Question

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)"`

Answer 1

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 `2.5%` ionized, then the fraction of dissociation is `alpha = 0.025`. That is not `x`, whereas you have assumed that `x = alpha`. It cannot be true, because `x` is an absolute quantity, and `alpha` is a fraction, a relative quantity.

Writing the traditional ICE table for any initial concentration `["HA"]_i`:

`"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 `K_c` is known yet. All we need to do is define

`x = alpha cdot ["HA"]_i`,

i.e. the extent of dissociation `x` can be written in terms of the initial concentration of `"HA"` and the fraction of dissociation `alpha`.

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 `K_c`.

`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.