Question

Question #a4e04

Answer 1

The `pK_a` of `HNO_2=3.25`

Let `x` L of 0.100M solution of.`HNO_2` be mixed with `(1-x)` L of.`NaNO_2` to make `1` L of buffer solution.

Now `x` L of 0.100M solution of `HNO_2` contains `0.1x` mol `HNO_2` in 1 L mixture

and `(1-x)` L of.`NaNO_2` solution contains `0.2(1-x)` mol `NaNO_2` in 1 L mixture.

Now Using the Henderson-Hasselbalch equation

`pH=pK_a+log_10(([NaNO_2])/([HNO_2]))`

`=>3.35=3.25+log_10((0.2(1-x))/(0.1x))`

`=>log_10((0.2(1-x))/(0.1x))=0.01`

`=>(2(1-x))/x=10^(0.01)~~1.02`

`=>1/x-x/x=0.51`

`=>1/x-1=0.51`

`=>1/x=1.51`

`=>x=1/1.51~~2/3`

So to prepare `1` L buffer `2/3` L of 0.100M solution of `HNO_2` should be mixed with `1/3` L of 0.200M solution of `NaNO_2`.