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?

1 Answer
anor277
May 9, 2017

#"% 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%#

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