Question

What is the [h3o+] at equilibrium of a 0.50 m weak acid (ha) solution if the ka of the acid is 4.6 × 10−4?

Answer 1

Answer : The concentration of
`H_3O^+` at equilibrium is, 0.015 M

Solution :

The balanced equilibrium reaction will be,

`HA+H_2O\rightleftharpoons H_3O^++A^-`

The expression for dissociation constant of weak aciod will be,

`k_a=([H_3O^+]* [A^-])/([HA])`

where,

`k_a` = dissociation constant of weak acid

Let the concentration of
`H_3O^+` and
`A^-` be 'x'

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

`4.6* 10^(-4)=((x)* (x))/(0.50)`

`x=0.015M`

The concentration of
`H_3O^+` =
`A^-` = x = 0.015 M

Therefore, the concentration of
`H_3O^+` at equilibrium is, 0.015 M