Question

What is the `pH` of a buffer composed of `0.145*mol*L^-1` `NaNO_2` and `0.125*mol*L^-1` `HNO_2`? How does the `pH` evolve if `2.75*g` of `HCl`, and we ignore any volume change?

Answer 1

See here for the background.....

Explanation:

A buffer solution consists of appreciable concentrations of a weak acid and its conjugate base. Because the acid and conjugate base compete for the proton, the `pH` of the buffer remains tolerably close to the `pK_a` of the acid........

And thus, from the link,........

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

You can read up on how this formula is derived, but the missing datum from your question is `pK_a` for `HNO_2`; this link tells me that `pK_a=3.398`, and we put these data into the equation:

`pH=3.398+log_10[(0.145*mol*L^-1)/(0.125*mol*L^-1)]`

Here of course, `"nitrous acid"=HNO_2-=HA`

And we plugs in the number and gets......`pH=3.46`, which makes sense given that the base component is slightly in excess. Does this make sense to you........?

And now we add a `2.75*g` mass of `HCl`, for which WE ASSUME NO VOLUME CHANGE. This represents a molar quantity of `(2.75*g)/(36.45*g*mol^-1)=0.0754*mol`......

This protonates the nitrite ion, and thus slightly increases `[HNO_2]` according to the following reaction......

`NaNO_2(aq) +HCl(aq) rarr NaCl(aq) + HNO_2(aq)`....

And thus the concentration of `HNO_2` slightly increases, and the `pH` of the buffer SLIGHTLY decreases.........

`pH=3.398+log_10{[0.145*mol*L^-1)/(0.125+0.0754*mol*L^-1)}`

`=3.398-0.1405=3.257`.

Are you happy with this? Please check for errata........all care taken but no responsibility admitted.........