A buffer solution is formed by adding 0.200 moles of solid potassium nitrite, KNO₂, to 500.0 ml of 0.600 M Nitrous acid, HNO₂ solution?

1.
Refer to: http://clas.sa.ucsb.edu/staff/Resource%2...
pKa for HNO₂ = 3.39
(The answer may be slightly difference due to different sources of pKa.)

In the solution:
[HNO₂] = 0.600 M
[NO₂⁻] = (0.200 mol) / (500/1000 L) = 0.400 M

The dissociation of HNO₂:
HNO₂(aq) + H₂O(l) ⇌ NO₂⁻(aq) + H₃O⁺(aq) …… pKa = 3.39

According to Henderson-Hasselbalch equation:
pH = pKa + log([NO₂⁻]/[HNO₂])
pH of the buffer solution = 3.39 + log(0.400/0.600) = 3.21


2.
On addition of Ba(OH)₂:
HNO₂(aq) + OH⁻(aq) → NO₂⁻(aq) + H₂O(l)

No. of moles of OH⁻ added = (1.5 mol/L) × (8.00/1000 L) × 2 = 0.024 mol
No. of moles of HNO₂ in the final solution = (0.600 mol/L) × (500/1000 L) - (0.024 mol) = 0.276 mol
No. of moles of NO₂⁻ in the final solution = (0.200 + 0.024) mol = 0.224 mol
Hence, [NO₂⁻]/[HNO₂] = (Moles of NO₂⁻)/(Moles of HNO₂) = 0.224/0.276

The dissociation of HNO₂:
HNO₂(aq) + H₂O(l) ⇌ NO₂⁻(aq) + H₃O⁺(aq) …… pKa = 3.39

According to Henderson-Hasselbalch equation:
pH = pKa + log([NO₂⁻]/[HNO₂])
pH of the final solution = 3.39 + log(0.224/0.276) = 3.30

There is nitrous acid coming out of your mothers v*glna