A 0.125 M solution of a weak acid is 2.5% ionized, what is the dissociation constant #K_c# of the acid?
Reaction is
#HCN"(aq)"+H_2O"(l)"\rightleftharpoonsH_3O^{+}"(aq)"+CN^{-}"(aq)"#
Reaction is
1 Answer
For the reaction of a general weak acid in water,
#"HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A"^(-)(aq)#
If it is
Writing the traditional ICE table for any initial concentration
#"HA"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A"^(-)(aq)#
#"I"" "["HA"]_i" "" "" "color(white)(..)-" "" "" "0" "" "" "" "" "0#
#"C"" "-x" "" "" "color(white)(....)-" "" "+x" "" "" "" "+x#
#"E"" "["HA"]_i-x" "color(white)(..)-" "" "" "x" "" "" "" "" "x#
Therefore, we set up the usual monoprotic acid mass action expression as:
#K_c = (["H"_3"O"^(+)]["A"^(-)])/(["HA"]) = (xcdotx)/(["HA"]_i - x)#
No approximations should be made here, because no
#x = alpha cdot ["HA"]_i# ,
i.e. the extent of dissociation
We are not introducing anything new; we merely have a simple conversion from percentage to decimal:
#alpha = "percent ionization"/(100%)#
Thus,
#K_c = (alpha["HA"]_i)^2/(["HA"]_i - alpha["HA"]_i)#
#= (alpha^2["HA"]_i^2)/((1-alpha)["HA"]_i)#
#= ((alpha^2)/(1-alpha))["HA"]_i#
As a result, we already have enough information to find
#color(blue)(K_c) = (0.025^2)/(1-0.025)(0.125)#
#= color(blue)(8.0_128 xx 10^(-5))# where subscripts indicate digits PAST the last significant digit, so we have 2 sig figs.
Another approach is to know that the percent ionization is:
#"Percent ionization" = x/(["HA"]_i) xx 100% = 2.5%#
That's what you were doing at first... so,
#x = (2.5%)/(100%) xx ["HA"]_i = 0.025 cdot ["HA"]_i#
#= ul(alpha cdot ["HA"]_i)#
just as we defined earlier. In this case, we had
#x = 0.025 cdot "0.125 M" = "0.003125 M"# .
From this not-all-too-different approach...
#color(blue)(K_c) = x^2/(0.125 - x)#
#= (0.003125^2)/(0.125-0.003125) = color(blue)(8.0_128 xx 10^(-5))#
...we get the same thing.