Question

The `K_a` of a monoprotic weak acid is `4.67 x 10^-3`. What is the percent ionization of a 0.171 M solution of this acid?

Answer 1

`"% dissociation = 15%"`

Explanation:

We address the equilibrium,

`HA(aq) + H_2O(l) rightleftharpoons H_3O^(+) + A^(-)`

And `K_a=([H_3O^+][A^-])/([HA])=4.67xx10^-3`

Now if `x*mol*L^-1` `HA` dissociates then..........

`4.67xx10^-3=x^2/(0.171-x)`

And this is quadratic in `x`, but instead of using the tedious quadratic equation, we assume that `0.171">>"x`, and so.......

`4.67xx10^-3~=x^2/(0.171)`

And thus `x_1=sqrt(4.67xx10^-3xx0.171)=0.0283*mol*L^-1`, and now we have an approx. for `x`, we can recycle this back into the expression:

`x_2=0.0258*mol*L^-1`

`x_3=0.0260*mol*L^-1`

`x_4=0.0260*mol*L^-1`

Since the values have converged, I am willing to accept this answer. (This same answer would have resulted from the quadratic equation, had we bothered to do it.)

And thus percentage dissociation................

`="Concentration of hydronium ion"/"Initial concentration of acid"xx100%=(0.0260*mol*L^-1)/(0.171*mol*L^-1)xx100%=15%`