A 500-mL buffer was prepared and it is 0.300 M in nitrous acid (HNO2 , Ka = 4.0 x 10-4) and 0.200 M in sodium nitrite, NaNO2. What will be the pH of this solution after adding 1.00 mL of 6.00 M HCl?

Answer 1

millimols HNO2 = M x mL = 0.300 x 500 = 150
mmols NaNO2 = 0.2 x 500 = 100
mmols HCl added = 6 x 1 = 6
----------------------------
.........NO2^- + H^+ ==> HNO2
I.......100......0........150
add..............6..................
C........-6.....-6........+6
E........94......0.......156

Now substitute the E line into the Henderson-Hasselbalch equation and solve for pH.