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?

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.........