Question

How many grams of solid sodium nitrite should be added to 1.00 L of a 0.226 M nitrous acid solution to prepare a buffer with a pH of 4.305 ?

grams sodium nitrite =  g.

Answer 1

Nitrous acid is a weak acid. The acid dissociation constant Ka value for nitrous acid (HNO2) is 7.2 x 10-4.

So its pKa=-logKa=-log(7.2 x 10-4)=3.143

Also by Henderson-hasselbalch equation, pH of an acidic buffer is given as

pH=pKa+log([conjugate base]/[weak acid])

Where pKa = -logKa for weak acid =3.143 in this case for nitrous acid.

[Conjugate base]=Concentration of conjugate base=concentration of NO2- in this case (which would come from potassium nitrite NaNO2)

[Weak acid]=Concentration of weak acid=0.226 M in this case for nitrous acid.

Substituting the values in Henderson-hasselbalch equation

4.305=3.143+log([conjugate base]/0.226 M)

4.305-3.143=log[conjugate base]-log(0.226)

(As log(a/b)=log a -log b)

1.162 =log[Conjugate base]-(-0.646)

1.162+(-0.646)=log[conjugate base]

0.516=log[Conjugate base]

So [Conjugate base]=[NaNO2]=100.516=3.281

Number of moles of NaNO2 in given buffer=Molarity x volume (L)

=3.281 Mx 1 L=3.281 mol

Molar mass of NaNO2=Molar mass of Na+Molar mass of N+2xmolar mass of O

=23 g/mol+14 g/mol+2x16 g/mol

=23 g/mol+14 g/mol+32 g/mol=69 g/mol

So mass of NaNO2 in given buffer=number of moles x molar mass

=3.281 mol x 69 g/mol

=226.389 g

So 226.389 g NaNO2 must be added to make the desired buffer.