Question

Calculate the acidic ionization constant?

Answer 1

We interrogate the equilibrium...

`HX(aq) rightleftharpoons H^+ +X^(-)`

Explanation:

Sometimes we represent `H^+` as the hydronium ion, `H_3O^+`..these beasts are equivalent....

We gots `pH=3.75`...and so, by definition, `[H^+]=10^(-3.75)*mol*L^-1=1.78xx10^-4*mol^-1`

AS for the table, you will have to construct this...initially we have `[HX]=0.80*mol*L^-1`...and at equilibrium we have `[H^+]=[X^-]=1.78xx10^-4*mol^-1`.

And clearly, we have `[HX]=0.80*mol*L^-1-1.78xx10^-4*mol^-1=0.7998*mol*L^-1`..

And we fill in the numbers for `K_a=([A^(-)][H^+])/([HX])`

`=(1.78xx10^-4*mol^-1)^2/(0.7998*mol*L^-1)=3.90xx10^-8`