Question

What is the `H_3O^+` and the pH of a buffer that consists of 0.24 M `HNO_2` and 0.68 M `KNO_2`? (`K_a` of `HNO_2 = 7.1 x 10^-4`)?

Answer 1

Well the buffer equation operates....

Explanation:

...which says that...

`pH=pK_a+log_10{[[A^-]]/[[HA]]}`

and so here....`pH=3.15+log_10((0.68*mol*L^-1)/(0.24*mol*L^-1))=3.15+0.453=3.60`

And so.....

`[H_3O^+]=10^(-3.60)*mol*L^-1=2.51xx10^-4*mol*L^-1`.

Solution `pH` is raised relative to `pK_a` given that the concentration of the conjugate base, the nitrite anion, is GREATER than that of the nitrous acid.

What would `pH` be, should `[NO_2^(-)]=[HNO_2]`?