# A 0.125 M solution of a weak acid is 2.5% ionized, what is the dissociation constant #K_c# of the acid?

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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** *initial concentration* of *fraction 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.