General Chemistry

Determine the ionization constant (Ka) of a weak acid if its 0.520 M solution has a pH of 2.40.

The plan is to determine the [H+] from the pH and use it to determine the equilibrium concentrations of HA and A.

pH = -log [H+], [H+] = 10-pH = 10-2.40 = 0.003981 M

 

Assuming a monoprotic acid HA, the concentration of A is also 0.003981 M because the mole ratio of H+ and A is 1:1.

 

HA(aq) → H+(aq) + A(aq)

or

HA(aq) + H2O(l) → H3O+(aq) + A(aq)

 

We can also set up an ICE table to determine the concentration of all species:

 

 

[HA]

[H+]

[A]

Initial

0.520 

0

0

Change

-x

+x

+x

Equil

0.520 – 0.003981

0.003981

0.003981

                                                     

\[{K_{\rm{a}}}\; = {\mkern 1mu} \frac{{[{{\rm{H}}^{\rm{ + }}}][{{\rm{A}}^{\rm{ – }}}]}}{{[{\rm{HA}}]}}\; = \;\frac{{{{0.003981}^{\rm{2}}}}}{{{\rm{0}}.{\rm{520}}\;{\rm{ – }}\;0.003981}}\; = \;3.07 \times {\rm{1}}{{\rm{0}}^{{\rm{ – 5}}}}\]

 

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