A 1.00 l solution contains 24.52 g of nitrous acid, hno2. what mass of sodium nitrite, nano2, should be added to it to make a buffer with a ph of 2.96

When moles of HNO2 = mass/molar mass

=24.52 g / 47g/mol

= 0.52 mol

[HNO2] = moles / volume

= 0.52 mol * 1 L

= 0.52 M

by using H-H equation:

PH = Pka + ㏒[salt / acid]

when we have Ka = 4 x 10^-4

∴Pka = - ㏒Ka

= - ㏒ (4x 10^-4)

= 3.4

by substitution in H-H equation:

2.96 = 3.4 + ㏒[NaNO2/0.52]

∴[NaNO2] = 0.189 M

when moles NaNO2 = 0.189 M * 1 L = 0.189 mol

∴Mass of NaNO2 = moles NaNO2 * molar mass

= 0.189 mol * 68.99g/mol

= 13 g