What is the pH of a 1.0L buffer solution which consists of 0.150 M NaNO2 and 0.070 M of HCl?

Question

Would the ph of this solution change if the volume was doubled/

Answer 1

HNO2 has pKa=3.34
Let's find out how much HNO2 is formed by the reaction of NaNO2 and HCl:
mole NaNO2=MV= 0.150*1= 0.150
mole HCl = MV= 0.070*1= 0.070
all of HCl reacts so we have forming
mole HNO2= mole HCl =0.070
and remaining NO2- mole =0.150-0.070 =0.080
since V=1L these values are also concentrations

Using the Henderson-Hasselbach equation, and doing the approximation the concentration of NO2-and HNO2 are equal to the concentrations just after the reaction:
pH = pKa+log[NO2-]/[HNO2]= 3.34+log(0.08/0.07) = 3.40

You can get a more exact solution if you set up an ICE table for the hydolysis of NO2- in the presence of the HNO2 formed by the reaction.

For a first level chemistry the approximation is OK.
Also you see in the equation that you have a ratio of concentrations [NO2-]/[HNO2] = (mole NO2/V) / (mole HNO2/V)=mole NO2 / moleHNO2.
Thus in the ratio volume is simplified, meaning that pH is the same regardless if you change the volume.

In practice you have changes in ionic strength because of the dilution, which change the activity coefficient and thus affect a bit the pH value-but I think that is too advanced for what you are looking for.