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# .
I'm asking this as a fun question. I know it's supposed to be
1 Answer
The dissociation of
#"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
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)#