Question

What is [H3O+] for a buffer solution that is 0.200 M in acid and 0.500 M in the corresponding salt, if the weak acid's Ka = 5.50 × 10−7?

Answer 1

Well, we use the old buffer equation....`pH = pK_a + log_10{[[A^-]]/[[HA]]}.`

Explanation:

And `pH=-log_10(5.50xx10^-7)+log_10{(0.500*mol*L^-1)/(0.200*mol*L^-1)}`

`=6.65`

Note since `pK_a=-log_10(5.50xx10^-7)=6.26`..the given buffer is SLIGHTLY on the base side in that the concentration of the conjugate base is greater than that of the conjugate acid, and of course this is reflected by the increased `pH`...