Question

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution is 2.46 2.46 . Calculate the Ka for the acid.

Answer 1

The value of dissociation constant of the monoprotic acid is
`1.099* 10^(-3)`.

Step-by-step explanation:

The pH of the solution = 2.46

`pH=-\log[H^+]`

`2.46=-\log[H^+]`

`[H^+]=0.003467 M`

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

Initially

0.0144 0 0

At equilibrium

(0.0144-x) x x

The expression if an dissociation constant is given by :

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

`K_a=(x* x)/((0.0144-x))`

`x=[H^+]=0.003467 M`

`K_a=(0.003467 * 0.003467 )/((0.0144-0.003467 ))`

`K_a=1.099* 10^(-3)`

The value of dissociation constant of the monoprotic acid is
`1.099* 10^(-3)`.