Question

A monoprotic weak acid, HA, dissociates in water according to the reaction: HA(aq) = H+(aq) + A−(aq). The equilibrium concentrations of the reactants and products are [HA]=0.160 M , [H+]=3.00×10^−4 M , and [A−]=3.00 ×10^−4 M. Calculate the ????a value for the acid HA .

Answer 1

Answer : The value of
`K_a` for the acid HA is,
`5.625* 10^(-7)`

Solution :

The equilibrium reaction for dissociation of weak acid will be,

`HA\rightleftharpoons H^++A^-`

The expression for dissociation constant will be,

`K_a=([H^+][A^-])/([HA])`

Now put all the given values in this expression, we get:

`K_a=((3* 10^(-4))* (3* 10^(-4)))/((0.160))`

`K_a=5.625* 10^(-7)`

Therefore, the value of
`K_a` for the acid HA is,
`5.625* 10^(-7)`