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?

Question

Cameron Andrews · Accepted Answer

#"% dissociation = 15%"#

We discuss the balance,

#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#

I'm willing to accept this answer because the values have converged. If we had bothered to solve the quadratic equation, we would have arrived at the same conclusion.

and consequently, the dissociation percentage.

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

Cameron Andrews · Answer

undefined The percent ionization of the weak acid is approximately 2.5%.