"] = ["MX"]_i/2 + ["MX"]_i/2 + ["MX"]_i/2 = 1.5["MX"]_i#

So, the van't Hoff factor is

#color(blue)(i) = "total concentration"/"undissociated concentration"#

#= (1.5["MX"]_i)/(["MX"]_i) = color(blue)(1.5)#

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