# 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 $H C N \text{(aq)"+H_2O"(l)"\rightleftharpoonsH_3O^{+}"(aq)"+CN^{-}"(aq)}$

Jul 25, 2018

For the reaction of a general weak acid in water,

${\text{HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A}}^{-} \left(a q\right)$

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 ${\left[\text{HA}\right]}_{i}$:

${\text{HA"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A}}^{-} \left(a q\right)$

$\text{I"" "["HA"]_i" "" "" "color(white)(..)-" "" "" "0" "" "" "" "" } 0$
$\text{C"" "-x" "" "" "color(white)(....)-" "" "+x" "" "" "" } + x$
$\text{E"" "["HA"]_i-x" "color(white)(..)-" "" "" "x" "" "" "" "" } x$

Therefore, we set up the usual monoprotic acid mass action expression as:

${K}_{c} = \left({\left[\text{H"_3"O"^(+)]["A"^(-)])/(["HA"]) = (xcdotx)/(["HA}\right]}_{i} - x\right)$

No approximations should be made here, because no ${K}_{c}$ is known yet. All we need to do is define

$x = \alpha \cdot {\left[\text{HA}\right]}_{i}$,

i.e. the extent of dissociation $x$ can be written in terms of the initial concentration of $\text{HA}$ 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,

${K}_{c} = \left(\alpha {\left[\text{HA"]_i)^2/(["HA"]_i - alpha["HA}\right]}_{i}\right)$

$= \left({\alpha}^{2} {\left[\text{HA"]_i^2)/((1-alpha)["HA}\right]}_{i}\right)$

$= \left(\frac{{\alpha}^{2}}{1 - \alpha}\right) {\left[\text{HA}\right]}_{i}$

As a result, we already have enough information to find ${K}_{c}$.

$\textcolor{b l u e}{{K}_{c}} = \frac{{0.025}^{2}}{1 - 0.025} \left(0.125\right)$

$= \textcolor{b l u e}{{8.0}_{128} \times {10}^{- 5}}$

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

$= \underline{\alpha \cdot {\left[\text{HA}\right]}_{i}}$

just as we defined earlier. In this case, we had

$x = 0.025 \cdot \text{0.125 M" = "0.003125 M}$.

From this not-all-too-different approach...

$\textcolor{b l u e}{{K}_{c}} = {x}^{2} / \left(0.125 - x\right)$

$= \frac{{0.003125}^{2}}{0.125 - 0.003125} = \textcolor{b l u e}{{8.0}_{128} \times {10}^{- 5}}$

...we get the same thing.