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]=0.250 M , [H+]=2.00×10−4 M , and [A−]=2.00×10−4 M . Calculate the value of pKa for the acid HA .

Answer 1

`pK_(a)` of HA is 6.80

Step-by-step explanation:

`pK_(a)=-logK_(a)`

Acid dissociation constant (
`K_(a)`) of HA is represented as-

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

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

`K_(a)=((2.00* 10^(-4))* (2.00* 10^(-4)))/(0.250)`

So,
`K_(a)=1.6* 10^(-7)`

Hence
`pK_(a)=-log(1.6* 10^(-7))=6.80`