The Ka of a monoprotic weak acid is 4.67x10−3. 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)+H2O(l)⇌H3O++A−

And Ka=[H3O+][A−][HA]=4.67×10−3

Now if x⋅mol⋅L−1 HA dissociates then..........

4.67×10−3=x20.171−x

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

4.67×10−3≅x20.171

And thus x1=√4.67×10−3×0.171=0.0283⋅mol⋅L−1, and now we have an approx. for x, we can recycle this back into the expression:

x2=0.0258⋅mol⋅L−1

x3=0.0260⋅mol⋅L−1

x4=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 ionInitial concentration of acid×100%=0.0260⋅mol⋅L−10.171⋅mol⋅L−1×100%=15%

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