One liter of a buffer composed of 1.2 M HNO2 and 0.8 M NaNO2 is mixed with 400 mL of 0.5 M NaOH. What is the new pH? Assume the pKa of HNO2 is 3.4.

Answer 1

I would convert to millimoles.
mmols HNO2 = mL x M = about 1200.
mmols NaNO2 = about 800.
mmols NaOH = about 200.

........HNO2 + NaOH ==> NaNO2 + H2O
I.......1200....0........800......
add...........200...............
C......-200...-200.......200.....
E.......1000...0.........1000

Plug the E line into the Hendrson-Hasselbalch equation and solve for pH

Answer 2

3.6