Question

The equilibrium concentrations of the reactants and products are [ha] = 0.220 m [h3o ] = 2.00 × 10–4 m [a–] = 2.00 × 10–4 m calculate the ka value for the acid ha.

Answer 1

The equilibrium equation showing the dissociation of weak acid is:

`HA (aq) + H_(2)O(l) <==> A^(-)(aq) + H_(3)O^(+)(aq)`

`K_(a)` is the equilibrium constant for this equation, which is referred to as the acid dissociation constant.
`K_(a)` value will determine the acidic strength of the acids. Greater the value, more will be the acidity.

Calculating the
`K_(a)` value from given equilibrium concentrations:

`K_(a) = ([H_(3)O^(+)][A^(-)])/([HA])`

=
`((2.00*10^(-4))(2.00*10^(-4)))/(0.220)`

=
`1.82 * 10^(-7)`