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] = 0.200 M , [H+] = 4.00 x 10^− 4 M and [A −] = 4.00 x 10^− 4 M .
a. Calculate the value of pKa for the acid HA .

Answer 1

Answer: The
`pKa` of the acid is 6.09

Step-by-step explanation:

For the given chemical reaction:

`HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)`

The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:

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

We are given:

`[HA]_(eq)=0.200M`

`[H^+]_(eq)=4.00* 10^(-4)M`

`[A^-]_(eq)=4.00* 10^(-4)M`

Putting values in above expression, we get:

`K_a=((4.00* 10^(-4))* (4.00* 10^(-4))/(0.200)\\\\K_a=8.0* 10^(-7)0`

p-function is defined as the negative logarithm of any concentration.

`pKa=-\log(K_a)`

So,

`pKa=-\log(8.0* 10^(-7))\\\\pKa=6.09`

Hence, the
`pKa` of the acid is 6.09